ELEMENTS OF PLANE TRIG ONOMETRY A SERIES OF MATHEMATICAL TEXTS
ED I T ED BY EARLE RAYMOND HEDRICK
THE CALCULU S By ELLER Y WILLIA MS DAVI S and WILLIAM CHA R LE S
B RE N K E . ANALYTIC GEOMETRY AND ALGEBRA B E X N D E R Z IW E T LO U I S E N O K S y A L A and A LL H P IN . ELEMENTS O F A NALYTIC GEOMETRY B A X D ZI WE T L I A O K S y LE AN ER and OU S LLEN H P IN . PLANE AND SPHERICAL TR IGONOMETRY WITH COM PLETE TABLE S B A LF RE D MO N RO K Y O I N GO y E EN N and LO U S I L D . P LANE AND SP HERICAL TR IGO NOMETRY WITH B RIEF T ABLE S B L F R E D y A D MONROE KE N YON and LO U I S IN G O L . ELE MENTARY MAT HEMATICAL A NALYSIS B J O HN E LE Y YO U N G K MI E T T MORGAN y W S and FRAN LL . COLLEG E ALGE BRA B R E S T BR W y E N O N S KINNER . ELEMENTS OF P LANE TR I GONOM ETRY WITH COM F LET E TABLES B A LF R M E K YO LO U I IN GO L y ED ONRO EN N and S D . ELEMENTS O F P L ANE TR I GONO METRY WITH BRIEF TABLE S B A F D E KE N Y N U IS IN GO L y L RE MON R O O and LO D . THE MACMILLAN TA BLES P re re h f E AR E R Y O D D pa d under t e directi on o L A M N HE RI CK . PLANE GE O METRY W O H S By ALTER B U RT ON F R D and C ARLE AMMERMAN . PLAN E AND SOLI D GEO METRY B W B OR H R E S A y ALTER URTON F D and C A L MMERMAN . SOLID G EOMETRY B W T E B U N F OR H S M M E R M N y AL R R TO D and C ARLE A A . CONSTR UCTIVE GEO METRY P u r h f E AYM D H RI K repared nd e t e direction o EARL R ON ED C . JUN IO R HIG H SC HO O L MATHEM ATICS
B W L B . EN T E M N . F W. y . . VOS URGH and G L A
Th is b o o k is issued in a fo rm identical With that o f the b oo ks anno unced abo ve ELEMENTS OF
PLANE TRIGONOMETRY
BY
ALF RED MO NRO E KENYO N
F E O F M AT H M AT I CS P U R U U VE S TY PRO SSOR E , D E NI R I
A N D
LO UI S ING O LD
ASSISTANT PRO F ESSOR O F MATHEMATICS T HE U NI VE RSITY O F M I S S O URI
N amEa th
THE MACMILLAN CO MPANY
1 91 9
All ria hta reserved PY I T 19 CO R G H , 19 ,
BY THE M ACM ILLAN CO MPAN Y .
e e Published A ril 1 1 . Set u p a nd lectro ty p d. p , 9 9
Nurim ob 19m m
in o Berwick Smith Co . a sh C . J . S . C g
A . N orwood, Ma ss . , U .S . PREFACE
THI S book ca rrie s o u t the chief motive s which guided the a uthors in their la rger work on Pla n e a n d Spherica l Trigon o m
r On a ha s n r r r a n d et y . the other h nd it been e ti ely ew itten ,
a r in a ra r ha s been ma de s till more element y ch cte . The new text form s a trea tment of P la ne T rigono metry which is q uite
bu t r ss a s s s s a brief, which neve thele de l with the mo t e enti l topics in more tha n the u sua l deta il . Thi s ha s been a cco mpli shed by o mitting o r cu rt a iling certa in topics tha t a re seldo m u s ed by the s tudent except in some
rk s a ll r a r s pecia l lin e of wo . Thu of Sphe ic l T igon om etry a n d m u ch of the deta iled dis cu ssion of T rigono metric Identities
u n s is o ra na a n d Eq a tio mitted . Such t ditio l topics a s De ’ M oivre s Theorem a n d in fin ite s e ries were omitted from the
’ a uthor s la rger work beca u se they ha ve few a pplica tion s with
’ s a re in the s tudent s pre sent g ra sp . The e of cou rs e omitted
r s k a s fro m the p e ent boo l o . Thu s thi s trea tment conta in s a minim u m of pu rely th eo reti
Its r za n is ca l ma tter . enti e orga ni tio intended to give a clea r v a s s s r iew of the im medi te u efulne of trigonomet y.
s o r a s r a s r a The lution of T i ngle e m in the p incip l motive . As s s r is a a k a a n d is s uch, thi p oble m tt c ed im medi tely it pu hed fi i to a de n te conclu s ion ea rly in the cou rse . More complete outlin e s tha n u sua l ha ve been given fo r the s u r r n h s ol tion of obliqu e tria ngle s by m ea n s of ight t ia gles . T i method of solution wa s empha siz ed recently in the Sylla bu s of
W r D r a r f r i r in S . A. T A the epa tment o n st uction the . C. ve y brief cou rse cou ld well clo se with thi s m ethod of s olving tri a n s gle . Other p ra ctica l p roblem s a re introduced to furn ish a motive fo r r a n ra a a r s the t e tm ent of the ge e l ngle , the ddition theo em ,
ra m . m s a ea sure etc. Am n r a a s di n , o g othe pplic tion , the co po i
V PREFACE
n a nd r s n s r s a n d a u a r s tio e olutio of force , p oj ection , ng l peed a re introduced pro min ently .
r n d s A is The ta ble s a e very complete a u a ble . ttention a a r a r a s a r s s a r s u s c lled p ticul ly to the t ble of q u e , q u e root , c be ,
b its u se P a r a r a n d s n la w etc. y the yth go e n theo e m the co i e
u se become p ra ctica ble for a ctu a l computa tion . The of the slide rule a n d o f four-pla oe ta bles is enco ura ged fo r p roblem s
r r On e k tha t do n o t dema nd ext eme a ccu a cy. edition of the boo
n a s - a a es M a who u se a co t in only the four pl ce t bl . ny th t edi tion fin d it a dvi sa ble to ha v e s tudents purcha s e a l s o the five pla ce ta ble s which a re publi shed s epa ra tely bo und under the
M a a a s title The cmill n T ble . The a uthors ha ve bo rn e in m ind con s ta ntly the n eeds of the beginner in trigonometry a n d ha ve a da pted the book to u se in s o a s ra a ec nd ry s chool s a s well a s in colleges . Illu t tiv e m te r a a o n s a n d a a s a e a r i l b u d , the expl n tion h ve b en c efully
rk in s f s u wo ed out g rea t deta il . The a mple form s o r the ol
r a n s is a s r k n s a s n tion of t i gle t i i g in t nce of thi te dency.
A M KEN Y O N . . .
LO U I S I N G O LD . CONTENTS
E A PART I . ACUT NG LES AND RIGHT TRIAN GLES
D CHAPTER I . INTRO UCTION M 1 . Subject atter
2 . Measurement o s to 3 . R elati n Other Subjects . Applications f 4 . Graphical S o lutio n o Triangles
min E . 5 . P reli ary stimate Check M 6 . easurements in the Field of E o 7 . Angles levati n and Depressio n P 8 . Squared aper ul o o 9 . R ectang ar C rdinates
II D F — O F RI GHT CHAPTE R . E INITIONS SOLUTION TRIANGLES
10 . Tables 1 n f o 1 . Defin itio s o the R ati s
i 12 . Right Tr angles Elementary R elations o o f 14 . Constructi n Small Tables
15 . Functions o f Co mplementary Angles
16 . Applicatio ns
17 . Directio ns fo r S o lving Triangles f 18 . The Q uestio n o Greater Accuracy of 19 . The Use the Large Tables
R CHAPTER III . TRIGONOMETRIC E LATIONS
20 . Intro ductio n 21 . P ythago rean Relations ° ° 22 . Fun ctions o f 0 and 90 ° 23 . Functio ns o f 60
24 . Trigonometric Equatio ns
25 . Inverse Functions
o o ns 26 . Pr jecti
27 . Applicatio ns of Projections
A C S O F R CH PTE R IV . LOGARITHMI O LUTIONS IGHT TRIANG LES
28 . The Use of Lo garithms
29 . Pro ducts with N ega tive Factors vii CONTENTS
P AG E
F B A M N F CHAPTE R V . SOLUTION O O LIQ UE NGLES B Y EA S O RIGHT TRIANGLES Decompositio n of Oblique Triangles into Right Triangles Cas e 1 : Given Two Angles and a Side Ca se II : Given Two Sides and the Included Angle Case 111 : Given the Three S ides Ca s e IV : Given Two Sides and the Angle Opposite One o f Them
B E A A D A E PART II . O TUS NG LES N OBLIQUE TRI NGL S
F CHAPTE R VI . F UNDAMENTA L DE INITIONS A N D F O R M U LAS
35 . Obtuse Angles R o o to A A 36 . educti n fr m Obtuse cute ngles
3 7 . Geometric R elations f ' 38. The Law o Cosines f i 39 . The Law o S nes f 40 . Diameter o Circumscribed Circle f 41 . The Law o Tangents
of - 42 . Tangents the Half angles f 43 . R adius o the Ins cribed Circle
Y O F B CHAPTER VII . S STEMATIC SO LUTION O LI Q UE TRIANGLES f 44 . Analysis o Data
45 . Case 1 : Given Two Angles and a Side wo A 46 . Case 11 : Given T Sides and the Included ngle o o of 47 . Logarithmic S luti n Case II
48 . Case 111 Given the Three Sides f 49 . Lo garithmic So luti o n o Case III
50 . Case IV : The Ambiguo us Case
A A P B CHAPTE R VII I . REAS PPLICATIONS RO LE MS
of 51 . Areas Triangles 2 A o two I A 5 . rea fr m Sides and the ncluded ngle o S 53 . Area fr m Three ides 4 I us E 5 . ll trative xamples o o f o o 55 . Co mpositio n and R eso luti n F rces and Vel cities
I us E 56 . ll trative xamples
N A A G PART III . T HE GE ER L N LE
E D E E D A S — R D ME S U R E CHAPT R IX . IR CT NGLE A IAN A
57 . Directed Lines and Segments A 58 . Ro tatio n. Directed ngles CONTENTS
P A o n R A 59 . lacing ngles ectangular xes of A 60 . Measurement ngles o f 61 . Radian Meas ure Angles
o f 62 . Use R adian Measure
63 . Angular Speed
64 . N ota tio n
O F AN Y AN G E CHA PTE R X . FUNCTIONS L
5 o of c s . P o o s 6 . Reso luti n For e r jecti n
fin itio ns . ri on o metrio 66 . General De T g Functi o ns of Any Angle
S o f o o o 67 . Algebraic igns the Trig n metric Functi ns ° 6 o f . o o f 0 90 0 8 . R eading the Tables Functi ns , , ° ° 0 180 i 0 , 270 i
o o f o o E o 69 . So luti n Trig n metric quati ns
I s E o n Co o o R o o of 70 . llu trative xamples mp siti n and es luti n F orces
CHA E HE AD D O U S PT R XI . T ITION F RM LA
7 1 . The Additi o n Fo rmul as
o l 72 . The Subtracti o n F rmu as B 73 . Reductio n o f A cos 05 i sin a
74 . Do uble Angles
of o r D 75 . Tangent a Sum ifference
76 . Applicatio ns
77 . Functions o f Half Angles
78 . Facto r Fo rmul as
E F C CHAPTE R XII . G RAP S O TRIGONOMETRIC FUN TIONS
a n d 79 . Scales Units
80 . Plo tting P o ints
81 . Graph o f sin a: f 82 . Mechan ical Cons tructio n o the Graph
83 . Inverse Functio ns 4 o n 8 . Graphical R epresentatio n o f the Inverse Functi s LOGAR ITHMI C AND TR IGON OMETR IC TABLE S
S ee Co x viii . [ ntents, page ]
ELEMENTS OF
PLANE TRIG ONOMETRY
T PART I . ACUTE ANG LES AND RI GH TRIAN GLES
CHAPT ER I
INTR ODUCTION
M tt r r r n m r s r 1 Sub ect a e . . j The wo d T igo o et y come f om
k r s a n n a su o f o r b a n s o f two G ree wo d me i g me re ment , y m e ,
es r a r s s s wa s a s tria ngl . The o igin l pu po e of thi tudy the me u rem en t of a ngles a n d dista nce s by indirect method s in ca s es in which direct mea s u rem ents a re inconvenien t o r impossible . Am o ng s u ch ca ses we ma y men tion the determin a tion of the
s a n d r z a s s s a a r ss height ho i ont l width of hill , the di t nce c o a a o r r r o r n s n a r s fields v lley ive , the le gth of the bou d ie of
o r un r r r a s a s on rough impa ssa ble gro d . T igonom et y t e t l o
r a s a m n s s a n d a s r a s a nd the el tion o g the ide ngle of t i ngle , the
a s s s a s a n d a r a s r a s a n d m e urement of the ide , ngle , e of t i ngle
o f r s n othe polygon which ca be sepa ra ted into tria ngles .
2 M ea surement r n i o r . . To mea s u e a y q ua n tity s t dete mine how ma ny time s it conta in s some conven ient u nit
s n r r r qua n tity of the a m e ki d . The exp e s sion of eve y mea su ed qua ntity con sist s of thes e two comp on en ts : the nu merica l
mea s ure a n d na me o the unit a s 2 s the f employed ; , inche ,
20 u n rs 3 u s a n d 10 un s 7 u rs a nd c bic ce timete , po nd o ce , ho
26 s a r s 36 r s er s n m inute , c e , deg ee , feet p eco d,
s a r s 1 10 s etc ohm , mpe e , volt , .
B 1 PLANE TRIG ONOMETRY 2
Sometimes we ca n m a ke direct compa rison of a qua n tity with
m a s r a s r n the unit of e u e, when we dete mine the le gth of a s n b a n r s k o r a s i egme t y pplyi g a ya d tic teel ta pe to t. On the
r a n a re u se s ti othe h d we often obliged to indirect method , e. to compute the nu merica l mea s ure of a q ua ntity by mea n s of its r a n r s m s r a s a ea su ed . s el tio to othe q u ntitie mo e e ily r Thu , we fin d the nu merica l mea s ure of the a rea of a tria ngle not by
r a sur ra b a k n - a r di ect me ement, but ther y t i g one h lf the p oduct
n of the nu merica l m ea su res of its ba se a d its a ltitu de .
R l i li t n s 3 . e a t n t h er S u ects ca i s A o . s o o O t bj . pp i It evi dent tha t trigo nometry is clos ely rela ted to pla ne geometry
a its u se s a s r a s a n d on ccount of of line , ngle , t i ngle other
s Ou r a s m a s r s s s polygon . the othe h nd , in ce the e u e of the ide , a s a n d a r a s a s a n d ra o s of s s a re ngle , e of tri ngle , the ti the ide , n umbers o o e r is a so r a a r a n d , trig n m t y l el ted to ithm etic ele
a a l eb a ment ry g r .
r r The a pplica tions of trigonometry a e ve y exten sive . Some
ook M a o s a re to of them will be given in thi s b . ny ther be
in s r a a o a s a r s found u veying, n vig ti n , tronomy, chitecture , de ign ,
r a n s a n d r ra s a ma s a n d g eomet y, mech ic , othe b nche of m the tic
s s a n d m a r a n d ri . phy ic , in ilit y civil enginee ng
4 G ra hica l lutio n o f ria n es . F o r s r tri . p S o T gl con t ucting
n s a n d a s r a s s u s a a a gle m e u ing their p rt , the t dent hould h ve
1 . FIG . 1 4 N RODUC ON 3 , 1 I T TI
f r su r n s a ro tra ctor fo r m a s r a n s a n d sca le o mea i g length , p e u ing gle ,
o m a ss fo r ra r s a a r s a n d u a s s a c p d wing ci cle , l ying off c eq l eg ment .
tria n les o etric fi u res a re sa Two g ,W m g , id to be co ngru en t when they ca n be su perimpos ed so a s to coin cide in a ll their pa rts . 4 T wo figures a re s imila r when their corre spon ding a ngles a re
n r rr s s s a re r r eq ua l a d thei co e ponding ide p opotiona l . Two tria n les a re s imila r i the a re mu tu a ll e u ia n ula r bu t s i g f y y q g , thi s
r s r a r d n ot n ece ssa ily true of polygon of mo e th n th ee si es . 0
2 . FIG .
To dra w a figu re to sca le is to ma ke a d ra wing which sha ll
s a r s a o r a r a s fo r a m a ma be imil to it but m ller ( l ger) , , ex ple, p
a a r o r a field o r flo o r of f m , the
a a n pl n of buildi g . The a dva nta ge of a s ca le dra wing is tha t the a ngle s a re the sa m e a s thos e of the figu re r r s n a n d b s a ep e e ted , y the c le r a ma rk ra el tion ed on the d w ing , a n y dim en sion of the origina l figu re ca n be rea d o ff on a sca le a pplied to the corres pond in g dimen sion o f the dra w
3 . m FIG . g .
’ A bu ilder u s es the a rchite ct s pla n s for this pu rpo se in con stru ctin g a building . ANE R G N ME R I PL T I O O T Y ( , 5
We kn ow fro m geo metry tha t the o ther three p a rts of a ny * a ctua l tria ngle a re d etermin ed if a n y on e of the followin g co m
bin a tion s is kn o wn
(1) two s ides a n d the in cluded a ngle (2) two a ngles a n d a ny sp ecified side ;
3 the three s ides ( ) ,
4 two s ides a n d the a n le o os ite o ne o them ( ) g pp f , but in the la s t ca s e there ma y be two solution s when the given i a ngle s a cute . When a sufficien t nu mbe r of pa rts o f a n tria ngle
a re k w rs ca n o b a n r a to no n , the othe be f und y dr wi g the t i ngle s ca le a n d mea suring the sides with the s ca le a n d the a ngle s
w r r ith the p ot a ctor . The p rocess of finding the unknown pa rts of a tria ngle from k a ny su ch set of given pa rts is ca lled solving the tria ngle .
EX M P E 1 . I o to o f fo r A L n rder measure the width a river, example , it is su fficient to measure the distance A B between two po ints o n the bank and the angles BA P and A BP made by AB with B to the lines j o ining A and , respectively , any k ll of b po int o n the o ther b a n . é i ese measure f v ments can be made fro m o n ebank o the ri er . Kno wing A B and the angles A B P and BAP the triangle PAB can be drawn to scale ; then the
G . 4 . FI perpendicu lar P R fro m P to A B can be drawn i P R of and measured, whence the w dth the stream can be determined by ° fi r AB = 98 ds 4 A = 3 8 actual measurement in the gu e. If yar , , and to o 56 d A B P R will be fo und be ab ut yar s .
Pr l h ck r r s min r t a te C . 5 . e Es i a y im . e In eve y exe ci e , the stu dent should ma ke a prelimina ry e s tim a te of the unkn own pa rts a nd he should keep this crude solution in mind to gu ide hi o him in s w rk.
A r r a n stu fte the unknown pa ts h ve been fou d , the dent
check a a s r should u se a ll mea n s a t his com m a nd to e ch n we ,
* The data c a n b e given so that it will be impo ssibl e to con struct a ny a re n m o triangle satisfying the co nditio ns . If s uch data give , the i p ssibility
Will appear when the attempt to co nstru ct the triangle is ma de . I 5 N ROD C N , 1 I T U TIO s in ce even expe rien ced person s a re lia ble to e rr o r in rea din g
k o s ca le s a n d in m a ing c mputa tion s . I n tria ngle s dra w n to s ca le obs erve the following checks (1) the s u m of the a ngles of a ny tria ngle shOn Zd be fi (2) the s u m of a ny two s ides sho u ld be grea ter tha n the o r third s id e ; (3) the grea ter of two sides sho u ld be opp o site the grea ter of the a ngles opp o site these sides ; (4) if two s ides a re u n equ a l their n u merica l mea su res sho u ld be un e u a l in the sa me sen se q , (5) the n umerica l mea su res of a n gles sho uld co rresp o n d to their
ma n itu des a s . a re a s g ; ngle of etc , e y to j udg e b y the eye . Thes e checks should revea l a n y gro ss error ; but the s tudent shoul d not expect thi s method of solution ( o r a n y o ther method of computa tion or mea su re ment) to give p recise a n swe rs in the s s a n o r r r r s s u en e of h ving er o wha teve . The pu po e ho ld be to obta in rea sona bly a ccu ra te resu lts a n d to detect errors tha t a re
s u nrea on a bly la rge.
EXERC E I — RAP AL O O O F TR AN E IS S . G HIC S LUTI N I GL S
o o o o u o S lve the f ll wing triangles by c nstr cti n and measurement .
° ° 1 Two 4 3 . angles are 7 and 5 and the included side is
A ns .
° ° Two 43 53 o 2 . angles are and and the side Opp site the latter is
An s .
° o 3 . Tw sides are and and the included angle is 57
Ans .
° 4 . An 1 The three sides are and s . 8
° 5 wo 4 ° is . T angles are 0 and 65 and the side Oppo site the latter 50 .
An s .
° ° 6 wo . . T angles are 30 and 105 and the included side is 7 feet 8 inches
ft. ft . An s . 5 , i 7 . Two sides are and and the altitude upo n the th rd side is
12 . Ans . 306 . Find the p en m eter and the area .
° ° 8 . o Tw angles are 30 and 100 and the sho rtest side is 8 . Find the
Ans . lon es s o it h . g t ide , the a ltitude up n , and t e area ANE R G N ME R I PL T I O O T Y [ , 5
9 o 3 . The sides are in the rati Find the smallest and the
a n An largest gle . s .
1 0 . o 3 : 4 : 5 o The angles are in the rati and the sh rtest side is 30 .
F o Ans ind the ther sides . . 37 , 41 .
1 1 . An The sides are 5, 7 , and 8 . Find the angles . s .
12 . n An 3 5 7 . a n . s . The sides are , , and Fi d the largest gle
1 Two 3 . sides are 8 and 10 and the included angle is Find the u of ir perimeter, the area, and the radi s the inscribed c cle . A ns .
14 f f . From which o the f ollowing sets o given parts is it possible to co nstruct a triangle D o any of the sets determine mo re than o n e ° ( a ) Two angles are 41 and the side o ppo site the latter is ° ( 5) Two sides are and the angle Oppo site the first is 66 ° c wo 30 ( ) T angles are and the included side is 7 . ( d) Two sides are 7 and the included angle is ( 6 ) The three sides are Two 6 7 o first is (f sides are and , the angle Opp site the
Ans . b e o ( ) and ( ) , imp ssible ; ( f two .
1 o 5 . Tw sides are 5 and 7 and the angle o pposite the latter is
a a . n 2 Find the perimeter and the re A s . 0 ;
M e sure nts n th l me i e F d . r 6 . a s a r rs ie In u veying l nd, ive ,
a k s a n d a r rs a n o u t r a s s o a o s l e , h bo ; l yi g o d , ditche , the f und ti n
r s s a n d r s r u r s a n d a r of b idge , building , othe t uct e in m ny othe
r s a n d m a n r n s a n s in p oj ect of civil ilit ry engi ee i g , di t ce the
field a re a s r cha in o r steel ta e. a s s me u ed with the , the p In c e
r a a is r a a o r o o n where ext eme ccur cy equired , long met l w de
s a e is s a n d is a r r a a n s a n d o rr c l u ed, c efully p otected g i t, c ected
fo r a s ra r . , ch nge in tempe tu e Angles in the horizonta l pla ne a re dra wn in position on the p la n e ta ble by mea n s of a pa ir of sights on a hea vy meta l
s ra o r m r r z a a n d r a t ightedge ; , o e often both ho i ont l ve tic l ’ a ngles a re sighted with the tele s co pe of the engin eer s tra n sit a n d their mea s u re s a re rea d off from the g ra dua ted circles of
s r the in t um ent .
I mi o s n deter ning distances and directi n in an extended survey , greater accuracy can be attained by measuri ng the angles of certain triangl es and
o t f . c mputing the leng hs o the sides, than by measuring these sides directly
PLANE TRIG ONOM ETRY [L 7
An les of Eleva on a n d D e ressio n . Ah s r r a t 0 7 . g ti p ob e ve
a o f el eva tio n o f a n o A r a n mea s ure s the ngle bj ect , highe th
s b s -a o z a i OH b a s him elf, y ighting h ri ont l l ne y me n of the level on the tele s co pe of the tra n sit a n d then eleva ting the i i h a H end of the teles cope u ntil he s g ts A . The ngle OA through which the tel es co pe ha s been turned in the vertica l
a a n d is ea o ff a ra a r pl ne , which r d from the vertic l g du ted ci cle o f ra s is a a A a the t n it, the ngle of elev tion of the obj ect bove
Ho rizo nta l Line
Ho rizo nta l Line
FIG . 7 .
a a s s a d e the observer a t 0 . Simil rly he me ure the ngle of
re ss io n a n B r a s b first s p of obj ect , lowe th n him elf, y ighting the horiz onta l line OH a n d depre ssing the end o f the teles cope
r u a H B s s B th o gh the ngle O until he ight .
EXER ISES — GRAP AL SO L TIO O F TRIA LES C II . HIC U N NG
o o a S o lve the f o llo wing exercises by c nstructi n and me surement.
w o f field o 1 . T o sides a triangular are r ds and ro ds and the
o is o f o it angle Opp site the latter Find the length the fence ar und .
Ans . o r
t o in two 2 . A a p int the street midway between buildings their angles ° ° o 30 60 . o of of elevati n are and respectively Find the rati their heights .
An s . 1 : 3 .
h f o 4 o 3 . T e hands o a cl ck are and 6 inches l ng respectively . Find
’ 5 1 0 c ck . n the distance between their tips at o lo A s . A B C A 4 . In the triangle , angle B and the included side
4 a o f is 1 . Find ( ) the angle at the center the circumscribed circle sub tended by the side AB ( b) the angle at the center of the ins cribed circle B 0 c of o 0 o AB subtended by ; ( ) the length the altitude fr m up n . An s .
5 f . The diagonals o a parallelogram are 10 and 12 and they cross a t of a n angle Find the sides . Ans . INTRODUCTION
r o f 10 in . o f 7 in . 6 . The steps o f a stai way have a tread and a rise ; at what angle is the stairway inclined to the fl o o r Ans .
6 is 7 . Two sides of a triangle are each and the included angle
Ans . Find the perimeter and the a rea .
i P o o 8 . Find the d stance Q acr ss the p nd 8 o o o AP ( Fig . ) fr m the f ll wing measurements , A n A s . 692 . 900 ft. A 780 ft. P , Q , Q
i AB o f i l o 9 . To determ ne the width a b l , a p int C is taken fro m which the po ints A and B o n op
o f . I A O 200 ft. po site sides the hill are visible f ,
A CB fin d FI G . 8 . B 0 223 ft. , and angle the width
A B .
o 1 : 2 : 3 1 0 . The angles o f a triangle are in the rati , and the altitude upo n th e lo ngest side is Find the perimeter and the a rea . A ns .
of a re ul a r five- o 1 1 . Find the angles and sides g p inted star inscribed
An s . 19 . in a circle of radius 10 .
d P a r is a n a a a ra ua re . 8 . Sq pe It often d v nt ge to d w the
r o n a r r i s a r s a s a r a r o r figu e p pe uled nto qu e , c lled q u ed p pe ,
FIG . 9 .
r ss-s n a n n s is a r u a r c o ectio pa per . The loc tio of poi t p tic l ly
a s on s a so a a m a fo r a is r a e y uch p per , th t p , ex mple, e dily ma b u in i B fi u re r r s t. s a a de y g y uit bly pl cing the g , eq ui ed
n s n r n le gth ca f equ ently be rea d off a t o ce.
s if of x . 1 4 con Thu , the triangle fo r the graphical so lutio n E , , be
- o n o o i P R F i . 9 ca b structed cr ss secti n paper, the required d stance, , g , n e se o en at nce to be a bo ut 56 y a rds . 10 ANE R G N ME R PL T I O O T Y [I , 9
'
9 . Recta n ula r C r ina tes n g oo d . If a y two pe rpendicula r r s O Y a n d OK s a a r see F i 1 uling of the qu red p pe ( g . 0) a re s s a n o elected, the po ition of y p int P in the pla n e is determined b a s s a n s r s two n s to y me n of the di t ce f om the e li e the point P .
The pa per ca n be so pla ced tha t the se dista nces a re vertica l a n d r z a r s s a u sua s ho i ont l , e pectively ; we h ll lly uppos e the
r in i pa pe th s po sition .
10. FIG .
s . 10 o o o O Y to o A is Thu , in Fig , the h riz ntal distance fr m the p int u o r nits . T o avo id conf us i o n between p o ints at the same distance abo ve (
o OX o n o o o f O Y s o to bel w) but pp site sides , it is cu t mary call distances
as r to o f O Y o to of O Y me u ed the right p sitive , distances the left negative t — r thus, B is said o be 1 unit fro m O Y. Similarly, distances measu ed
o o 0 X fo r D o d wnwards fr m are called negative example , is fr m X 0 C 1 o 0 X a n d s o 1 o O Y. , and is fr m al fr m
The two dista n ces to a n y point P fro m O Y a n d 0 X a re ca lled
’ recta n ula r co o rdina tes P a n d a re u o the g of , freq ently den ted I 9 N R D C N 1 1 , ] I T O U TIO
d r b rs a: a n s . z y the lette y, e pectively The hori onta l dista nce as is ca lled the a bs cis s a of P ; the ve rtica l dista nce y is ca lled the
s s a s i o rdina te of P . In giving the e di t nce it s genera lly under
first is x s a a s . tood th t the one m entioned , the l t y
u A . 10 b riefl o B Th s , Fig , is y den ted by the numbers is —1 — 1 D deno ted by , b , by
s 0 X 0 Y a re a a xes o co ordin a tes s The line , c lled the f , or imply
- h - X is a x a s Y t e a s . o a xes . 0 0 0 the c lled the xi , y xi The p int is ca lled the o rigin . The fou r portion s in to which the pla ne is divided by the
s a re a rs t s e co nd third a n d o u rth u a d ra nts a s fl a xe c lled the fi , , , f q ,
1 0 . in F ig . To lo ca te a poin t is to des cribe its po sition in the pla ne in
i r r 2 is r s ts s a s o a a s e. . 5 te m of di t nce f om the co din te xe ; g , ) a point 5 un its to the left of the y- a xis a n d 2 unit s a bove the cc- s lo t a is a rk r r s a xi . To p point to m it in p ope po ition with
r x es re spect to a pa i of a .
EXERC E — UA D' PAP IS S III . SQ RE ER
1 Lo o o f o ow o w to o . cate and pl t each the f ll ing p ints ith respect s me pair o f axes a 1 b 2 C 4 d — 5 6 — 7 ( ) ( , ( ) ( , ( ) ( , ( ) , ( ) , 7 8 73 — 5 6 (f ) ( , ( 9 ) 0 0 ( , ( ) , ( j) ( ,
2 . o o 5 4 5 4 is s Sh w that the line j ining ( , ) and , ) bi ected by the o rigin .
3 . O u d o o 1 2 3 0 what lines all p ints ( , ( , , ) he
u d o o 0 0 0 0 0 2 O what line all the p ints ( , ( , ( , ( , ( , ) lie o o Make a general statement ab ut such p ints .
f Ex . 1 . o o to o o 4 Find the distance fr m the rigin each the p ints in , o f by us ing the f lded edge o ano ther piece of squared p a per . Compute the same distances by regardin g each of them as the length of o o f of o the hyp tenuse a right triangle , the lengths wh se sides can be
fi r E o f o s read directly fro m the gu e . ach these meth ds can be u ed as a
o n o An s . a d e check the ther . ( ) ( b) ( c) ( ) ( ) 13 h vs (f ) ( g) , ( ) ( ) ( f ) 10 5 . Co o 6 8 nstruct the triangle wh se vertices are ( , ( , and ( ,
An s . 6 . Find its perimeter and its a re a . ANE RIG NOME R I PL T O T Y [ , § 9
6 . of o en d o a 2 4 Find the lengths the segments wh se p ints are ( ) ( , ) 5 8 b 4 —3 — 1 c 1 — 2 4 and ( , ) ( ) ( , ) and , ( ) ( , ) and ( ,
An s . 5, 5 .
o f w 7 . Find the sides and diag nals o the parallelo gram ho se vert ices are _ f 5 4 1 A n s . 2 x 10 4 . ( , ( , ( , and ( , ,
8 P o o ints A : 1 B : — 3 C : 1 D : 7 3 . l t the p ( , , ( , ( , ) and i A B D determine the angle at which the l ne crosses the line C . Ans .
- — 9 . P o A : 2 l B : 6 C : D : 2 3 find l t ( , ) , ( , , ) and the angle at which A B cro sses CD als o find the area of the triangle who se
AB CD BD . A sides are , , and ns .
10 . P o A : 5 B : 14 C : 2 3 fin d o l t ( , ( , ( , ) and the distance fr m
A to B C o find of AB C. Ans . als the area the triangle 75 13 ,
1 1 fa rm 1 . A N . E . E . of N . is described in the deed as 3g and % IV. o o ns i o o n d Secti n 5, Wayne T w h p, Tippecan e C unty, I . Taking the center
of i o o o o : A lines th s secti n as axes , make a map fr m the f ll wing data o o o —80 80 160 136 ditch cr sses the farm thr ugh the p ints , ( , ( , )
as in s . o s 152 distances being me ured rod The h u e is at ( , There are fields o o A 0 0 seven wh se c rners are , 8 , 8 , 16, 16 B 0 - 1 6 1 6 — , , 8 , , , 80, C 80 0 0 80 D — 1 6 80 , , ( , ( , , , , ( , 80 16 160 E 80 160 160 80 ( , , ) , ( , ( , ( , ( , F 80 160 O 160 80 80 G 0 80 80 , ( , ( , ) , ( , ( , ) , ( , ( , ( ,
0 of field o of n c . ( , Find the area each and the t tal length fe e
' An s . 25 26 54 35 3 68 s . , , , , ( acres) miles rod
1 o o o n o o i 2 . P siti ns a rectangular farm are given by their c rd nates in o e to two of a s o o o s 10 r ds , r ferred sides the farm as axes , f ll ws h u e ( , 6 of 60 A l o a o barn ( , gate pasture ( , rai r d passes between the h use o o 3 D o w n and barn, with a cr ssing at the p int ( , raw a map sh i g these D h o w o o to o bjects . etermine much farther it is fr m the h use the barn by f o w way o the cr ssing than along the straight line connecting them . Ho much farther is it fro m the barn to the pasture gate by way of the cro ssing than alo ng a straight line Ans .
1 o n i o o 3 . A certain city park is b u ded by a ma n street, tw cr ss streets
i to it a nd . i to perpend cular , a stream The d stances , in feet, the stream measured perpendicularly fro m the main street at 100 ft. intervals are
o un to 680 650 525 450 450 460 540 . D of a n d f d be , , , , , , raw a map the park
o i its a a . An s. 7 9580 s . ft. determine appr x mately re acres, q
o f 0 A i field 14 . To determine the height a tree stand ng in a level the
f o 0 o f to o in distance 0 B 100 t . fr m the base the tree a p int B the
field f o O BA . , and the angle o elevati n are measured Find the
A ns 75 ft. height o f the tree . . CHAP T E R I I
I T N S S O U O N O F R G H R AN G ES DEF NI IO . L TI I T T I L
les h l s fo r s n r a n s x Ta b . e 10 . W i e the m ethod olvi g t i gle
a r a re su fficien t for a ll a s s a re r a pla ined in Ch pte I c e , they e lly n n n r r a a ra is s s n fo r s o t conv e ie t whe e g e t ccu cy de ired , i ce thi pu rpo se the figu re woul d need to be dra wn o n a very la rge
a m u su a s r s r a er s c le . The ethod lly employed when one de i e g e t a ccura cy tha n ca n be conven ien tly a tta ined by the method of
r n d r i a con st uction a mea s u ement s the method of t bles . Ta ble s a re con stru cted which give a pp roxima tely the ra tio s fi
a of ea ch pa ir of side s for a ll right tria ngles . To obt in the ra tio of a certa in pa ir of sides of a right tria ngle with a given a cute a ngle it is then on ly n ecessa ry to con sult the ta ble .
F o r o o o n e example , it is kn wn by ge metry that if angle o f a right trian gle is the side o gpo site this n e- a ngle is o half the hypo tenuse . Hence if the hy o ten use is i n p g ve ; that side , and hence als o the o ther
o n e d . I . 1 1 AB A , can be determine f in Fig , and
F . AA then the side B 0 ( 1 IG 11 .
If fo r a n a u a r r r a n ra , c te ngle of eve y ight t i gle, the tio of the
s s n s r kn n u s oppo ite ide to the hypote u e we e ow to , then we
s r r r a sa a could olve eve y ight t i ngle in the m e m nner.
s w a r a a ll r a n s ca n It will be ho n l te th t obliq ue t i gle be gu t.‘ u p in to right tria ngle s in su ch a wa y tha t the sa me ta bles ca n
u s a ll a s s fo r s u r a be ed in c e olving obliq e t i ngle s . Since a ny tria ngle ca n be en la rged (o r reduced) in size by
ra n on a a r r o r sma r s a nl the a d wi g it l ge ( lle ) c le , o y r tio s of the 4 s ides a re r a m r a n e lly i po t t .
11 . Defin iti ns o of the Ra tio s . As n a in 10 i dic ted § , the ra tio of two side s o f a tria ngle doe s n o t depend u pon the siz e 13 14 ANE R G N ME R 11 PL T I O O T Y (II , §
r a a s . s in o f the t i ngle , but only upon the ngle Thu the right ’ ' ” ” s M P N MP N M P N F i . 12 in P l V tria ngle , , of g , which , ’ ’ ” ” P N P N a re r a , pe pendicul r to
MN ra o s N P M P , the ti / , ’ ’ M r o r N P M P a re a ll e u a l . q o e ve , if
is ra w a r M P d n p erpendicul to , ea ch of the ra tios j u st mentioned is equa l
12 . FIG . to (Why The se ra
s on l o n the a n le a a t M is co en t tio , then , depend y g It nl eg i to pla ce the a ngle on a pa i r of a xe s so tha t the v ertex fa ll s a t the
r n 0 o n e s s a rf a x is to a n d Z} o igi , ide lie long the , the right , the
n s s e a k a n other s ide fa ll s in the first qua dra nt . O thi id t e y
o n P a t a 0 a n d dro p i t r ndom , except , p
the perpendicula r P M to the w-a xis
F i P : b ( see g . Let O r ; then y geometry
where a a n d y a re the coordina te s of 1 3 T f F IG 0 the p o int P . he v a rio u s ra tio s o
a s a s as r a re sa or a ll oints P p ir of the three q u ntitie ,y, the me f p
k s s a re 5 a n n OP of a ng cc. ” t e i the ide the le The e r (I V I l y 1 a the s in e o the a n le a e s1n a . ( ) c lled f g , w ritt n r
r co 2) a the co s in e of the a ngle a n s a . ( f, c lled , w itte
y 3 a the ta n ent o the a n le a e ta n a . ( ) c lled g f g , writt n
The recipro ca l s Tof thes e ra tio s a re a l s o o ften u s ed
4 r is a co s eca nt o the a n le a r csc a . ( ) /y c lled the f g , w itten
5 r wis a seca n t o the a n le a sec at. ( ) / c lled the f g , w ritten 6 w i s a cota n ent o the a n le a r a . ( ) /y c lled the g f g , w itten ctn
d c n The ra i al sig is used to deno te th e pos itive sq uare ro o t . ja i l ’ v Th e f n r ro r recipro cal o a umber is un ity d ivided by th e n umber . The ecip
‘ ' h ' ' cal o t a co mmo n fracti o n is the reshlt of in verting it ; thu s the recipro cal o f
r r . E t 0 n o t. y/ is / y very p , which has
16 PLA IG ONOM ETRY [II , 13
a r stu ca n a s s Simil ly, the dent e ily how tha t
sita 1 . 1 7 ctn a ( ) ’ co s a ta n a 1 18 s 1 s ( ) ec a ’ ( 9) c c a co s a
r r a s wil r Othe el tion l be given la te . The following exa m ple s illu stra te a method of con stru cting
n i a a ngle when one o f its ra tio s s given .
EXAMPLE 1 . Cons truct a n acute angle whose sine is 2 7 . T o co ns truct such an angle draw a right triangle who se hyp o tenuse is 7
and on e whose side is 2 . This can
n - o eas ily be do ne o cross secti n paper . With a radius of 7 draw a circle and mark its intersectio n with the ho ri
zo nta l ruling 2 units abo ve the center . FIG . 14 . The angle between the ho rizonta l di a i ameter and the r d us to this intersectio n is the angle required . 2 EXAMPLE . Construct an acute angle whose tan gent is This is m o st easily done by drawing a tri a n gle wh ose base is 8 and wh ose altitude is The angle between the hypo tenuse and
As in base is the angle required .
E 1 o o xample , it will be f und c uve nient to draw the figu re o n cross
o . 15 . secti n paper FIG .
E — T XERCISES IV. TRIGO NO M ETRI C RA IO S
O h — 1 . cro ss section paper construct angles who se sines are ( a ) ( b) ( c) ( d) ( e) ( f) ( g) ? ? 2 . Is there an acute angle whose sine is any given po sitive number
c 3 . Construct an gles wh ose tangents are ( a ) ( 5) ( ) ( d) 1 ; ( e) ( f ) 2 ; ( g ) ( h) ( i)
’ i 4 . Is there an a cute a ngle wh o se tangent is any g ven number
5 H o 1 . o w in th e large , degrees , is acute angle wh se tangent is
Ho w n 2 o n 6 . do es the angle who se ta gent is c mpare with the a gle fi ure whose tangen t is 1 Check your an swer by drawing an accurate g . 11 ig DEF N N 17 , g ] I ITIO S
m l les A 14 C nstruction of S a a b . a a s . o l T pproxim te v lue of the trigon ometric fun ction s of a giv en a cute a ngle ma y be
50
16 . FIG .
O n a s o f s a r foun d by mea su rem ent a s follows . heet q u ed
a r n s ru a u a r r r its ra u s 100 a nd p pe , co t ct q te ci cle with di , C 18 ANE R G NOME R 1 PL T I O T Y (II , § 4
w its r a t in rs a rul in ith cente the te ection of two he vy gs .
Dra a a s r r u r w t ngent to thi . ci cle pe pendic la to the horizonta l
ru n s . G a n a a a la off a li g iven now y cute ngle , , y it bove the
r z a a s its r a t r o f r ho i ont l xi with ve tex the cente the ci cle . Ca ll the poin ts where its side cro sse s the circle a n d the ta n P d a n r s . r n a gent Q, e pectiv ely Then the o di te (y) of the
P ca n r a a t a s s a nd s b point be e d le t to unit , thi divided y
r O s a sin a a a s . g é LQ give the v lue of to two decim l pl ce
a r a s ssa x P ca n r a t u s a n d s Simil ly, the b ci ( ) of be e d o nit , thi
b 1 00 s co s k divided y give a . Li ewise the ordina te of Q ca n ’ > b e r a u s a n d s b 1 n 00 s ta a . e d to nit , thi divided y give
F n a l a sec cc cs c a ca n a s r r a s i l y, ctn , , , be computed the ecip oc l
ta n a co s a sin a r s s hn d of , , , e pectively. The tudent will it
' n s r s wa r m F i 1 6 a s fill i t uctive to com pute in thi y, f o g . , v lue to
o ut n a the followi g t ble .
ment An l s a ll s 1 uncti ns f Com e a r . 5 . F o o pl y g e If of thi
a is filled rr n t ble out co ectly, it will be fou d tha t every nu mber in it occur s twice ; o nce fo r
° a n a ngle le ss tha n 4 5 a n d once fo r a n a ngle A g rea ter tha n This res ult indica tes tha t
F I ' 1 G 7 ‘ the s in e of a ny a ngle is the cosin e of its com p lemen t; a n d the ta ngen t of a ny a ngle is the co ta ngent of its
e nt comp l me .
a The se rela tion s will n o w be p roved fo r a n y a cute a ngle .
° : 9 0 a e a a n d 8 a re a a s o f a Let B ; th n , the cute ngle
r D s s o o s e a a n d b a a n d b right t ia ngle . enote the ide pp it B y , 16 N OF R G H R ANG E II , § ] SOLUTIO I T T I L S
b 1 2 a n d s b 0 . n re spectively ; the hypotenu e y The y ,
hypotenu s e
hypotenu s e
ta n a side a dj a cent b
ctn B side oppos ite b
° r r ha 90 a whence , emembe ing t t [3 ,
° 20 sin a co s 3 co s 90 a ( ) [ ( ) , ° n (21) ta a ctn B ctn (90 a ) . In the sa m e wa y it ca n be shown tha t
° s (22) ec at csc (90 a ) .
i n 1 A ica t o s . a s r m r ra s 6 . ppl The v lue of the t igono et ic tio
a n m a r a fo r a ll a a n s a n d h ve bee co puted pp oxim tely cute gle ,
r o r in o n a s . s a s r ec ded c venient t ble The e t ble , togethe with the
rm a s s n a u s to s a ll a s s r fo ul j u t give , en ble ol v e c e of ight
On a 21 is r a a u s tria ngle s . p ge p inted ta ble giving the v l e of
s r a a s s a r a ra is the ra tio to th ee decim l pl ce . If till gre te ccu cy
- r r a ur a five a a s . eq ui ed , fo or pl ce t ble hould be em ployed In
- the following exa mple s the three pla ce ta ble is u sed .
° o f 3 a nd o EXAMPLE 1 . One angle a right triangle is 8 the hyp tenuse f f is 12 ft. Find the lengths o each o the o ther sides .
D fi ure raw a g , mark the given parts , and indicate
to o i a: . the parts be f und by su table letters , say and y The sides (1; and y are then respectively the side a d
n o fin o jace t and the side o pposite . T d cc, n te that o 12 the hyp tenuse is given hence by , FIG . 18 a: 12 co s ° 7 88 The value o f the co sine o f 38 fro m the three place table is fo und to be . Using this value we fin d 20 PLANE TRIG ONOMETRY [IL § 16
i o 12 Sim larly by equati n ,
° 12 SID 3 8
° o - of is and fr m the three place table the sine 38 f o und to be 616 . Using this value we o btain y 12
. 616 12
y
As P o o a check , the ythag rean the rem may be used , particularly if a
o f is . o o h tabl squares available Thus, den ting the hyp tenuse by , we s ho u ld have
o l This agrees reas nably wel With the given value h z 12 . Ano ther check o o o o fi ure that is m re practical is given by measurement fr m a g d g .
EXA MPLE 2 . One side of a right triangle is 17 and the angle o ppo site th1s side is what is the length o f the hyp ote nus e of the o ther side Deno te the hypo tenuse by u and the unknown i side by 0 . No ting that the side opp os te the given
fin d id a d a c nt 0 12 . angle is given, the s e j e , , by
1 . FIG . 9 o fin d o 12 T the hyp tenuse , use v 1 7 ctn 27 ° 17
17
0
° u 1 7 sin 27 1 7
P erf o rming the divisi o n we find u fi Check these answers by drawing an accurate gure.
o s of 41 o ne EXA MPLE 3 . The hyp tenu e a right triangle is and side is
13 find the o ppo site angle .
o o Deno te the pp site angle by a , then by
o 12 equati n ,
a 13 4 1 . 3 1 7 sin z FIG . 20 .
° ° . 326 . 2 09 19 Fro m the table ( p 1 ) we see that sin 18 . 3 and that sin , ° so that sin a is very nearly halfway between sin 18 and sin We ° judge therefo re that the angle a is abo ut halfway between 18 and ° hence a 18 . 5 16 N OF R G H A E II , § ] SOLUTIO I T TRI NG L S
TRIG ONOMETRIC F UNCTIONS TO THREE PLACES OF DECIMALS
sin a sec a ta n a ctn a csc a co s (it
cos a csc a ctn a tan a sin a 22 ANE R G N M E R PL T I O O T Y [II , § 16
- EXAMPLE 4 . The two perpendicular sides of a right triangle are 23 83 i and determ ne the acute angles and the hypo tenus e . Deno te the hypotenuse by h and the an gle o pposite the smaller side a o 12 12 by then by equati n ( ) ,
a 2 tan 3 83 .
After performing the divisio n it is fo und that
tan a . 277 As in the example abo ve it is no ticed that tan a lies very nearly halfway ° ° 1 5 16 o o x between tan and tan we have , theref re , very appr imately ,
Ot
1 irections or ol in ri n l s 7 . D f S v a e g T g . In the solution of
r a n s u se n r r t i gle , the followi g p ocedu e
a Dra w a a ra a r a s a a ( ) di g m pp oxim tely to c le , indic ting the
n a r s . M a rk k a r s b su a rs a n d give p t the un nown p t y it ble lette ,
e stima te their va lu e s .
0 I on e o the iven a rts is a n a cute a n le s r re ( ) f f g p g , con ide the
a n kn n a r s on e is s r find l tio of the ow p t to the which it de i ed to , a n d a r r r u a s 1 0 12 pply the p ope one of fo m l ( ) .
c I two sides a re iven a n d o n e a a n s is ( ) f g , of the cute gle
s r n k d efin ition a un a de i ed , thi of the of th t f ction of the ngle
which employs the two given sides .
k r s (d) Chec ea ch e ult .
— EXERCISES V. SOLUTIO N O F RIGHT TRIANGLES
1 of 21 n . One side a right triangle is the adjacent a gle is de termine the remaining side and the hypo tenuse . Check .
° of 21 o o 42 2 . One side a right triangle is and the pp site angle is de
C . termine the remaining side and hypo tenuse . heck
n of 28 o n e D 3 . The hypo te us e a right triangle is angle is eter
C . mine the two perpendicular sides . heck
is of n o o f o o 4 . What the angle incli ati n a r f which has half pitch 1 3 pitch
[ No w The pitch of a ro of is equal to the height of the co mb above the eaves divided by the total distance between the eaves ]
I o o h o o A 5 . n the f ll wing triangles den tes the hyp tenuse ; the angle is oppo site the side a a n d the angle B is o ppo site the side 0 . Use the
C . ta ble to compute the unkno wn parts fro m the given parts . heck
24 PLANE TRIG ONOMETRY 19
- 19 . Us e of the a r e a bles F L g T . ive pla ce ta ble s a re u sed in r s s m m a a nn r a s s a a 21 . p eci ely the e e the m ll t ble of p .
° ’ EXAMPLE 1 One angleof a right triangle is 42 20 and the hypo tenus e
28 ft. 6 in . n . i is lo g Find the remain ng sides and the other angle . Draw ia to o a d gram illustrate the pr blem, indicating the given parts . Deno te
k n o n a b in the un w parts by the letters and , as Fig . 22 .
T o fin d b o is side a d a cen t to , n te that it the j the given angle , and that
o s . the hyp tenu e is given Hence , by 12
° cos 42 20 ’ x . 73924
No te that a is Opposite the given angle ; hence
by 12.
° a 42 20' sin x . 67344 ° the sine and the c osine o f 42 20’ being fo und in FIG . 22 . 4 the Tables, p . 3 . ° o of 42 is 47 The angle 13, being the c mplement
of EXAMPLE 2 . The perpendicular Sides a right triangle are 22 ft. 6 in .
ft . o th e and 54 . , respectively Find the hyp tenuse and angles . t i s Draw a diagram , indica ing the g ven part and lettering the parts to find o o . 23 . T o a be f und , as in Fig , n te that the given parts are the sides o o a n to de n itio n o ta n ent pp site d adjacent it hence by the fi f g , we write 4 tan a 54 . 1667
o . Fr m the Tables, p 33 ,
° ’ ° ’ tan 22 37 . 41660 and tan 22 38 . 41694 whence
° ° a 22 o 67 and 6, its c mplement, is
o of o us By the P ythago rean the rem plane ge metry , ing
of o o . 9 4 a table squares and square r ts , Tables p , 2 h? z 571 B h . 1 03 . whence , Tables, p 225
F I G 23 Another meth o d of finding h is the fo llowing : Having ° ° ’ 2 fo und a 22 h 54 / cos 22 37 by ( 15) § 1 . o hi o o to o o o H wever, t s meth d is pen the bjecti n that any err r made in
o a o fo r h . I co m c mputing vitiates the resulting value f und n general, t do n ot use com uted pute each un kno wn part fro m the given parts ; . e. p p a rts a s da ta if it ca n be a vo ided.
I n s n r r a s s r a r r s olvi g ight t i ngle , ob e ve c efully the di ection of — 1 7 . 22 a n d u se five a a s un n s a s , p , pl ce v l ue of the f ctio (T ble , — — . 22 44 n d 94 11 1 s ra r pp a pp . ) a s illu t ted in the p eceding
a s ex mple . I 19 N OF R G H R ANG E I , § ] SOLUTIO I T T I L S
— R R A E EXERCISES VI . IGHT T I NGL S
o o o . o s 1 . S lve the f ll wing right triangles The hyp tenu e is deno ted by o h , o ther sides by ther small letters, and any angle by the capital letter to o correspo nding the small letter that den tes the side o ppo site it.
° ° A 1 b d M : 49 h : 24 . 6 = = a 6 18 . 2 50 . ( ) ( ) ( g) p , q ° A 32 a 33 . e b h 10 . h = ( b) ( ) ( ) u h 25 . ° : = 2 4 . U 63 4 . m 4 .2 = c A h 5 0 i 3 h 100 . ( ) ( f ) u ( ) , ° ° a b 6 14 75 Ans . ( ) ( ) ( ) ( 6 ) ( f ) ( g) ( h) 27 ° 62° ( 7)
ol o find n ot 2 . In the f l wing right triangles the side given
I of o o find s not 3 . n each the f ll wing right triangles the three part given and the area . ‘ ° ° ( a ) a h Ans . 24 65
A b . Ans . ( b) 100 3 501 .
c h 43 B An s . ( ) ,
d h z 17 6 A Ans . ( ) , ° 6 h 425 b 304 . Ans . 45 297 45144 . ( ) , ,
4 . of iso is 324 ft. The base an sceles triangle , the angle at the vertex is ° 64 Find the equal sides and the altitude . An s .
o of o 200 ft. i ft. n . is th e 5 . The shad w a t wer h gh is lo g What f ° angle of elevati o n o the sun Ans . 38
o of ft. 6 . A ch rd a circle is , the angle which it subtends at the
of An center is Find the radius the circle . s .
wi A B of B o 7 . To determine the dth a river, a line C 100 r ds long is laid o ff at right angles to a line fro m B to some o bject A o n the Opposite ° B A is o to 43 A B . bank visible fro m B . The angle O f und be Find
Ans .
f f o o 8 . What is the angle o elevati on o a m untain sl pe which rises — ° f o n e o f u o Ans . 21 238 t. in eighth a mile ( p the sl pe) ANE R G N ME R 1 PL T I O O T Y [II , § 9
9 Two o . ships in a vertical plane with a lighth use are o bserved fro m
to 2 ft . f its p , which is 00 . abo ve sea level The angles o depressio n o f the ° two i 1 5 1 7 ’ 1 1° 22’ sh ps are and Find the distance between the ships .
Ans .
10 . A fl a sta ff o n to f o . t f f g stands the p o a h us e A a po int 100 t. ro m the ho use the angles of elevatio n o f the b o tto m and to p o f the staff are ° ’ ° f respectively 21 50 and 33 Find the height o the sta fl . An s . 25 .
1 1 A 24—o o o . f t ladder can be so placed in a street as t reach a windo w f i o n o n e o n o o w 16 t. h gh side and by turning it o ver its f t it ill reach a w 4 f f i o 1 t. hi o n o o . w nd gh the ther side . Find the width the street
Ans . 1 f 2 . The length o o n e side of a regular pentagon is 24 ft. Find the lengths of the radii of the inscribed and circumscribed circles and the
n area. A s .
1 of u in . n . of 3 . The side a reg lar decagon is 10 lo g Find the radii the inscribed and circumscribed circles and the area .
Ans .
o o o o f 14 . A r und silo feet in diameter subtends a h riz ntal angle
Ans . Find the distance from the o bserver to the silo .
1 5 I o o n o . n an is sceles right triangle sh w that li es fr m either base to o of o of o o ff angle the p ints trisecti n the Opp site side cut respectively, o n e-fifth and o n e- half the altitude from the hypo tenus e to the vertex of i the r ght angle . CHAP T E R III
TRIG O N O M ETRIC RELATIO N S
I r u o n A s m r n m r r a n s 20 nt o d ct . . i few i ple t igo o et ic el tio
a in 1 2 13 a n d 15 . s a r s a h ve been given , , I n thi ch pte we h ll
rs s u s u first r w s a r a obta in othe . The t dent ho ld evie tho e l e dy given .
21 P tha orea n Rela t ons a be . y g i . The following equ tion A
a s s sa co a a nd ra s r is ru tween the b ci , the ordin te y, the diu t e
’ fo r every point in the pla n e
? 2 2 (1) at y r .
D b r2 a ividing y , we obt in
? 932 y 7 $2
b 1 1 a t a s n a is a but y , le t whe cute w r co s a r sin a / , y/ ; hence Em , 24 .
2 2 (2) sin a + co s a = 1 ; A
e. th th s u a res h i i t t. e su m of e q of t e s n e a n d cos ne of a ny a cu e a ngle is equ a l to u n ityfr
D n 1 b a n d n b a r s ividi g ( ) y the y y , we obt in e pectively 2 2 3 1 ta n a s ec a ( ) ,
4 1 c 2 2 ( ) tn a c s c a .
F ormu la s a n d (4) a re exa mples of trigon ometric
s . An a n ua n a is a n a n identitie identity in y q tity, , eq u tio con
F o rm u las a nd (4) are called the P ytha g orea n rela tio ns because o n o m u o n P n o m o f they are btai ed fr this eq ati , which is the ythago rea the re plan e geo metry .
m n a s w 3 a nd l o un to o TThis state e t , ell as ( ) (4) be o w , will later be f d h ld fo r n fo r n all a gles , the ge eral d efinitio ns o f si n e a n d co sin e . 27 28 ANE R G N ME R 2 PL T I O O T Y [III , 1
. ta inin g a which is sa tisfied by every va lue of a for which both
r r fin d M a n a m s f membe s a e de e . y other ex ple o identities will
be found in the pa ges tha t follow . The s e form ula s a n d tho se of 1 3 a re often u s eful in simplify
in r s o r in u O r r s n g exp e sion s v erifying eq a tion s . the inte e ti g
r r s s a ela tion s a e given in exerci e th t follow .
o o 4 a co s4 a 2 a cos2 a EX AMPLE 1 . T sh w that sin sin . The expressi on o n the left is the difference of two squares and can therefo re be facto red hence we have sin4 a co s4 a (sin2 a cos2 a ) 2 a cos2 a i to 2 a co s 2 a 2 a co s2 a ( sin ) wh ch is equal sin , since sin 1 . The fo rmulas may also be used to co mpute the value of o n e of the o o o s o of o trig n metric functi n fr m that an ther .
2 . 0 to fin d co EXAMPLE Given tan s 0. 2 2 2 An a l tic M ethod . B 1 0 0 y y tan sec hence , sec 0 1
o r see 0 co s 2 Hence , 0 since cos 0
1/sec 0.
G eo metric M ethod. o o i o o The f ll w ng meth d is much m re practical , and is e a sily applied to any example of i o th s s rt . Draw a right triangle who se base 2 o is 1 and wh se altitude is 5 . The
o o o hyp tenuse is easily f und t be 1 3 . It follo ws that side adjacent cos 0 hypo tenus e 25 . FIG .
— EXE ISES P AGOR AN R LA O N . IDE TITIES RC VII . YTH E E TI S N
1 I a 3 of . n exercises ( ) ( ) determine the values the remaining func
o s of 0 of o of E 2 o . ti n the acute angle by each the meth ds xample , ab ve
: sin 0 : 1 3 . c co s 0 : 1 3 . ( a ) sin 0 3 / 5 . ( b) / ( ) /
= e ta n 0 = 3 4 . ( d) sin 0 5 / 13 . ( ) ( f ) /
z l s ec o z 2 . ( h) sin o b/ c. ( )
P ro ve the fo llowing relations fo r any acute angle 0
0 cos 1 2 0 cos 0 . 3 . co s 0 0 s : 0 . 2 . (sin z sin tan sin
0 0. 5 . 0 9 0 . 4 . tan 0 ctn 0 sec csc sin sec tan
9 0 0 0 1 . 6 . (sec tan ) (sec tan )
7 3 0 co s3 0 0 co s 1 0 co s . (sin ) (sin 0) ( sin 2 — 2 — ? 2 — 8 . co s 0 sin e z l 2 si11 9 = 2 cos 0 1 .
' 2 2 2 2 9 . sec (9 csc 0 tan 0 ctn 0 2. 23 R G NOME R C RE A N III , ] T I O T I L TIO S
° ° 22 unctions of 0 a nd a n a n 0 a on . F If gle of be pl ced
n s ru a 1 4 ma co ordina te a xe s a d the con t ction of p ge be de, the
x-a s a nd s a a p o int P will lie on the xi , we h ll h v e
a: r O. , y
° o s si o s n a a n d s a n 0 a re The functi n ne , c i e , t ngent, ec t of d efin ed by the sa me ra tio s a s a re the co rresponding fun ction s< o f a cu te ce a s a n d a 14 m in p ge ,
° ° 59 ° in = 2 : 0 co s 0 1 ta n 0 0 sec s 0 , , , r r cc
The defin ition s of cota ngent a n d cos eca nt given fo r a cute
° s a n a 0 a s 0 a n d r r a ngle c n ot be pplied to bec u e y , the efo e the
s s x a n d r r s defin itio n s a re divi ion /y /y, which occu in tho e , impossible . ° Simil a rly if the a ngle of 90 be pla ced o n the coordina te
n d s r a 1 4 ma n P a xe s a the con t uction of p ge be de, the poi t
o n -a s a n d s a a will lie the y xi , w e h ll h ve
a: 0 r. , y
° s n s n a n a n d s a 90 a re defin ed The i e , co i e, cot nge t, co ec nt of by the sa m e ra tios a s a re the corre sponding fun ction s of a cute a ngle s hence by the defin itio n s
° 3 ° ° in 2 = 1 o s 90 1 90 0 csc 9 0 Z s c 0 1 . , , ctn , r r y y The defin ition s of ta ngent a n d s eca nt given fo r a cute a ngle s
a a a u s a: 0 a n d s o s x c nnot be pplied to bec e , the divi i n y/
° s h s n o r a n d r/w a re impo ss ible . We a y tha t 0 a cota ngent o ° “ s a a n d 90 ha s n o a o r seca n tfi co ec nt, t ngent
unctions of r i 23 . F In pla ne geomet y it s shown how to con struct a right tria ngle in which on e a cute a is o r F ro s r a s sin ngle or m the e t i ngle the e,
s n a . s a n s ca n u . co i e , t ngent , etc , of the e gle be comp ted
I o n n n o f fo r m in nite t is fte said that the ta ge t exa ple , is fi ; this ex pression do es no t give a n y value to the tan gen t at but merely describes the fact that the tangen t beco mes a nd remai n s larger than a ny number we m a n m y a e as the a n gl e appro aches Sim ilar statements ho ld fo r the o thers . 3 0 ANE R G N ME R 2 PL T I O O T Y [III, 3
To fin d the fun ctio n s of con struct a n iso scele s right
r a a l s s s n n t i ngle with the equ ide ome conve ie t length m. By the Pytha gorea n Theorem compute the hypotenus e m\/2
b definition s 1 0 1 1 12 Then y the ( , ) ,
° m 1 sin 45 _ m \/ 2 x/ 2
° m 1 co s 45 _ mx/ 2 V2 2 . FIG . 6
b a s r a s 16 1 7 1 8 12 whence y m e n of the el tion ( , , , § ,
° ° ta n 45 1 a n d sec 45 csc ctn ,
° To fin d the function s of 30 a n d con stru ct a n equ ila tera l
tria n le s m a n t in to two ri ht tria n les b a er g of ide , g g y p
en dicu la r r s s A p f om one o the oppo ite ide . pply the
s 12 a n a u s s definition , to obt i the v l e of the function
° ° n d in of 30 a 60 given the following ta ble .
s a s s z s a s The e v lue hould be memori ed , ince the ngle
° is a s s a ll a n d 90 occu r freq uently . It e y to how th t a of
r a s r in 13 1 5 o fo r a u s the el tion p oved , , h ld the v l e given hi in t s ta ble .
ri m i u 24 . ono etr c a t ons An o t a n T g Eq i . Aidentity 21) is sometime s ca lled Thu s the equa tion sin a co s a 1 1 there a re ma ny va lues of a for which it is not true ; there a re
a s o f a r sa s a : fo r v lue , howeve , which do ti fy the eq u tion
32 LANE R G ON ME R 2 P T I O T Y [III , § 5
“ 1 a l s o be w ritten in the form a sin (2/7 Thi s eq ua tion is to r a a a s s is 2 7 be e d, the ngle who e ine / It shoul d be ca refully n oted tha t the 1 ) o f this nota tion I S n o t a n expon en t a lthough it is w ritten in the position
su An r r r u a lly occupied by a n exponent . y othe cha a cte w ritten in the sa me position would be rega rded a s a n o rdina ry
s ss sin 2 8 rs exponent ; thu the expre ion , would be unde tood to
a s a r si a 8 . me n , the qu e of the ne of the ngle ,
Ma r n a n a rcsin a; a ny p efer the ot tio to the one given bove , a n d s a h n ot so r a s thi not tion , t ough f equently employed the
o r is r ss u s a s ra . s a the , neve thele ed to con ide ble extent We h ll therefore thro ughout thi s bo ok u se either nota tio n in order to
a a r z s u n o f mili i e the t de t with b th .
E ERC E — P TR O O E R E T O X IS S VIII . SIM LE IG N M T IC Q UA I NS 1 f . S o lve the fo llo wing equatio ns by co nstructing a figu re oreach . 63 ( a ) sin a: ( 9 ) co s a; .
( b) sin a; ( h) co s cc V3 / 2 .
i a: 0 . ( 6 ) sin a; . 8 ( ) sin
d a: 66 co s a: 0 . ( ) sin . 8 (j)
sin 48 k as 1 . ( 6 ) a; z . ( ) sin
= z l co s az l . (f ) co s m 1 / 2 . ( )
o o o f o 2 . P ro ve that there is always an acute angle s luti n the equati n
x z c c u 0 1 . sin , if is any n mber between and
is o o o f o 3 . P ro ve that there always an acute angle s luti n the equati n
a; c c is o m . tan z , if any p sitive nu ber whatever
- 1 2 5 i l . 4 . Find sin ( ) graph cal y
T Co EX . [ HIN . mpare
5 E to o f l a to l l . xpress the answer each the exercises ( ) ( ) by — — s o f o o sin 1 or co s 1 or o mean the n tati n ( arcsin, arcc s,
"1 s in l o 1 2 . 6 . Find and a s tan ( ) graphically
“ 1 7 . Find arcsin and als o tan by the Tables .
o f So lve each o f the fo llo wing equati ns o r x .
2 - 8 . 2 sin m+ sin c 1 .
Hm r o hi for 33. of o n o o [ S lve t s quadratic sin There are , c urse , s lu tio ns co rrespo nding to values o f sin a: greater than
a 2 sin2 m 9 . ( ) 25 R G N ME R C E A ON III, § ] T I O O T I QU TI S
1 0 . ( a ) tan ( d) tan a: ( b) ta n fc ( 6 ) tan a:
c ta n mz 2 . ta n z ( ) (f ) w o.
2 2 2 1 1 a a; 3 . b 0 6 . 6 . ( ) tan ( ) tan 4 ( ) tan 0 6
2 2 2 12 . 2 a: cos a; 1 . 1 . co s a; sin 3 sin a) .
2 = 2 = 14 . 5 sin m+ 2 cos zc 5 . 1 5 . sec c + ta n az 3 .
1 I o f c o 6 . a b A f and are the sides a right triangle , the hyp tenuse , and o a o of to the angle Opp site , sh w that the area the triangle is equal either o f the expressio ns a o cos A be sin A 2 2
1 A 7 . Two straight pieces of railro ad track M and N B are to be co n
n ected ul AKB of 500 ft . by a circ ar track with a radius and center, 0 t A tangent o M and N B . The straight po rtio ns of the track pro d u ced intersect at a po int V at an angle of
' ( a ) Ho w far back fro m Vsh ould the track begin to turn ( b) How far fro m V alo ng the bisecto r O V of the angle A VE is the center 0 ( 6 ) Find the sho rtest distance F IG 27 o o from V to the curved p rti n .
1 fi u re to of Ex 17 A A VO n 8 . If , in a g similar that . is any a gle , and
a 0 A r o A VOA is denoted by , and , sh w that
( a ) A V z r ta n a ; ( b) KV z r exsec a ; AB = in ( c) 2 r s a .
1 b of Ex . 16 o A to 9 . The side the triangle in is extended bey nd a AD 0 so A BD o o po int D , making , that is is sceles . Sh w that
B ( a ) A AD B z A/ 2 ; ( b) BD 2 6 co s ( 6 ) From the right triangles D OB and A CE sho w that
c sin A = a = 2 c co s A 2 q A b C ( / ) sin
28 . FIG . hence sin A z 2 sin co s 2 d Li o 6 co s A b 2 2 6 co s A 2 c ( ) kewise , sh w that ( / )
‘ hence cos A z 2 cos2 ( A/ 2) 1 co s2 ( A/ 2) sin2
D 34 LANE R G N ME R 2 P T I O O T Y [III , § 7
Pro ections 26 . ro ection j . The p j of a line segment AB upon a lin e I is defin ed to be‘ the portion M N of the line l between
r n a rs ra n r A a n d B s pe pe dicul d w to it f o m , re pectively . The length of thi s proj ection is ea sily found if the length of AB a n d the a ngle a which the line AB m a kes with l a re
kn own . F o r ra a a a l r u , d w p r llel to th o gh
A n BN a t 0 . A E is a , meeti g Then C right tria ngle a nd the a ngle a t A is a ;
29 . FIG . b 12 hence y ,
M N = AB co s a
o r the ro ection o a se ment u on i , p j f g p a g ven lin e is equ a l to the p ro du ct of the length of the segment a n d the cos in e of the a ngle the se men t ma kes with the tbf en lin g g e. The proj ection s of a seg ment upon the coordina te a xe s a re freq uently
n u s ed . If the s eg ment m a kes a a n a r z a ro gle with the ho i ont l , the p
ectio n s w a n d a s a re j on the y xe ,
r s e pectively,
5 P ro AB AB co s a ( ) jz , FIG . 30 . P ro AB AB sin a j, ,
r P ro AB a n d P ro AB n r s AB on whe e jI jy de ote the p oj ection of
x—a s a n d -a s r s the xi the y xi , e pectively
2 li i f Pro e ion s n 7 . A ca t ons o ct a s a d r a pp j . In mech nic el ted
s s r s a n d s a re r r s ra a b ubj ect , fo ce velocitie ep e ented g phic lly y
A 1 l is r r s b s s s . sa 0 b . a e line egment force , y of , ep e ented y g
r A velo c ment 1 0 units in length in the direction of the fo ce .
b a s m 20 un s in 2 f er sec . is r r s n ity of 0 t. p ep e e ted y eg ent it
r o f length in the di ection motion .
n a n I a s r r s The proj ection upo given li e , of eg ment ep e ent
in a r r s s r in r l s g force , ep e ent the effective fo ce the di ection thi
is ca lled the comp on ent o f the given force in the direction I. 111 27 PR JEC N , § 1 O TIO S
A o f 50 lh . EXAMPLE 1 . weight is placed u pb n a smo o th plane in ° o f 27 o o o clined at an angle with the h riz ntal . What f rce acting directly up the incline will be required to keep the weight at rest Draw to so me co nvenient scale a segment 50 un its in length directly do wnward to represent
r ct this the f o rce exerted by the weight . P oje I seg ’ ° ment upo n a line inclined at an angle of 27 with
of o o W the ho rizontal . The length this pr jecti n Q , ° 3 1 50 co s 63 . re re Fig . , is nearly This p FIG . 31 . f o o sents the co mponent o the f rce d wn the plane .
o o o f lh . i . Theref re , a f rce acting up the plane will be requ red
A 2 . A 30 ft. o o o o EX MPLE ladder l ng, when lying h riz ntal supp rted at
its s o of 150 lb . o n un . Is end , will carry a safe l ad its middle ro d it safe to for a man weighing 190 lh . mo unt it when it to f is so placed as reach a window 18 t. abo ve the gro und
to find o o er endicu We have the c mp nent, p p ‘8 ’ to of lar the ladder, the man s weight when he
o n un . L WP stands the middle ro d et , drawn
C vertically do wnward fro m the middle po int
of AB . 32 190 , Fig , represent ( which need n o t be o n the same scale as A B which repre sents Then the compo nent perpendicul ar to AB is Em . 32.
WQ 190 co s P WQ 190 cos CAB.
co s CAB A C/ A B
l 9o x 4 WQ 1 52 , 5 which is greater than the safe lo ad .
A l o nifo r EXA M PLE 3 . trave ing crane m ves with u m speed do wn a
0 ft. 1 n e I o o 6 mi . 41 se . o sho p 297 ft . l ng and wide in t carries a l ad fr m o of o n e co rner along the diago nal t the o pposite co rner . Find the speed the crane and of the car which 29 7 runs o n it. Let AP the speed of the lo ad 6 alo ng the diago nal which by the of data the pro blem 3 ft. per 33 . FIG . AP no t o f o r o n sec . ( need c u se be B A A f the same scale as A and D) . Then Q the speed o the crane s A P o f A + 3 co P Q and Q the speed the car 3 sin P Q . 59 ANE R G N ME R I 27 PL T I O O T Y ( II , §
E E P RO EC O XERCIS S IX . J TI NS
1 o i o a o o of . Find the h r z nt l and vertical pr jecti ns the segments ° a t 42 o f 37 o o . ( ) leng h , making an angle with the h riz ntal i of 50° ( b) length mak ng an angle with the vertical . A ns . ( b)
2 . A straight railro ad cro sses two no rth and so uth ro ad ways a mile
of o 1 . A apart . The length track between the r adways is 4 mi train f o of travels this distance in 2 min . Find the components o the vel city
the train parallel to the ro adways and perpendicular to them . Find the ° o An s . 53 angle between the track and either r adway . g, 3,
as yelo cit f 3 . The e tward y o a certain train is 24 mi . per hour . The
o 2 . o no rthward vel city is 3 mi per h ur . Find its actual velo city along the
track and the angle the track makes with the east and west direction . ° An . s 40 , 53
4 . A is s f I o o f 5000 lh . car drawn by mean o a cable . f a f rce exerted o is to ll o r al ng the track required pu the car , what f rce will be requi ed f ° n when the cable makes an angle o 15 with the track A s .
5 . o o o o o f o of 30 1b Find the h riz ntal and vertical c mp nents a f rce . ° a n o f 40 An . making angle with the ho rizo ntal . s
6 . Find the ho riz ontal and vert ical projecti o ns of the segment which
0 1115 o 8 3 2 Ans . 10 10 . 3 the p ints ( , ) and , ,
for t 2 ft. in . n . 7 . The stringers a s airway are 0 lo g The steps are
12 in . s hi 1 in . o . to have 7 in . risers and tread (w ch includes verhang) of o ro ec Determine the number steps, using the ho riz ntal and vertical p j
An . . tions o f the string er to check the result . s 19 A — AB A AD AE 8 . o on o : 4 0 viz . : O Five f rces act the p int , ) , , , , ' AF o s A B O D E F o f a , and the p int , , , , , are the vertices a regular hex
o n o r . o com g , center at the igin Sh w that the vertical
“ o n nts fin d of o iz o 1 W p e balance , and the sum the h r ntal 1
An s . 24 . co mpo nents .
of for 9 . Determine the width and height a crate
4 . An s . the chair sho wn in Fig . 3
1 0 I o o of i o n . n surveying , the pr jecti n a l ne a P ” n o rth and so uth line is called the la titud e of the line and the projectio n o n an east and west line is called
“ > the dep a rtu re o f the line . Find the latitude and de
F I G“ 34 ’ parture o f the fo llo wing lines
° ’ a 41 od N 26 15 E . An s . ( ) length r s , bearing ° ’ A ns . b 487 E 32 30 S . ( ) length feet , bearing ° ’ 6 o N 40 45 W. ns . ( ) length r ds , bearing A CHAPT ER IV
L O G ARITHM IC S OLUTIO N S O F RIG HT TRIANG LES
2 The se o Lo a r thms r s ma s f . a 8 . U g i Log ithm y be u ed to s r u a s n mu lti lica tion s division s ra s ho ten comp t tion involvi g p , , i A in o wers o r ra ro ots a n s o r g to p ext cting , but not involving dditio
u r s In r a rk ws s bt a ction . m uch of the nu me ic l wo which follo , the u se of loga rithm s is very a dv a n ta geou s in sa ving tim e a n d
a r s u s a r in a a m s a re l bo , but the t dent hould be mind th t log rith d 4 n o t ssa r . a re r n a n nece y They m e ely co venient, they belong M M n e u s no more to trigonometry tha n to a rithm eti c . O of the q e tion s which a compute r ha s to decideis whether or not it will be a dva nta g eou s to u s e loga rithm s in a given problem . At the end of this book will be found a ta ble of the
a r ms n rs a s . a n d a a log ith of u mbe (T ble , p t ble of the W a r h s r n n s a s . log it m of the trigono met ic fu ctio (T ble , p ith
r u v a s a r expla na tion s of thei se (pp . In c e eview of
s a s is s r s a a n s the principle of log rithm de i ed , thi expl n tio hould
r be studied before proceeding with the re s t of this cha pte .
° ' The nota tion l o g ta n 62 51 mea n s the loga rithm of the
° ’ ° ’ a n n 62 5 1 a 62 51 is a r t ge t of the t ngent of nu mbe , a n d the loga rithm of thi s nu mber is a s m a y be seen
n I s a s r s is un by looking u p log i Ta ble . Thi l t e ult fo d
3 na s u s a vo id a r a . 7 in T ble III , p , which e ble to the l bo of
k a s a n d I in s s s . loo ing in T ble II , ucce ion A formula which ha s been a rra nged so a s to involve only produ cts a nd q uotients of p o wers a n d roots of qu a n tities
r kn n a n s either kn own o r ea sily computed f om the ow qu titie ,
wit bri ta les o n o u - In the editio n o f this bo o k h ef b , ly f r place tables are
o u n o n o u to THE M C LL B L E S given . Th se si g that editi sh ld refer A MI AN TA ,
to which a ll page referen ces made here apply . 37 38 LANE R G N ME R 29 P T I O O T Y [IV, § is sa id to be a da p ted to loga rithmic comp u ta tion
u fo rmula h 2 w Th s the b , hich gives the hypo tenus e h o f a o s of a is n o right triangle in term the sides and b, t adapted to lo garithmic
o o . O n o o c mputati n the ther hand, the f rmula
which gives o n e side in terms of the hy o ten us e o p and the ther side , is adapted to logarithmic computatio n because ( h a ) and ( h a ) are easily o btained fro m h and
a . u if o us Th s , the hyp ten e is and o n e side is the o ther side is
35 a: FI G . . z log log log log a: cc 2 Tables, p .
rmu a s 1 0 12 13 a re a ll a a a The fo l ( to , , d pted to log ri hmi t c computa tion .
° a 46 23 EXAM P LE 1 . Find sin log log sin 46 ° 23 ! 10 log a
a Tables, p . 4
a o a EXA MPLE 2 . Find fr m tan
log log log tan a 0: 57 ° Tables, p . 7 8
h EXAMPLE 3 . Find co s 1 5 ° 20 ’ log 10 ° lo g cos 15 20 ’ 10 log h 8 h Tables , p .
w h e a e n u 2 Produ cts t N t v a ctors . fi d b se 9 . i g i F To y of loga rithm s the p roduct of severa l fa ctors some o f which a re
a e r f sa a rs a ll ta ken os itivel neg tiv , the p oduct o the m e f cto , p y, i first o a a n d s n is n rm n u sua s bt ined , the ig the dete i ed in the l
ANE R G ONOME R IV 29 PL T I T Y [ ,
A ( g) ns . h ( ) Ans . 2 i ( ) Ans .
A ns .
2 o o w o d . The f ll ing f rmula is used to determine the
d of of co efficient o f o c diameter , water pipe in terms the fricti n , the
l fl ow h . Co d 6 l 500 length , the f, and the head mpute when ,
: h 1 0 . f 5 , An s . A 3 . wire cm . in diameter and cm . long is stretched of 4 4 f cm . by a weight 5 grams . Find the mo dulus o el a s ticity by the
o e of o o s rmula , in which l length a area cr ss secti n , and f 32 ,
7 o o o a A ns . 1 the el ngati n pr duced by weight w. x 0 .
i fio w o f o o 4 . The water ver a weir is given by the f rmula
Find f when k g h Ans .
5 o o . A steel bar cm . l ng between supp rts cm . wide and
fl cted . cm . deep is de e cm by a weight of 5000 grams at the middle . wl3 o ul s of o r 6 Find the m d u elasticity by the mula ’ in which I f 4 bd3h
fl c i w b d h de e t o n to . length, breadth , depth , and the due the weight A n 9 s . x 10 .
o 0 of o 6 . The pressure p and the v lume a gas at c nstant tempera
0 73 ture are connected by the relatio n p 0 Is. Find p when
Ans a z k 12600 . .
o o f o o 7 . The peri d a c nical pendulum is given by the f rmula
a w 4 Find T when m t z 3 0 .
An s .
8 o a l . of o i of h ih . . The v lume ( g ) a c n cal tank height ( ) and vertical 3 2 of o angle 2 a is v 1rh tan a 693 . Find the capacity such a tank wh se ° ’ is 2 ft 5 in angle at the vertex is 42 30 and who se height 1 . . An s .
If of i s r o r of a di s R 9 . a ball rad u is r lled inside a spherical su face r u , z( 16 the time o f o scillation is given by the ormula T 21r f 5
in oscilla p the ra dius of a co ncave mirro r in which a g . steel ball makes an
Ans . tion in sec . Take g 384 . 29 R G H R ANG E BY OG AR HM 4 1 IV , § ] I T T I L S L IT S
10 o s o f o o o . S lve by mean l garithms the f ll wing right triangles, where h o o deno tes the hypo tenuse , ther small letters the sides , and the c rre spo nding capital letters the angles o ppo site tho se sides .
n ( a ) A h A s . n ( b) P p A s . ° ( 0 ) A 28 h An s . ° : n ( d) U 28 v A s . ° n 60 ( 6 ) a h A s . ° An 55 ( f ) r 8 s . ° n 59 ( 9 ) a h A s . ° ' ( h) G 36 21 h Ans .
1 1 o o o . S lve the f ll wing right triangles having given ° An 17 ( a ) hypo tenuse side s . ° An s . ( b) angle 43 side a dj. ° O Ans . ( 6 ) angle 55 side pp . ° h Ans 37 ( d) yp . side . ° d s: A ns . ( 6 ) angle 49 side a j. ° Ans . 32 ( f sides 126 and 198 . ° An 7 s . ( g) angle 5 side Opp . ° 2 Ans ( h) angle 3 side opp . .
1 . o n o o f o o 2 A tree stands the Opp site side a small lake fr m an bserver. At the edge of the lake the angle of elevatio n o f the to p of the tree is ° 10 ft d fo und to be 30 The observer then measures 0 . irectly away ° from the tree and finds the angle of elevati o n to be 18 Find the f n height of the tree and the width o the lake . A s .
o o 250 ft. o of o o n 13 . Fr m a p int fr m the base a t wer a level with the 2° base the angle of elevatio n of the to p is 6 Find the height .
Ans .
14 . T o o f o o is determine the height a t wer, its shad w measured and o t ft o A ten- o o o th en o o f und o be . l ng . f t p le is held in vertical p siti n o to f of o and its shado w is f und be t. Find the height the t wer and ° the angle of elevatio n of the sun . An s . 61 1 5 . Find the length of a ladder required to reach the top o f a building
50 ft. hi o o 2 ft o o f i o gh fr m a p int 0 . in fr nt the bu lding . What angle w uld ° h o o n W Ans 6 the ladder in t is p siti make ith the gro und . 8
1 6 . o f o f o 4 ft o f o The width the gable a h use is 3 . the height the h use o ft of o f ab ve the eaves is 15 . Find the length the rafters and the angle ° o of A ns 4 1 inclinati n the roof . .
. A f fin d 17 ssuming the radius o the earth to be 3956 mi . the distance to the remotest po int o n the surface visible fro m the top of a mo untain 2 5 mi . high . An s . mi . CHAPTE R V
SO LUTIO N O F O BLIQ UE TRIAN G LES BY M EAN S O F RIG HT TRIANG LES
30 Decom osition of O bl ue ria n les in o Ri h T i . p iq T g t g t r
les A ra fo r s u r ll a ng . gene l m ethod olving obliq e t ia ngles in a ca ses con sists in dividing the tria ngle into two right tria ngles by a perpendicula r from a vertex to the opposite side ; thes e right tria ngles a re then solved by the methods of the p revio u s
r a ll r s a s r r cha pte . In except the th ee ide c e the pe pendicul a ca n be dra wn so tha t one of the res ulting right tria ngle s co n
m s ta in s two of the given pa rts . It a y om etimes ha ppen tha t
r ul r a u s r the pe pendic a will f ll o t ide the given t ia ngle .
3 Tw n les nd de i 31 Ca s e 1 : G ven o A a a S . s im . i g i It
ma r a s is s n ir a ca n te i l which ide given , i ce the th d ngle be found fro m the fa ct tha t the su m of the three a ngles is Dr o p
i h it o f i n s i the perpendicul a r from e t er extrem y the g ve de.
An o o n e to o EXA MPLE 1 . blique triangle has angle equal an ther n o n t equal to a d the side o pposite the unkn w angle equal o 51 . De
termin e the remaining parts . It is immediately seen that the third angle is ° 180 T o s o lve this triangle draw the figure approximately to scale and dro p the perpen 5 ‘ ' dicul a r 0 s from o n e extremity C of the kno wn
to A B o o C. D o o side , the side pp site en te the unkn wn B A CD h o t side C by a . In the right triangle , the yp
o n o 12 A enuse and e angle are kn wn hence by ,
F I G 36 . ° p z 5 1 sin 67
A n o o B OD n o w angle and the side pp site , in the right triangle , are
o n 12 kn w ; hence by ,
° a z 1) / sin 70
B o . C a s 5 . 4. The side A may be f und in the same manner heck in , p 42 ° If in the equa tio n a p /s iu 7 0 we s ubstitute the v a lue ° 5 1 sin 67 r sl u n a n fo r a a n p p eviou y fo d, we obt i the eq u tio
51 sin 67 _ a , ° sin 7 0
s r a r A Thi fo mula is a da pted to log ithmic com puta tion . pply in g the principle s of loga rithms we obta in
° l og a = ‘lo g 51 lo g sin 67 lo g sin
Remembering tha t s ubtra cting a loga rithm is eq uiva lent toé a n co - o a r sa n r ma a rra n ddi g the l g ithm of the me u mbe , we y ge the n u merica l work a s follow s
7 lo g 51 ° lo g sin 67 10 ° co lo g sin 70 log a
I n s s o wa s a . E ua n s thi oluti n , p elimin ted ven if the eq tio a re s a i a a a n n o t u ed without elimin t ng p , the ctu l v lue of p eed
un s l o is s u . be fo d , ince only g p needed to complete the ol tion
32. Ca s e : G ven Two Sides a n d the Included An le II i g .
r a ca n n r r a n s o f The t i ngle be divided i to two ight t i gle , one
a n s k a r s b a r n u a r r which cont i two nown p t , y pe pe dic l f om either extremit u n kn own side s O s y of the to the ide ppo ite .
Two o f EXAMPLE 1 . sides a triangle are and the included ° is 52 angle Find the remaining parts .
In the figure let AB A 0 : ° and the angle at A 52 Dro p a per en dicula r o t p p fr m B o the o ppo site side . Deno te the unknown side by a and the seg of A 3 ments O by a: and y as in Fig . 7 then 9 a) 1 , , y, and tan 0 can be co mputed in the fo llo wing o rder 2° ’ p sin 5 18 x . 79122 (1: co s x . 6 11 53 y cc
ta n 0 : p y 2 X 44 PLANE TRIG ONOMETRY
o Hence fr m the tables , 6°
a y cos 0 . 24101
The s e form ula s a re not well a da pted to loga rithmic compu
i n a s a n d as ma s a ra ta t o . The v lue of p y be computed ep tely
b a r s a a n d ta n 0 ma foun d . y log ithm , fter which y y be
u se r a s 0 sin A x c -co s A b x We the fo mul p , , y ,
z rk ca n n a a in ta n 0 p y. The wo be co veniently rr nged
mn s a s s : two colu , follow
log 2 log log sin A log cos A log p log cc log y 23 2 log tan 0 ° 0 76 log y a y co s 0 log co s 0 a log a
5 h Three des i 33 Ca s e 111 : G ven t e S . s a s s . i i In thi c e it not pos sible to divide the tria ngle into two right tria ngle s in s uch a wa y tha t one o f them conta in s two of the given pa rts ; how
X . r a r ul a r is ro to l s ide r eve , if pe pendic d pped the ow s; f om
r o f a n le O osite s s s the ve tex the g pp , the eg ment into which thi
r side is divided by the perpendicula r a e ea sily computed .
' of r EXAMPLE 1 . The sides a t iangle are a b and 6 Determine
the angles . Draw a figure and dro p a perpendicular C o o D o s of fr m B up n A O . en te the segment a: 38 the base by and y as in Fig . then
a 2 2 p 2 m y ; hence 71 3 ° y? is :c a: that , ( y) ( y) Since a: y b we have a: y a: whence , adding,
and , subtracting , y V 4 N OF B E R ANG E 45 , § 3 ] SOLUTIO O LIQU T I L S
n o w o a; A un . Since we kn w and y , the angles and C are easily fo d The student ma y co mplete the so lutio n by us ing the f o rmulas
: cos C y + 36 . 4
Loga rithm s ma y be u sed a s in the previou s ca se to compu te
n d is a . the s epa ra te products a qu otients . The following con ven ien t a r ra ngement :
; r “ 3 2 y? 6 2 K Facto ring bo th sides gives — — — (w y) y) ( C a ) — — c ( o a ) b 6
a c a log ( 0 a ) 6 a log ( 6 a ) (5 y b co log b 3: y log ( CD y) 33
cos A z a: c co s C = y + a log a: log y log 6 lo g a log cos A log cos 0 A 27 ° C 40° B 11 1°
34 Ca se I : iven Two ides nd the n le o site . V G S a A g O pp
O ne Of Them r i . The t ia ngle s s olved by dropp ing the p erp en a tca la r rom the vertex h n l h i f of t e a g e in clu ded by t e given s des .
° ’ EXAMPLE 1 . One angle of a triangle is 37 20 o n e side adjacent is o and the side Opp site is So lve the triangle . First co ns truct the given angle A and o n o n e side o f A lay off AB With B as center and radius describe an arc o f a circle meeting the Opposite side in two po ints 0 and Either of the tria ngles A B C " , AB C sa tisfies the given co nditions ; the case is o n this acco unt called the mbi a uous ca s . g e Em . 39 . 46 ANE R G N M E R V 34 PL T I O O T Y [ ,
The student sho uld note that the triangle B 0 0 ’ is is osceles and that the interio r angle of A B C at C is equal to the exterio r angle o f AB C’ ” ’ o a nd f at C ; hence the interi r angles 0 O are supplements o each o ther . To so lve AB C draw the perpendicular BD p fro m B then determine p fro m the right triangle A BD .
° p sin 37 20 ’
N ext determine 0 fro m the right triangle BB 0 ; p S‘D C . 75223 ; a
° " u te o is . 7 22 t . : 4 4 hence C is the a c angle wh se sine 5 3 . e C 8 7 The student can co mplete the so luti o n as fo llo ws
° B : 1 80 ( 21 + Als o for triangle A B C’ ’ 0 C 1 80 C ; ’ ° B : 180 ( A A ’ AD D O C .
F o r the loga rithmic solution we u se the fo r mula p _ c sin A sin 0 a a
Then the work ma y be a rra nged a s fo llow s log 6 log sin A co log a
log sin C : 0’ 48° C’ 13 1 ° B 9 3° B ’ 1 1 ° b a sin B/ sin A b’ a sin B’ / s in A log a log a lo g sin B log sin B ’ co lo g sin A co log sin A ’ log 0 log b b b’
I i in a r s s a is , given p oblem , the ide oppo ite the given ngle
ss n r r a k s le tha the pe pendicula let f ll upon the un nown ide ,
r is n o s a n d is r a r a e the e olution , if it g e te th n the oth r giv en
’ s r o a side there is one solution o nly . The con t ucti n indic ted in
x 1 l r s o s . E . will in a l ca s es show the nu mbe of lution
48 ANE R G N ME R v 34 PL T I O O T Y [ ,
f of to o f o n 1 1 . The angle o elevatio n the p a m u tain is o bserved at a ° po int in the vall ey to be 50 o n go ing directly away from the mo un tain ° l in clin ed 30 to o o o f o ne hal mile up a s o p e the h riz n, the angle elevatio n f ‘ o to of o i o f the top is f und be Find the height the m unta n .
Ans . ft.
1 of o of as 2 . The base an is sceles triangle is and each the b e ° angles is 68 Find the equal sides and the altitude .
Ans .
1 f o r is f 3 . The altitude o an is sceles t iangle and each o the base ° of Ans . angles is 32 Find the sides the triangle .
° ’ 1 o f 100 ft. o 4 . A cho rd a circle is l ng and subtends an angle o f 40 42
s of . n at the center . Find the radiu the circle A s .
1 5 o o dir o of i i 150 o . Fr m a p int ectly in fr nt a bu ld ng and feet away fr m ° it o f i of 36 Ho w o , the length the bu lding subtends an angle l ng
An is it s . f 16 . Find the perimeter and the area o a regular pentago n in
of di 2 . scribed in a circle ra us 1 Ans .
17 m of ul o o o r . Find the peri eter and the area the reg ar ctag n f med by f cutting o ff the co rners o a square 1 5 inches o n a side .
Ans .
of 1 8 . Find the perimeter and the area a regular pentago n whose
n . Ans diago nals are inches lo g .
of 1 9 . Find the perimeter and the area a regul ar do decago n inscribed
o f i s 24 . An in a circle rad u s . 1 728 .
° ° wo o of 72 144 20 . T ch rds subtend angles and respectively at the center
of a circle . Sho w that when they are parallel and o n the same side of
s o is o n e- l the center, the di tance between the ch rds ha f the radius .
1 D is o ul fo r o 1 1 2 . ev e a f rm a s lving an sos ce es triangle when the base and the bas e angles are given when the base and o n e of the equal sides are given ; when o n e of the equal sides and o ne of the ba s e angles are
given . ES AND BL UE PART II . O BTUS E AN G L O I Q TRIAN GLES
CHAPT ER VI
FUNDAMENTAL DEFINITIO NS AND FO RM ULAS
l h s r a n s i 35 btuse An es . T e n . O g olution of obliq ue t i gle
F o r s r volves obtus e a s well a s a cu te a ngles . thi ea son we n eed to be a ble to determine the v a lu es of the trigonom etric ra s fo r su a s is n o t n ssa r r n a r tio ch ngle it ece y, howeve , to e l ge ou r a sfo r s r s fo a s s wn ever ra tio t ble thi pu po e , r , will now be ho , y for a n o btu se a ngle ca n be exp ressed in terms of so me ra tio of a n a cu te a ngle. ? 4 Let a n obtu se a ngle a be pla ced on the co brdin a te a xe s wi th the vertex a t the origin a n d one s ide a long the cv-a xis to the right ; then the other side will fa ll in
in s u a ra . ra s s a the econd q d nt The tio ,
in r w co s a . a re d efin ed s , etc , te m of , y,
? a n d r x/ w + y2 p recis ely a s they were
ul for a cute a ngles in 1 1 . It sho d be n r a sin a: is n a oticed , howeve , th t ce eg tive
a nd r a re s r ra F IG 40 while y po itive , eve y tio which in volve s a: is nega tive fo r a n obtu s e a ngle ; thu s w/r co s a w ta n a a n d r r r a s see a a nd ctn a a re , y/ , thei ecip oc l , ,
ll n f r a ega tiv e o obtus e a ngles . We now proceed to obta in eq ua tion s simila r to the equ a tions
° sin 90 a co s a r in a u s ( ) , etc . (p oved which en bled to hn d the va lu es of the ra tios of a cu te a n gles g rea ter tha n
° ° 45 in terms of the ra tios of a ngle s le ss tha n 45
ili ° ° An obtuse an gle is a n angle which is greater than 90 a nd less than 180 E 49 50 LANE R G N ME R P T I O O T Y [VI , 36
R m b s t 36 . n fro O tu e o Acute An les m g . Let a be a or na a s a s s r a a nd pl ced on co di te xe de c ibed bove , let the
a n b is n supplement of be de oted y B(which a a cute a ngle) .
La ofi 8 first a ra s a x- y , in the q u d nt with one ide long the a xis . F rom a point P in the side of a (in s econd qua dra nt) a n d ' a poin t P in the side of B(in first qu a dra nt) a t the sa m e dis
a r r r n ra t nce f om the o igi , d w the ’ ' a rs P JII P M a s perpendicul , , in
i 41 Th f r F g . . e va lue of x o the point P will be nega tive S ince P is
s r its 0 0 in the econd q ua d a nt . Let
a s a b n s n ordin te be , ) ; the , i ce
F I G“ 41 ° ’ ’ the tria ngle s GP M G P M a re ' r r r A 1 1 s a s P a e a b . s a ym met ic, the coo din te of ( , ) in , we h ve
o s a — 9 _ — co s c B, r r
° o r S i 1 80 a , nce 3 ,
° (1) sin a = sin (1 80 a )
° 2 co s a co s 1 80 a ( ) ( ) .
In a simila r ma nner it ca n be shown tha t
° 3 n a ta n 180 a ( ) ta ( ) .
It follows tha t if a is a n o btu s e a ngle we find its sin e by lo okin or the s in e o its su lemen t which is a n a cute a n le a n d g f f pp , g , s m a r for r n s a wa s a r a r fo r i il ly the othe fu ction , l y h ving eg d the
r r s p ope ign .
EXE E — UN N O F O B E AN L RCIS S XII . F CTIO S TUS G ES
n 1 . Fro m the acco mpa ying figu re pro ve the f o llo wing relati ons :
° ( a ) sin ( 90 a ) ° ( b) cos ( 90 a ) ( 0 ) tan ( 90° a ) ° ( d) ctn ( 90 a ) 2
FIG . 42. VI 38 LAW or C NE , ] OSI S
o s o s o f o o o 2 . C n truct btu e angles wh se uncti ns have the f ll wing values
: b o ( a ) sin 9 1 / 3 . ( ) tan ( c) 0 0 8 0 ( 01) sin e ( 6 ) sin a ( f ) sin 9
o f ni o s o f o f Ex . 2 . 3 . Find the values the remai ng functi n the angles
E o o wi on of 4 . xpress the f ll ng as functi s an angle less than and lo o k up their values in a table .
( a ) sin ( b) co s ( 6 ) tan 168 sin ( 6 ) ctn ( f cos ° ° ° ( 9 ) cos 133 ( h) tan 144 ( i) sin 92
o 6 cos2 a; cos 2 5 . So lve the equati n 7 a: 0 . [ To so lve an equati o n of this type o n e sho ul d first regard it a s an algebraic (quadratic) equatio n in which the unkno wn is co s ao replacing 2 co s a: by the letter t we have the equatio n 6 t 7 t 2 0 . The so lu tio n s o f this equati o n are t ( or co s <5 ) or t : Then fin d fro m the tables the angles a: satisfying the equatio ns co s x and ° ° co s x they are a: 120 o r a: 13 1
o w o x o o o c 6 . Sh that the equati n tan c has an btuse angle s luti n if is any negative number .
7 o a n o . Sho w that the equati n sin a; c has bo th an acute and btuse
o o c o 1 angle s luti n if is any p sitive number less than .
° 8 o co o . Sh w that the equatio n s a; c has a s luti on between 0 and ° 180 c 1 1 o o a n if lies between and , and that this s luti n is acute o angle if c is p sitive and an o btuse angle if c is negative .
° ° o f o o fo o o 9 . Find all the s luti ns between 0 and 180 r the f ll wing equations
2 — 2 — ( a ) 3 sin az 2 sin az ( b) 4 sin :c 3 sin z 2 n in 2 — — z ( c) G si a + s x ( d) 6 s in m sin x l o.
m i Rela ion s r Geo etr c t . In n s s a 37 . the followi g ection ce t in fun da men ta l geom etric a n d trigonometric rela tion s connecting
s is the side s a n d a ngle s o f a n y tria ngle a re given . U pon the e
a s a s s a o s o o r a s b ed y tem tic meth d of luti n of oblique t i ngle ,
is o o a r which given in the f ll wing ch pte .
38 The La w Of Co sines In a n tria n le the s u a re o a n . . y g , q f y side is equ a l to the s the squ a res of the o ther two sides min u s twice their p ro du ct e co sin e of their in clu ded a ngle.
D s s a r a b a b c a n d a s enote the ide of t i ngle y , , , the ngle
s b A B a n d r ss s a r s d a r s oppo ite y , , O exp e the qu e of i e in te m VI 38 52 PLANE TRIG ONOMETRY l ,
’ o f b c a n d O a s o s . D r a r a ro B , , foll w op pe pendicul r , p , f m to the oppo site side a n d denote the segments of this side by a:
. 43 F IG . 44 . FIG .
d a n . B 13 14 12 a in F i . 43 y y ( , ) , we h ve g ,
‘ = c sin A rc = c co s A = b — x = b — c co s p , , y A 2 2 2 — 2 2 2 a = y + p = (b c co s A) + 0 sin A
2 — 2 2 - 2 b 2 bc co s A + o ( co s A sin A)
2 2 n s n sin A co s A : 1 whe ce , i ce
a 2 = bz 2 — 2 bc A (4) c co s .
i 44 s a to o is s n If a s in F g . the ide be f und oppo ite a obtu s e
° a A = b 33 b 2 3 6 x : 0 co s 1 80 A ngle , y + ; but y ( ) , ( ) — ° — 6 co s A ; hence y = b 0 co s A a n d p : 0 sin (1 80 A)
0 sin A a a s a s s d r a . , ex ctly in the c e con i e ed bove The la w of cosine s ca n be u sed to co mpute one s ide of a
r a r two s s a n d o n e a a re kn t i ngle when the othe ide ngle own , a n d a s hn d a s s s a re kn own l o to the ngle when the three ide .
° o f 66 25 ’ EXA MPLE 1 . One angle a triangle is and the including sides are 3 and 5 . Find the third side .
5 2 3 2 52 4 30 22 a; z 6 ,
Two o f 7 8 o o EXAMPLE 2 . sides a triangle are and and the angle pp site the fo rm er is Find the third side .
7 2 3 2 + 82
two o o whence 11: z 3 o r a: 5 and there are s luti ns .
3 . o f 3 5 7 . EXAMPLE The sides a triangle are , , and Find the greatest angle . — 30 co s a: ° whence co s a: and a: 120 I 39 LAW or NE V , § ] SI S
— E CO E AW EXERCISES XIII . TH SIN L
f a nd a nd 1 . Two sides o a triangle are their included angle is
i n Find the th rd side . A s .
o f n a nd 2 . T wo sides a tria gle are 5 8 and the included angle is
Find the third side . Ans .
wo o f 4 o o 3 . T sides a triangle are 3 and and the angle pp site the
Ans o r fo rm er is Find the third side . . v5
o f 3 5 6 . l . 4 . The sides a triangle are , , and Find the smal est angle ° An s . 29
5 of 10 14 1 . The sides a triangle are , , and 7 . Find the angles . ° ° ° A ns . 36 55 88
wo of 1 1 1 a nd o o 6 . T sides a triangle are and 7 , the angle pp site the o of o f rmer is Find the third side by the law c sines .
o fo r fin din t o i w o 7 . Devise a meth d g the angle between w l nes ith ut f o of an instrument or measuring angles . C uld the law cosines be used for this purpose
w o in es 3 The La f S . An two s ides o a tria n le a re to 9 . y f g ea ch o ther a s the sines of the a n gles opp o s ite.
D s s a n d a n s a a b a b c A B C enote the ide gle of tri ngle y , , , , , , a s P r a a bove . ove th t
a s foll o w s Dro p a perpendicu la r from O (the a ngle in clu ded by the s s n d i 4 r a a b s s In F . 5 ide ) to the oppo ite ide . g , whe e the
. 45 . 46 . FIG FIG . a s A a n d B a re a u ngle both c te, 12
a sin B a n d a s = b sin A p l o p , when ce a sin B b sin A 54 ANE R G N M E R PL T I O O T Y [VI , 40
a n d ro b b sin B dividing th ugh y ,
a sin A (5) b sin B
F i . 4 6 r a n s is tus In g , whe e one of the given gle ob e,
' ° — p = a sin B = a sin (180 B) : a sin B
a n d a s b sin A a a s a . l o p , ex ctly bove
r e l a is a n r o f s If the pe p ndicu r dr w f om one the other vertice , sa r A a ro a s to y f om , the bov e p cedure le d
b sin B (6) 6 sin 0
F ro m eq ua tion dividing ea ch side by sin A a n d m ulti
a s b b we see ha plying e ch ide y , t t
a b
sin A sin B
F o 6 see s l a r a a s ra s is u a r m ( ) w e , imi ly, th t e ch of the e tio eq l
ws a to c/sin 0 . It follo th t we ha ve
a b 0 (7 ) sin A sin B sin C
i t o ircums rib ed le 40 . D a me er f C c Circ . It can be sho wn that
of o 7 a b 6 for each the rati s in ( ( where , , , stand the numerical measures of the sides) is equal to the nu merical mea s ure o f the diameter of the cir s ib d u cum cr e circle and this f rnis hes ano ther pro of o f the law of sines . Circumscribe a circle abo ut the triangle ’ A B C BA : co n , draw the diameter (1, and
‘ ect A ’ ' n 0 . Then angle A G B is a right angle ’ and A = A since each is measured by o ne-h
a re B 0 o 1 2 the . Theref re by ,
a a
sin A ’ s in A
b c S im ila rl and y d ’ sin B sin C
If the angle A were o btuse we sho uld have
47 . FIG . ’ ° A 180O A A 1 80 , but since sin ( )
A o in o . o in the sin , the same result h lds this case als Theref re general , dia meter of the circle circum scribed a bo u t a tria n gle is equa l to a ny
s ide divided by the sin e of the opp osite a n gle.
56 LANE R G N ME R I 4 1 P T I O O T Y [V ,
a a s b sho rter iven sides a n d r 0 With r diu , the of the g , cente ,
e e inclu ded a n le s r a r r u the v rt x of the g , de c ibe ci cle th o gh A
5 c r A 4 3 c 1; pA U m ( O l d e/ D pp ) 0 ( ° L CHEl 1 ‘ $ C o 4 : 6 09 4 L ( z fi fl l
( SM M 5 10 8 5)
- ° A CD! ” 50 : 4 5 4 L B 96 0 150 6 3 111 6 9 A F I G . 48 . A1: n d which cu ts the . s ide GB in a p o int D between B a n d 0 a
D A n d a t r a s a t a s n E o 0 . a E a B l o econd poi t bey nd r w , e ect
hi A r On DF a s a a perpendicula r w ch meets E p oduced a t F . dia meter con stru ct a circle ; this circle will p a ss through A a nd
B fo r F AD is a r is s DAE 7 , ight the upplement of which is in s cribed a n d F BD is a right a ngle
s n s ru n is s s fo r a n r a n by con stru ction . Thi co t ctio po ible y t i gle n which a b.
A BF E A B si is a ngle 7 ( ) nce it the complement of ngle ° OEA = O a n d O = 90 s n su m of § ; , i ce the the a ngles of a tria ngle is An gle DF A B since ea ch is > mea s u red by o n e-ha lf the a re AD therefore BF D EF E — = —— — B A B . 2 ( )
r r a DBF a n d E BF b 12 In the ight t i ngle s y 9 ,
a — b BF ta n -— A — B g( ) , B ) ,
ta n A — B a — b %( ) a + b
130 ° 4
A l z f VI § 42 LAW OF ANG EN 57 , ] T TS
—‘ s r a is s r f a b s in tha t ‘f i Thi fo mul till t ue but , ince
a s a s r s z r b r s l ob c e e ch ide edu ce to e o ; , the e ult wou d vio u sly be — — b a ta n (B A) (9) % b + a
A B is o f C 8 ca n re Since % ( + ) the complement 4 , ( ) be du ced to the form ! (1 0) ten (A B)
- - 42 Ta n en s of the Ha a n les . a o f . g t lfi g The t ngent one ha lf a ny a ngle of a tria ngle ca n be expre ssed in term s o f the sides
s s a follow . Bi sect the a n gle s of the tria ngle AB C a n d dra w the in
s r b a s s a t P a n d . r c i ed circle t ngent to the ide , Q , R Let be the ra diu s of this circle a n d let 3 s ta nd fo r one-ha lf the perime
n r a i. e. ter of the give t i ngle ,
2 3 a + b + c.
AP = AR BR = B = P , Q, 0 Q C ,
A S “ P 2 AP = 2 s — 2 a
— AP = AR = s a .
’ — BR = B Q s b
P CQ O 8 c.
r r a AP b In the ight t i ngle O, y
12 FIG . 49 . ’ 7 ta n 5 A
s H A simila r re sult hold s fo r the other two a ngle . ence we ha ve the three for mula s
1 1 1 ta n a ta n -- C ( ) g 2 5 = a 8 58 ANE R G N ME R VI 43 PL T I O O T Y [ ,
43 Ra dius of the n s crib ed C rc e r . I i l . It ema in s to expre ss r in r s s s r a n te m of the ide of the t i gle . I n the tria ngle AB C p roduce the sides AB a n d
B s AO. i ect the a ngle A a nd the exterior a ngles a t
B a nd s s rs C. The e bi ecto meet in a poin t I which is the center of a circle which
s s a a n d touche the ide , the
s s a n d r ide b 0 p oduced .
50 . FIG . This circle is ca lled a n ’ es rib a D its ra s b r a n d c circle of the tri ngle . enote diu y
e n c P R . a pw y , Q, Then we h ve
A = AF B = BR P : Q , Q , O OR , ther efo re AB BR = A = s + C + CR ,
r o s a r o f i a whe e 8 den te h lf the perimete the g ven tri ngle . It foll o w s tha t A Q s a n d
B Q A Q AB 8
I n r a B I the ight tri ngle Q ,
° a ngle IB Q B) : 90 4 B a n d r r a n le ‘ BI B b 13 12 a n d the efo e g Q % ; then y ( ,
42 in a E I , tri ngle Q ,
n d r a A I a in t i ngle Q ,
s ta n é A s — a
' i s two a s o f r a n d so i fo r r a Equa t ng the e v lue l v ng , we h v e,
(12) W105 r
s a s r o f s r s a b 0 s s . a The ym met y thi e ult in , , how th t we h ll
r s s 6 a n d a o r a a n d b. get the sa me re s ult if w e p oduce ide , 43 U AS 595 VI , ] M L
o f E XAMPLE 1 . The sides a triangle are and o f Find the radius o f the inscribed circle and the angles the triangle .
fir o o f a s b s c We st c mpute the values s , s , , and
— z s s b 7 , s
Substituting in the fo rmula fo r r we o btain
tan A tan % B tan % 0 3 a s hence fro m the tables we find (A/2 30 ° B/2 23 ° 35 °
f EXA MPLE 2 . Two sides o a triangle are 12 and 8 and the included angle is Find the remaining angles . Deno ti ng the unkno wn angles by A a nd B we have
A B 1 80° 60° then by the law o f tangents we have
12 8 tan 60° hence ° tan ; (A and y ( A B) : 19
° ° A hi to A B 60 o A 79 6 ’ 4 dding t s result % ( ) we btain , and sub tracting we get B 40°
EXERCISES
1 of 1 1 . f . The three sides a triangle are 7 , 2, and 5 Find the radius o the ins cribed circle and the angles .
D e o f o o 2 . et rmine the angles the f ll wing triangles ( C)
D of o o 3 . etermine the angles and third side the f ll wing triangles
a a 4 b 8 C C a 10 b 12 C ( ) , , ( ) , , ° ° = z : = b a 4 b 8 O 40 . b z 17 C 44 . ( ) , , ,
o 4 . T determine the distance between two o bjects A and B separated
s A O 4 rd B 4 rd . to by a barrier, the di tances 0 . , C 8 are me a s ured a ° o A B third p int C. The angle C 68 is then measured . Find the dis n AB o o f B ta ce and the ther angles the triangle A C. CHAPT E R VII
SYSTEM ATIC S OLUTIO N O F O BLI Q UE TRIAN G LES
l i Da ta 44 Ana s s of . . y I n the solution of obliq ue tria ngl es the following ca ses a ri se
Ca se I . G iven two a n les a n d a side g .
Ca se 11 . G iven two s ides a n d the in clu e a n l d d g e.
a s e 111 . G iven the three s ides C .
Ca se IV. G iven two sides a nd a n a n le o o site on e o th g pp f em.
” The directio n S o lve a tria n gle tacitly assumes that a su ffi cient f f number o parts o an actual triangle are given . A pro posed pro blem may violate th is ass umptio n and there will be n o so lutio n . Thus there is
n o ri o 1 4 24 40 . A n to o t angle wh se sides are , , and attempt s lve such an o o v to o o for the imp ssible pr blem gi es rise a c ntradicti n such as , example , An ri n whi n sine o r cosine of some angle greater than 1 . y t a gle ch ca be constru cted ca n be s o lved.
I ven Two An les a nd a Side In s a s 45 Ca se . G . . i g thi c e
is a r a s is s r a ca n it im m te i l which ide given , ince the thi d ngle be found fro m the fa ct tha t the su m of the three a ngles is
is a n d nl s n r su m There one o y one olutio , p ovided the of the
° given a ngles is less tha n 1 80
w des c n be o u nd o n e a t a time b the la w o The o ther t o si a f , , y f s in es
E XAMP LE 1 . Given on e side of a triangle a a nd two of the ° ° angles B 79 C 33 15 ’ find the
remaining parts . A 180° ( 79 ° 40 ' 33 ° 67 ° 5 ’
By the law o f sines b sin 79 ° 40 ’ ° ’ F IG . 51 . sin 67 5 VII 45 N OF B E R AN G E , 5 ] SOLUTIO O LIQU T I L S
Many o f the co mputatio ns in the so luti o n o f triangles are of the fo ll o w
n rm r n wh . To n d o e te of a ro o tio en the other three ing type fi p p , g 5,
re kn own n o o f o o o o . a , matter in which the f ur p siti ns the unkn wn stands
o . o o The student sh uld master this pro blem The f ll wing rule applies . m in the mea n s a n d a lso the extremes to be co n n ected b stra i ht lin I a g e , , y g es
r r n n crossin g a t the s ign . M u ltip ly to gethe the p a i of k ow s thus con
n h u n n wn n ected a n d divide by the kn ow opp o site t e k o . i o f b Applying th s rule to the computation , the wo rk may be written down as fo llo ws ° ’ ° ’ sin 79 40 . 983 78 sin 67 5 2 76321 295134 92703 885402 92107 196756 59634 b
s rk ca n s r n b u se a r m Thi wo be ho te ed y the of log ith s . I n a ll ca s e s where the p ro duct of two o r m o re nu mbers is to be divided by other nu mbers w e ca n u se the following prin ciple
a x u in the lo a rithm o a n umber is e u i s . . S btra ct na (T ble , p ) g g f q
n h le t to a dding its co loga rit m .
The computatio n of b by logarithms may be written as f o llo ws log ° log sin 79 40 ’ 10 ° co lo g sin 67 5 ’ log b b
The side 0 is fo und similarly fro m the pro po rti on
° a sin 33 15 ’ sin 67 ° 5 ’
k a la w s s o r La w o f To chec , pply the of ine the n n 1 ta ge ts 4 ) to the co mpu ted s ides b a nd c.
E — EXERCIS S XV . CASE I
So lve the f o llo wing triangles . Small letters represent sides and co r o resp nding capital letters the angles o pposite .
° ° 1 B 50 o 122 a 72 . An s . . z
° ° s: 2 . F 82 G 43 f 48 . An s . M ° ° 3 . 9 N 7 44 p 477 . An s . ANE R G NOME R VII 46 PL T I O T Y [ ,
° ° P 37 Q 65 r Ans .
° ° ' A 70 K : 52 9 a Ans .
° 1 ° A 1 4 B 6 c Ans . 5 7 , 6
° ° A 48 B 54 c Ans .
° ° B 38 C 2 61 a An s .
° 7 ° U 46 1 : 124 w 1001 . Ans .
° ° A 10 . B 21 o 113 d ns .
° ° M A . 1 1 . B 62 52 a ns
° B G A n 12 . 58 g s . 1 4 ° A 3 . G 3 50 4 Q c ns .
14 . G K : k A ns .
wo o kil o 1 5 . T o 3 bservers, facing each ther meters apart and at the ° find o f o o f Z i to 57 20’ same altitude , the angles elevati n a eppel n be and ° 64 respectively . Find the height . A ns .
° ’ 16 . A diago nal o f a parallelo gram is and it makes angles 26 30 ° ' and 38 40 with the sides . Find the sides and the area o f the parallelo
'
a Ans . gr m .
° 1 7 to W. a A i o s o o N . 16 . l ghth u e was bserved fr m a ship be ; fter sail ° o N . 48 W. ing due east miles , the lighth use was Find the dis tance
o to in o o . A fro m the lighth us e the ship bo th p siti ns ns .
° 1 of i a n o f 22 ’ 8 . The side a h ll is inclined at angle 3 7 to the ho rizo n . ° A fla gsta ff at the to p of the hill subtends an angle of 13 1 7 ’ fro m a po int ° ’ o o o f of 18 2 o o 100 ft. at the f t the hill , and an angle fr m a p int directly of fl a sta ff An up the hill . Find the height the g . s .
1 9 . T o fin d o o A to o B the distance fr m a stati n an inaccessible p int , a A ft ° ° ’ 500 . A CB A B base line O , and the angles 68 C 58 28 are i AB meas ured . Find the d stance .
20 . T o find of R i the height an inaccessible o bject A , a base l ne CD f 250 t. is meas ured directly to ward the o bject : als o the angles of eleva ° ° o AD B 2 ’ A B ti n 48 0 and C 38 Find the height AB .
11 ven Two des a nd the ncluded l 46 . Ca s e . G S An i i I g e .
r is a a s n d s u n The e l w y one a only one ol tio . The obvious meth o d o f s o lution is to fin d the third side by the la w of cosine s a n d then the other two a ngles by the la w o f sines
1 . T wo o f 10 1 1 EXAMPLE sides a triangle are and , and the included ° ' angle is 3 5 24 Find the o ther parts .
ANE R G ON ME R 47 PL T I O T Y [VII ,
M E 1 . I n a M P EXA PL triangle T, side m ° side t a nd the included angle P 59 o Find the ther parts .
A n la w n pplyi g the of tange ts to the given sides , o m n ting that t > ,
— 1 t m ta n ( T — M) 2 . t + m
In this pro po rtio n three terms a re kno wn sin ce T M 1 ° P 80 . Th e wo rk m a y b e set do wn a s
fo ll ows .
t :
171. 5 m log ( t m) t m co lo g ( t m ) ° T M) P ) 60 log ta n T M ) — — ° — ( T M) 22 log ta n %( T M) = 83° 1 7 ’ M 37 ° 30 ’
The side 1) ca n n ow b e f o und by so lving the pro po rtion
° p sin 59 ° sin 37 30 ' log ° log sin 59 10 co lo g sin 3 30 ’ log p
'
o hi Co E a 2 46 . fr m w ch p mpare x mple ,
— A EXERCISES XVI . C SE II
1 o o o b s la w o . S lve the f ll wing triangles y u ing the of c sines
° ° a a 22 b 12 C An s . 106 3 1 ( ) , , ° ° ° b a 14 c 1 6 52 A ns . 56 7 1 ( ) , z , B ° ° c l 28 m 36 N : Ans . 23 3 1 ( ) , , ° ° ° b 24 c 2 . Ans . 46 56 76 a 21 , , 8 ° ° ° n 21 43 1 14 ( e ) a b c z A s . ° ° ° l 13 m 16 n 20 . Ans . 40 52 86 ( f ) , , ° ° u 41 5 1 W An s . 69 49 ( g ) , v z , ° ° ’ - b c A Ans . 47 99 58 7 ( h ) z ,
Two a a re a nd a nd is 2 . sides of triangle the included angle ° ° Ans . 36 53 Determine the remaining parts . VII 48 O N or B E R ANG E 65 , t ] S LUTIO O LIQU T I L S
° ro d ds of 0 o mt i 3 . Ho w long is a which subten an angle 6 at a p wh ch
f o o ne of ro d 8 ft. o o An s . 7 ft. is 5 t. fr m end the and fr m the ther
° rod o f 12 o f 4 . Ho w long is a which subtends an angle 0 at a p int 3 t.
f o o Ans 7 ft. from o n e end and 5 t . fr m the ther .
5 . o of o o o S lve each the f ll wing triangles, using l garithms ° ° ( a ) a b C 124 Ans . 17 ° ° An 47 ( b) b c A 16 s . ° ° 6 l 13 1 m 72 N : 39 An s . 3 1 ( ) , , ° ° d a 35 21 o 48 Ans . 36 ( , , ° f 6 u = 604 Ans . 22 9 . 5 ( ) , ,
° (f ) a b D 81 50’ ° Ans . 56
° . 6238 . 2347 C 108 ( 9 ) u , v , °
An s . 53 1 7
6 wo of a . T sides a triangle are and the included ngle ° is 43 Determine the remaining parts . ° ° An s . 40 96
s A 7 . T o determine the distance between two object and B separated A A 1t. B C 27 7 ft. CB by a hill , the distances O 300 , , and the angle ° find 65 are me a sured . Fro m these measurements the distance A A B . ns . A 8 . Two o B . bjects , , , are separated by an impassable swamp A statio n 0 is selected from which distances in a straight line can be meas ured to of s o to A 2 41 each the o bjects . These di tances are f und be C 3 ' °
t . in . B f h . 7 C 237 t. 5 i . A CB o to 53 , , and the angle is f und be i A B Find the d stance . An s . A B 9 . Two o . T o bjects , , , are separated by a building determine
o of o i o C o o the directi n the line j in ng them , a p int is taken fr m which b th A B i C f in A 1t 13 7 t. 9 . and are v sible and the distances O 200 . , B , A B 52° 25 ’ l and the angle C are measured . Determine the ang e which A B ° A . Al o 4 makes with O s the distance A B . Ans . 3
1 o s 0 . T determine the di tance between two o h ects A B as i CD ft j and , a b e l ne 350 . in the same as A plane and B is measured , and the angles B CD 40° A CB 30° ADB 51 ° AD C
° 32 o are bserved . Find the dis tance A B. A ns .
4 s i en h 8 a e 111. v t e T ee des C hr . r i . G Si The e s on e a nd
on e s n r sum o f a n n only olutio , p ovided the y two of the give s s is r a r a r s ide g e te th n the thi d ide .
F 66 LANE R G NOME R VII 49 P T I O T Y [ ,
law s s a s s r u r The of co ine , pplied to the ide oppo ite the eq i ed a a a s a s n a n d s s a re s a ngle , will l w y give olutio ; if the ide m ll ,
or n a n is r r is n s if o ly one gle equi ed, it ofte the be t method .
1 EXAMPLE . Find the angles of the triangle o 5 wh se sides are , 7 , 8 . By the law of cosines
- - 2 — - - A 1 8 2 7 8 0 0 s , — 2 - - CO E 5 8 S , 2 — - - 7 2 o 5 -7 cos 0 56 . 1 FIG . ,
COS CO 1 ‘ COS A B S C 1, 14286
A 32° B C 81 °
° CHE CK : A B C 179
f n 24 1 124 nd 4 EXAMPLE 2 . The sides o a tria gle are 3 , 3 , a 23 1 . Find
the largest angle . 2 2 2 - 3 124 23 14 2431 cos a , 2 2 23 14 243 1 3 124 COS “
o o Ca ll the numerator a: and the denominator y . Then the s luti n carried o ut by lo garithms as fo llows log 23 14 log 2431 log 3 124 2 2 2
5354500
‘ log 2 1 1264 100 log 23 14 log 243 1 2: 1504850 log y
log a: log y ° log cos a 10 a 82
4 o a ri hmic olution o a se t S f C 111 . u 9 . L g To comp te by the a id of loga rithms the three a n gle s of a tria ngle who se s ide s
a re k w first fin d ra u s n s r r b no n , we the di of the i c ibed ci cle y the formul a of 43 4 O U N OF B E R ANG E VII , § 9] S L TIO O LIQU T I L S a n d then compu te the a ngle s by the formula s of 42
r T ta n l A ta n l B _ ta n l C 2 2 ’ 2 s s — b 3 — 6
of o a re 2 14 24 EXAMPLE . Find the angles the triangle wh se sides 3 , 31 ,
a nd 3 124 . Th e work ma y b e arranged a s fo llows Computatio n of log r a 23 14 s co lo g s 10 b 243 1 s a log ( 8 a ) : 6 3 124 s b log ( 8 b) : 2 s 7 869 s c log ( 3 c ) 2 s ( Check) log r2 log r log r log ( 3 a ) log ( 3 b) log ta n 5A 10 log tan ; B 10 % A 23 ° B 25° log r log ( 8 c) log tan % 0 10 C 4 1° Then A 47 ° B 50° C 82°
° CHE CK : % (A B C) 90
— A E EXERCISES XVII . C S III
1 In o ow a re . . each of the f ll ing triangles, the three sides given Find the small est angle .
° Ans . 0 a 1 2 3 . ( ) , , ° Ans . 6 b 3 5 7 . 8 ( ) , , ° Ans . 6 c 3 4 5 . 3 ( ) , , ° d 13 14 15 . Ans . 53 ( ) , , ° e 35 4 1 47 . Ans . 46 ( ) , , 3 ( f ) Ans . . ° An s . 48 ( g ) 16 ° ( h ) An s . 49° ( i ) Ans . ( j ) Ans .
o of o o s o a ri s 2 . S lve each the f ll wing triangles , u ing l g thm ( a ) a b c ° ° ° An s . 32 51 95 ( b ) a b a ° 37 ° 78° Ans . . 63 ANE R G ONOME R PL T I T Y [VII , § 50
( c ) a b 0 ° ° ° Ans . 29 3 1 1 18 ( d ) l m n ° ° ° Ans . 35 76 68
( e ) u 0 w ° ° ° Ans . 34 45 100
( f ) p b d ° ° A ns . 57 35
m . 126 n . 3226 c . 253 ( g ) , , ° ° ° A n s . 21 1 1 12 4 , 1 6
h a . 0506 b . 1234 c . 0936 ( ) z , z ° ° A ns . 2 1 43
i u = 167 w : 23 1 . ( ) , ° ° ° A n s . 29 106 43 4 ° 2 ° ( j ) u 11 w A ns . 6 9
’ b a ft . o a t o 3 . Find the angle subtended y rod l ng the bserver s
is ft. o o n e end a nd ft o o . eye , which fr m . fr m the ther ° ’ An s . 73 44
o i o a n 4 . T determine w th ut instrument for measuring angles the angle
a t C CA 2 500 ft. a nd CB between two lines meeting , the distances AB i o A B . f a re a r s b e 633 ft. A C 700 t. me su ed then f und to Find ° An s . Abo ut 6 1
5 A i o n b . piece of land s b u ded y o n three intersecting streets , which the
o ha s a o 3 12 ft. 472 pr perty fr ntage of , ft f a n d 1 1 t. . . , 5 respectively Find the
angles a t which these streets cross . ° f A n s . 37 51 . 4
6 . In . 57 A B ft. B o Fig ,
F 1° ° 57 °
f A = ft . a nd AD t. , O 43 1 . 6 ,
B An . ft. Find D . s
s e I The mb u ou s a s e H r we n A C . a 0 Ca V. 5 . ig e e h ve give
wo s s a n d th e a s o n e of m t. e. a n a a t ide ngle oppo ite the ; ngle , h s a a a n d t e s s . ide dj cent, ide oppo ite
The n r of s o s o is s r u mbe oluti n (two , one , r none) be t dete mined by the geometrica l con stru ction of the tria ngle from the da ta .
s a n a A G ua to the a n Con truct ngle Q , eq l given gle which we sha ll a t first s uppo se to b e a cu te ; on on e of its sides la y VII 50 N OF OB E TR ANG E 6 , ] SOLUTIO LIQU I L S 9
off G A equ a l to the given a dj a cent side a n d drop a perpen
dicula r AP r s G . n A a s n r , to the othe ide Q The with ce te
nd a ra us ua n s s d r n r a with di eq l to the give oppo ite ide a w a a e.
Ii a s in F i . 58 a s a re s n o t r a G r is , g ( ) , thi doe e ch Q, the e
n o solution is a n G a s in F i . 58 b r is ; if it t nge t to Q, g ( ) , the e
one so lu tion u s G a s F i . 58 r a re ; if it c t Q twice , in g the e w i t o solu tion s u s G a s F . 58 r is on e ; if it c t Q on ce, in g the e solu tion a n d fin a ll a n is u s r is n o ; y if the given gle obt e , the e solu tio n when the ra diu s of the a rc is less tha n GA a nd one so u ti n is r a r l o when it g e te .
FIG . 58.
The resu lts ma y be collected fo r referen ce a s follow s : Let
G a a a s o . the given ngle , the given dj cent ide , ( pp the given opposite side ; then
h n is a cu te c m u te th n i I . e G o s in e W , p G ; f
o there is n o so lu tion i o : o ne so lu tion ( pp ) p ; f ( pp ) p , ; i o a d two so lu tions a n d i O a d o n e f p ( pp ) ( j ) , ; f ( pp ) ( j)
solu tio n .
. Wh en G is ri ht or o btu se i o a d there is n o II g , f ( pp ) 2 ( j ) ,
lu i n t i o n e lu io so t o bu f so t n .
ra a m r a s a n The p ctic l ethod , howeve , in the c e of y given
r i u th i n o s p oblem s to constr ct e tr a gle a pp r xima tely to ca le.
Ha r n u r s s kn n ving dete mined the mbe of olution , the un ow
n b l w n pa rts ca be computed y the a of si es . ANE R G N ME R 50 PL T I O O T Y [VII ,
EXAMPLE 1 . Two sides are and and the angle o pposite the latter is ° 64 o S lve the triangle . Construct the angle G 64° 20 ' and lay o ff GA and draw AP . A glance
4 o at the tables ( p . 3 ) sh ws that sin
.9 . 9 Q G , whence p x 1 1 . o n o o 59 , o . Em , Theref re , s luti n
E AB C a EXAMPL 2 . In the triangle , 6 and A ° ni 33 Determine the remai ng parts . ° o A 33 17 o ff AB B C nstruct angle lay and draw P . ° o o 1 With ut any tables whatever, we kn w that sin 33 7 . 7 and there
o . 7 x 360 260 o two o o s . f re p , and theref re there are s luti n
sin 0 sin 33° 17 ’
° log sin 33 17 ’ 10 log
log sin 0 10 A C 39 ° 53 6 F I G 60
° There are two angles less than 180 having a given s in e ; therefo re C; ° ’ ° 39 53 and 0 2 140 o o n to o wo ri From this p int we have s lve t distinct t angles , Viz . AB C; AB C A C b A O b AB C B and C; . all 1 l , and 2 z angle l , 1 and angle B B B 106° 0 ’ B 6° A Cg , 2. Then 1 5 and 2
° ’ ° ’ b; sin 1 06 50 bg sin 6 36 sin 33° 1 7 ’ sin 33° 1 7 ’
log log ° ’ ° ’ log cos 16 50 10 log sin 6 36 10 ° ’ ° ’ co log sin 33 1 7 co lo g sin 33 17
log bl log 02
b; 02
f n o o EXA MPLE 3 . Two sides o a tri a gle are 5 and 7 and the angle pp
site the latter is So lve the triangle . Construct the angle G lay o ff o us G A 5 . It is at o nce o bvi that a A of s 7 ll circle center at , radiu , wi cut
o o o n B . G Q nce and nly ce, at o o Let the student co mplete the s luti n,
findin of B g by the law sines , angle and ° A . 3 . the side G B. ns 38
CHAPT ER VIII
AREAS —APPLICATIO NS PRO BLEM S
51 Area s of Tria n les is s in m r g . It hown pla ne geo et y * a a r a A a r a n is a - a r th t the e , of t i gle equ l to one h lf the p oduct
a n s a n d a of y ide the ltitude from the O pposite vertex . (1) The a rea of a tria ngle is equa l to on e-ha lf the p roduct of
the ba se a n d a ltitude.
2 Area from Two Sides a nd he l d d n l 5 t Inc u e A . . g e If we
a s s a n d u a n a b a n d C a nd r h ve two ide the incl ded gle, , , , d op
b C
3 . . 64 . FIG . 6 FIG
r n ul a r o n e s s a s b a pe pe dic upon of the given ide , p upon , then p a sin 0 a n d by ( 1) 4 b (a sin 0 ) whence (2) T he a rea of a tria ngle is eq ua l to o n e-ha lf the p ro du ct
n luded a n le a ny two s ides into the s in e of their i c g .
m hree des A Area fro S . 53 . T i If
r s s a re ra s the th ee ide given , d w line from the v ertices to the center of the in s cribed circle dividing the tri a ngle in to three tria ngles ha ving a
mm a u r. B 43 co on ltit de , y ,
65 . FIG .
The area is deno ted by the bo ldfac e type Ain disti nctio n fro m the a ngle A . 72 4 A EA VIII , 5 ] R S 73
su m a s s r r a n s is a b 2 The of the b e of the th ee t i gle c 3 .
r r r mi a r a is b The efo e thei co b ned e , y (3) Hen ce we ha ve the ru le (3) Add the three sides a nd ta ke ha lf the s um ; fro m the ha lf s um su btra ct the three s ides severa lly ; ta ke the p rodu ct of the ha lf
n d rs a nd extr t h sum a n d the three rema i e a c t e squ a re ro o t.
llu stra tive xa m les a r a is m s n n . I e 54 E p . The e o t co v iently foun d in other ca s e s by solving the tria ngle su fficien tly to
r a a r r b r r s n a secu e the d t equi ed y one of the th ee ule give bove,
a r a ll of which a re a da pted to log ithmic computa tion .
o f EXAMPLE 1 . One side a i 50 o triangle s , the angle Opp ° site is 10 and ano ther angle ° is 46 Find the area .
A . 66 The third angle , , Fig , 8 is then If we knew
F I G“ 66 ° r o o two a o b, we sh uld kn w
n g sides and the included a le . 0 log 50 16 9897 f By the law o sines ° log cos 33 12’ 10 ° ’ 123 12 ° ' a sin co log sin 10 12 50 10° 12 ’ sin log a 2 lo By rule ( ) the area, g 25 ° log sin 46 3 6 ' 10 ° A 4 50 a sin 46 log A A
wo f EXA MPLE 2 . T sides o a triangle are 35 and 50 and the angle o ppo ° is site the latter 28 Find the area . h O n o . 67 i is o c nstructing the triangle , Fig , it is ev dent t at there nly
n e o o B is . B o f o s luti n and acute y the law sines , sin B 3 5 ° ° sin 28 30 ’ 50 log sin 28 30 ’ 10 log 35 co log 50 10 log sin B 10 B 19 ° whence A 131 ° FIG . 67 . 74 PLANE T RIG ONOMETRY [VIIL 54
We n o w kno w two sides a nd the included angle a n d ° ’ 3 2 Az 3 5 sin 13 1 59 . log 5 log 35 ° log cos 41 10 A log A
E 3 . The a a re 13 3 7 a n d 40 . . XAMPLE sides of triangle , , Find the area
Using we have, — a ) ( s b) ( s co mputati on m a y b e made a s fo llows log — 8 a z 32 log — S b : 8 log — s c 5 log Check = 90 log A A 2400
A A EXERCISES XIX . RE S Find the area of the fo llo wing triangles 2 1 . a 829 , b 59 , C Ans . 12 a 7 13 , b 987 , C 55 . Ans .
n B C b 392 . A s .
° P 47 An 1) 23 1 , q 195, s .
: An 8 5 W s . u , v , K ° M 6 ° A k 52 3 5, 3 ns .
m n 42 An l . 582, . 601 , . 7 s .
° b c D 41 Ans . A it 41 11 401 w 408 . ns . 1 . , , 8 60
” An M?"p q 7 s . An Mf A b 30, a 70 . s . ° ° A 12 . a B 34 C 66 ns .
1 3 . p R Ans .
a n field o n e 1 o 14 . Find the area of tria gular having of its sides 5 r ds ° ° in a n d 70 a n d 69 length , the two adjacent angles , respectively,
Ans .
1 5 The a o is o ne . Two of its . area of triangular plat of gr und acre
re 12 d . a nd 1 50 d . . . sides a 7 y y , respectively Find the angle between them ° An s . 30
The o o ne of a n o 1 6 . length of the bisect r of of the acute angles is sceles
4 . right triangle is . Find the area Ans . 4 . AREA VIII , 55] S 75
m si ion a nd Res olu n o 55. Co o t t o f orces a nd Velocit es p i F i . We sa w in 27 tha t force s a n d velocities ma y be repres ented
r b s ra s g a phica lly y t ight line egments . The length of su ch a
s r r s s ma o r o r eg m ent ep e ent the gnitude of the f ce velocity, a nd its r n r n r o r di ectio the di ectio of the fo ce v elocity .
hn d f s a u s s u s su To the ef ect of two imult neo velocitie , let p po se tha t a body moves a lon g a s tra ight tra ck with a velocity of 4 u nits per s econd a n d tha t ea ch point of the tra ck move s with a vel o city of 3 u nit s per second a long a line m a king a n ° r k a is a ngle of 60 with the t a c . Wh t the position of the body a t the end of 1 s eco nd To a n s wer thi s questio n dra w a seg m ent 4 units long to r ep res ent the m a gn itude a n d direction
o a o a a n d r of the vel city of the body l ng the tr ck, f o m the end s
s s a s s A O BI) a 3 s in of thi eg ment dr w eg ment , , e ch unit length
° a n d m a king a n a ngle of 60 with AB to repres ent the m a gnitude a n d direction
o s k of the vel city of the end of the tra c . The tra ck will then ta ke the position
1 s Bu s CD a t the end of econd . t ince 68 . FIG . the body move s a long the tra ck a t the ra 4 un s er s r a n D a t te of it p econd , it will e ch the poi t the end
i r n ha d 1 s n . a s a sa a s of eco d Th t , it will e ch the m e poi t if it moved a long the dia gona l AD with a speed rep re sented by the
r r s n b AD is len gth of the dia gona l . The velocity ep e e ted y ca lled the resulta nt of the velocities rep re sented by AB a n d
com on en s n AD AB a n d A a re a t . AO. O c lled p The le gth of
n r ca be compu ted by solving the t ia ngle AB D . The res ulta nt of a n y two velocitie s ma y be foun d by d ra w in r a n A s s AB AO r r s g f o m com mon poi t , eg ment , to ep e ent the given velocitie s in m a gnitu de a n d direction a n d then complet in r r a a AD r r s n s g the pa a llelog a m AB CD . The di gon l ep e e t le o a m la w r s a is n a ra r . the e sulta nt . Thi f ct ofte ca lled the p l l g The res ulta nt of two force s is fou nd by a simila r con strue
m i kn a ra llelo ra m of orces . s r f tion Thi dia g a s own a s the p g . 76 PLANE R G NOME R T I O T Y [VIII , 56
llustra e l 56 tiv xa m es 1 . . I E p . EXAMPLE The angle between the
i o of two o f 1h 26 1b . d recti ns f rces o 19 . and is Find the magnitude o f and directi n o their resultant . The fo rces may be represented by segments 1 9 units long and 26 un its ° o of 54 o I l ng, respectively, and making the angle with each ther . f the parallelo gram is completed which h a s these seg f t o of di ments o r w its intersecting sides , the ag o nal extending fro m their intersectio n to the o pposite co rner will represent the resul tant bo th i in magnitude and in d rection . This diagonal is a FIG . 69 . side o f a triangle having two sides equal to 19 ° 26 a n of 126 and , respectively, with included angle ( the supplement of n o f Hence we can fi d the magnitude and directi n o the resultant .
o o of 5 1 lh . 73 lb . o f 80 lh . EXAMPLE 2 . N f rces and have a resultant
Find the angle between them .
I a o r of o o two n this case , in the p rallel g am f rces , the diag nal and inter secting sides are kno wn ; the angle Oppo site the diago nal is determined C 111 o f o n e by ase . The required angle is the supplement this .
EXAMPLE 3 . A weight o f 10 lh 0 . is supported by two o A 7 f 14 3 t . o c rds , l ng, B IV f 5 t. o a and , l ng, t ta ched to a horizontal A B f beam at and , 7 t. apart . Find the tens io ns
0 . FIG . 7 s in A W a n d t in B W a Since the weight acts vertically, we need the angles and Bwhich A B M A W and B W make with the vertical WP . So lving the triangle 5 ° ° we fin d A : 38 B 21 whence a : 51 and 48 68° 1 A O T o construct Fig . 7 draw making the angle a 5 1 with the vertical and similarly draw AD making 6 Take A V 100 o n so me co nvenient scale and draw VE parallel to AD and VF
to AE re re parallel A C. Then p s A F t o sents and , ; because a f rce o f lb A 100 . acting upward at must 1 . FIG . 7 be the resultan t o f s and t sin ce the
A is n n A E A F o point at rest . I the tria gles V and V we have en ugh data to find 3 and t 56 FORCES AND E C E VIII , ] V LO ITI S
EXERCISES XX . VECTO R S
A ° 1 . o E 1 o . ns . 22 w AB S lve xample , ab ve ith .
° o o o of E 2 . An . 2 . C mplete the s luti n xample s 101
3 . Co a s t of E 3 . mpute , B, , and xample
4 . C to E 3 findin a of heck the answers xample by g , ( ) the sum the o o of s t b o o o o vertical c mp nents and ( ) their h riz ntal c mp nents .
5 . o of 13 lh . 22 lb . 28 lb . in i Three f rces , , and , respectively, are equ librium . Determine the angles which they make with o ne ano ther .
° ° ° ' n 1 30 [ HJ N L Study Example A s . 76 153
f wo o o f 1b . 4 lb 6 . Find the resultant o t f rces 30 and 0 . acting at a n ° o f An angle 60 with each o ther . s .
o o fl o r f o 7 . A ball r lls al ng the diago nal o f the o o a car fr m the back to o f o o the fr nt with a speed o f 30 t. per second . The car is m ving f rward f with a speed o f 40 ft. per seco nd . Find the actual speed o the ball if
ft . ft n the car is 7 wide and 30 . lo g .
A ns .
100 Lbs . 8 . Two fo rces are acting o n a blo ck resting o n the gro und as
o fi u r sh wn in the g e. What h o ri z o nta l fo rce co uld replace them A n s .
9 . A po int is kept at rest by 2 . FIG . 7 ‘ o s of 6 8 1 1 lh . F in d the f rce , , angle ° ’ ° n . 4 . between each pair . A s 7 7 1 7 50 6, 134 47 . 6 1 0 . A bo at is ro wed acro ss a river at the rate of mi . per ho ur the
h o o f o river ws at the rate o f mi . per h ur . Find the speed o the b at ° ’ f n i o . o o o A s . 36 6 and the directi n its m tio n . w th sh re
1 1 A o . ship is sailing 10 mi . per ho ur and a sail r climbs the mast
200 ft. in direc high 30 sec . Find his speed relative to the earth, and the ° ’ f . . o o o o . A . ft. min . 24 26 6 ti n his m ti n ns per , with the vertical
12 . A is o o train g ing 15 mi . per ho ur northward ; a man cr sses the ft h t o car eastward 12 . per second . Find is speed relative o the gr und , ° o . An . . 28 E . and his directi n s N ,
13 . o o fl o o r ft o n so A ball r lling al ng the 10 . per sec d is struck that its
ft o f o o speed is increased 2 . per seco nd, and the directi n o m ti n is changed What speed and directi o n of mo tio n is due to the stro ke alo n e °
A n t. o 80 s . f per sec nd ,
14 . A fl o ws 4 . o a nd o o o o 6 . o . river mi per h ur, a m t r b at g es mi per h ur In what directio n must the bo at be p o inted to g o straight across the river ° n d . o . a what will be its speed Ans . 63 mi per h ur 8 ANE R G N ME R 56 7 PL T I O O T Y [VIII ,
1 5 n o hi d u e o a n o r i . A arsman ro ws s bo at n rth 5 miles h u . There s a
D ul breeze of 12 miles a n ho ur fro m the so uthea s t . etermine the res ting Speed a n d directi o n of the bo at if the resistance of the water damps the — ° ’ o n e i . A ns . . o r N . 27 58 W. effect of the wind th rd mi per h u ,
16 . A 4 lh is n o weight of 00 . draw al ng o b a o of 0 lh the gr und y f rce 50 . attached
a s o in . W sh wn Fig 73 . hat pressure do es it e ert o n the gro und ? If the . x resisting fo rce R ( due to fricti o n) is 1 F I G 73 r o n o re of the pressu e the gr und , what s ulta nt f o rce is effective in mo ving the weight fo rward
Ans . 150 1b . , 408 lh .
EXERCISES XXL — M ISCELLANEOUS P RO BLEM S
S o lve the f o llo wing triangles
° ( a ) a B 5 ( b ) a 5 ( 6 ) l m ° ( d ) a B 108 ( e ) u 0
a . 75632 b . 62751 ( f ) , , ° ( 9 ) C J : 1 1 ( h ) a b ( i ) a 6 ( j ) d 1) 25° ( k ) a b ( 1 ) u 5
m 01023 1 c . 0047233 ( ) a . , ° / ( n ) a P 40 17 ( o ) b a ° b 43 124 a . 53467 A 99 ( p ) . , ,
Answers to the p recedin g exercis es . ( b) ( d )
N o solutio n . ( f ) ( h ) ( j ) ° f 16 ° 2 20 . 4 I 5 , ( ) 1 15° ( 19 )
80 ANE R G NOME R PL T I O T Y [VIII , 56
1 5 A u o o of . o o . tall b ilding stands at the f t a hill Fr m a p int o n the side o f the hill the angle of depressio n of the base of the building is oh ° ° served to be 14 and the angle of elevation of the to p is 21 A
u n f in . level lin e fro m the instrument meets the b ildi g 19 t. 7 abo ve the
f i . b as e . Find the height o the bu lding Ans .
1 6 . A o o o o o o ball n is bserved , at the m ment it passes ver a level r ad , wo o o f f from t p ints in the ro ad an eighth a mile apart . The angles o ele ° ’ ° vation fro m the two po ints are 33 17 and 42 Find the dis tances of
l o o b . A n the bal o on fro m the tw servers s .
17 . I i n survey ng, it is sometimes desired to extend such a line as A B in the figu re beyo nd B an o bstacle . If at a right turn of ° BE 12 ft. E of 6 , and at a left turn 1 10 are o ff o E C laid , c mpute , and the angle (right FIG . 75 . G . A turn) at ns .
1 o find is P . 76 8 . T the d tance Q in Fig , a i At A base line A B is measured 5 18 t. ° ' ° ’ th e angles P A Q 43 18 and QA B 48 32 are measured and checked by measuring ° B ' ABP 38° P AB 9 1 and at , P B Q 41 ° A E Q 80 ° Find P Q
Ans . by two metho ds .
1 9 n A O . 7 7 o . Fi d the distance , Fig , thr ugh FIG . 76 . AB o a thicket, having measured r ds , ° o s AB C 55 B 0 r d , angle A ns .
A B f n wo o s 300 t. o 20 . Fro m t p int and , apart
B o f i o S o . the deck a sh p , a sec nd ship , , is bserved The angles A B S 85° and BA S 83 ° 47 ’ are
is measured . What the distance between the ships
. 77 . An s . 2496 2502 a v. 2499 . FIG , ,
Ho w to of 1300 ft. o 2 1 . far the side a target away sh uld a gunner aim f f o 15 . o if o 2000 t. fro m a ship g ing mi per h ur, the speed the bullet is A n per seco nd and he fires when he is directly o ppo site s .
o o 50 . o r fired 1000 ft. 22 . Fr m a railway train g ing mi per h u a bullet is f ° per seco nd at an angle o 7 5 with the track ahead . Find its speed ° An ft. o . s . 7 1 and directio n . per sec nd
o 45 mi . o o 23 . A man in a railway car g ing per h ur bserves the rain ° dro ps falling at an angle of 30 with the vertical . Assuming that the
l find . Ans . ra indro ps are actually fal ing vertically, their speed III ISCE ANEOUS PROBLEMS 81 V , 56] M LL
4 of two o is 10 lb . o n e o f o 8 lb . 2 . The resultant f rces ; the f rces is and ° o f ul of he makes an angle 36 with the res tant . Find the magnitude t
An s . o ther fo rce .
5 o l s o ro e w i of 2 . A h rse pu l a canal b at by a p h ch makes an angle ° ' f o o o 25 35 with the to w path . What size o engine w uld pr pel the b at at the same speed ( Assume th at the horse is do ing o n e horse
Ans .
° l l o n 32 o 26 . A man c imbs a hill inc ined ( the average) with the h ri
nta l o o o of 2 hr. zo . His p cket bar meter sh ws that at the end 4 he has
a s f m e . incre ed his elevatio n 2750 t. Find his average speed up the p
Ans .
h f ld ods o s 27 . T e sides o a triangular fie are r , r d , and
s D e mi of field th e . rod . et r ne the area the and angles between the sides ° ° ° Ans . A . , 49 57 73
f of o n 28 . Find the area o a triangul ar piece gr u d having two angles, ° ' ° 73 10 90 o o s . respectively, and and the side pp site the latter rod
A A . ns .
o f o o w s 29 . Find the areas triangles which have the f ll ing given part ° a a b 100 C 1 18 ( ) , b b = 100 O = 40° 5 ’ ( ) , ° ° ( c ) u 11 13 U : 53
d a 408 b 41 401 . ( ) , , c = e a . 9 ( ) ,
Ans . a 5 01 16 54 ( ) ( ) ( 0 ) ( ) 8 0 ( e) .
o e 30 . Three circles wh se radii are 2 , 3 , 10, respectiv ly, are tangent of o o n externally . Find the area the triangle f rmed by j i ing their centers .
Ans . 30.
1 P o th e of o o 3 . r ve that area the triangle f rmed by j ining the centers o f any three circles which are tangent externally is a mean pro po rtio nal o o f between the sum and the pr duct their radii . See 53 .
- f f 32 . Pro ve that o n e hal the pro duct o the three sides of any triangle is equal to the pro duct o f its area into the diameter o f its circumscribed 4 circle . See 0 and 52.
. P o f 33 . r ve that the area o any triangle is equal to the pro duct of the radii o f its inscribed a nd circumscribed circles into the sum of the sines o f its angles . See 40 and 53 . PART III . THE G ENERAL AN G LE
C HAPT E R IX
DIRECTED AN G LES — RAD IAN MEASURE
e ed Lin s a nd e ments As a Dir ct e S . 57 . g expl ined in ele
n a r a ra is to s o n e r n me t y lgeb , it often convenient elect di ectio on a stra ight line a s the p os itive directio n ; the other is then
wo r a n e a tive directio n . s t s a ct a c lled the g Thu , if fo ce long the sa s s is n ni a me line, but in oppo ite direction , it co ve ent to c ll
n on e positive a d the other nega tive . Two s eg ments a re sa id to ha ve the sa me sense if they lie on
sa o r a ra in s a n d o a re s o r the me line on p llel l e , if b th po itive
n s r s s both a re nega tive . Two segme t a e a id to be of opp o ite sense e o n sa o n a ra s a n d if th y lie the me line or p llel line , if
o ne is s a n d r is o o n a . A B c o p itive the the eg tive
s F i 7 8 A B : EF Thu , in g . , , while F
— : — A O G E a n d 0 B F G .
FIG . 78 . The nu merica l mea su re of a directed seg ment is the nu mber of units in its length with the sign o r
r n s s is s o r n a cco di g a the egment po itive ega tive .
. i n re ed An les In r 58 Ro ta t o . D ct . s n r a n i g de c ibi g ot tio , it
’ is convenien t to rega rd a n gles a s positive o r n ega tive in a
nn r na u s a a 57 fo r n - m a e a logo to th t expl ined in li e segments . An a ngle is thought of a s genera ted by the rota tion of on e of its sides a bout the v ertex a s center ; its first p o sition is
a in itia l side fin a l s is a termin l c lled the , the po ition c lled the a
si An a n ra b a r a s n de. ngle ge e ted y ot tion oppo ite to the motio
a n s a k coun terclockwise is sa ositive of the h d of cloc ( ) , id to be p ; 82 IX 59 D REC ED NE AND ANG ES 83 , 1 I T LI S L a n an ra b a clo ckwise ro a o is sa gle gene ted y t ti n , id to be * n ega tive.
An s ma a n a u s o r a gle y be of y m gnit de , po itiv e neg tive .
s in F i . 7 9 a 8 a re s a s is a Thu , g , , B, po itive ngle ; 7 neg tive ; B is grea ter tha n a stra ight a ngle ; a nd 8 is grea ter tha n o r a complete
' I n r a i a r s ma revolution . ot t ng p t of
n r su a s a a r chi e y, ch ngle h ve ve y vivid
a . u s a r a s me ning Th , wheel which ot te 1 \ ° 37 0 per secon d ha s a very difi eren t speed from tha t of a wheel which rota te s
° 10 er s n . p eco d FIG . 79 .
59 . Pla c n An les on ecta n ula r Ax s a a n i g g R g e . To pl ce y given a ngle on a pa ir of recta ngul a r a xe s in the pla ne of the
a u t r a t r n a n d ngle , p the ve tex the o igi the — in itia l s ide o n the x a xi s extendin g to the right ; the termina l side will then fa ll in one
u r a ra s o r a n is a of the fo qu d nt ( , if the gle
u a r a n o n o n e a s . m ltiple of ight gle, of the xe ) If the termina l side fa ll s in the first qu a d
ra a is sa a n a n le in the nt, the ngle id to be g FIG . 80 . i is s rst u a dra n t . I n F . 80 a a fi q , etc g , po itive a n first a ra is a n a a r gle in the q u d nt, B eg tiv e ngle in the fou th
ua d ra 8 is a s a n in r a ra . q nt, po itiv e gle the fou th qu d nt
— D E AN AN EXERCISES XXII . DIRECTE LIN S D GLES
n o f o 1 . What angle will the mi ute hand a cl ck generate in 2 hr. 24 min . 10 sec . ?
2 . A flywh eel is running steadily at the rate o f 450 revo lutio ns per
o o n e of o 2 se I minute . What angle d es its sp kes generate in a ? n sec . ?
Either o f these directio ns m a y o f co urse be cho sen as the po sitive dirco
n o f n n d o n . o tio ro tatio n , the o ther is the the egative irecti The ch ice here m u o m o n fo r n b u t in m n n o f m n he ade is the c st ary e a gles ; a y ki ds achi ery , t
o e n o f o o n on o in o f c ck . th r se se r tati is c sidered p sitive . as the case a lo LANE R G ON ME R IX P T I O T Y ( . 59
s m o r resulta nt of two o 3 . u s Find the , , f rces that act in the ame line o o 5 wh se intensities ( measured in p unds) are and 10, respectively. r Draw a figu e to represent the s o lution .
4 . If o o f s 7 15 2 i three f rces inten ities , , respect vely,
o n o find o . D act a b dy in the same line, the resultant f rce raw a
figure .
5 of . . If a man walks with a speed 4 mi per ho ur to ward the rear o f
o r find a train go ing 3 5 mi . per h u , his actual speed . Draw a figu re .
’ n o 6 . A man s gai s and l sses ( indicated by in business in succes
a 250 1 18 35 7 12 15 . i o sive months are $ , 3 , $ , $ , $ F nd the t tal gain and the average gain per month . Draw a figure.
7 . B of o o o u o lo i y means a ruler and a pr tract r, c nstr ct the f l w ng angles and their sums check by adding their numerical mea s ures .
° ° ° ° ( a ) 7 5 and ( b) 66 a n d ( 6 ) 45 and and 7 0 ° ° ° ( d) 60 and ( 6 ) 485 and ( f 750 and
o two of i a a . 8 . With s me the angles just given ver fy B B ° ° a Co 27 85 9 . ( ) nstruct ° ° ° ( b) Co nstruct 150 9 6 24
° 1 I o 120 o h o w o o ns o 0 . f a wheel is r tating per sec nd , many rev luti d es it make per minute ho w many per ho ur Ho w many degrees d o es it turn thro ugh per minute
1 o f o o o n 1 . Express an angular speed rev luti ns per sec d in degrees per seco nd in revo lutions per minute in degrees per minute .
1 fl wheel o of 40 e o on . 2 . A y r tates at the rate r v luti s per minute Through what angle do es o n e of its spo kes turn in a seco nd ,
of o o ns o to 1 3 . Reduce an angular speed rev luti per sec nd degrees o per seco nd to degrees per minute to rev lutions per minute .
1 of o o o f r o n x s 4 . Find the angular speed the r tati n the ea th its a i
( a ) in revo lutions per minute ( b) in degrees per seco nd .
o ns 4 o 1 5 . C truct a right triangle who se sides are 3 and c nstruct an angle which is 3 times the smaller angle of this triangle .
16 . Co ns o o o n truct the f ll wing angles a nd place them the axes , ( a ) ( b) ( c) ( d) ( 6 ) ( f ) 1 7 . In what quadrant is each of the fo llo wing angles ° 4 4 4; 4 4 4 4 355 3
18 a ° . fi . Taking B 7 6 3 10 draw a gure ° o a o o o 6 360 sh wing that differs fr m B, and als that 7 differs fr m by 1 ° ° f 9 . Find the angle between 0 and 360 which differs from each o the ° fo llo wing angles by a multiple of 360 ° ° ( a ) 42 ( b) ( 6) 364 ( d) IX 2 AD AN EASURE , § 5 1 R I M 85
M ea surement of An es Ah m 60 . a a n d gl . ngle y be na med a u s belo re is r s s a n s s a s r ed it exp e ed in y y tem of m e u ement .
s ma r a n a A a a s Thu , we y refe to ngle of right tri ngle who e
r 1 6 ih . a n d 24 ih r r a s s a e . s a n d pe pendicul r ide , e pectively ;
n t n A 24 1 6 o ca a . a s we compute etc , with ut m e uring
n G r k l A in term s of a y unit a ngle . enera l theo e m s li e the a w
’ s s r a in a n s s a s of ine em in true y y tem of me u re ment . The unit a ngle ( see 2) chiefly u s ed ih ' G eom etry a n d
r is de ree its s s o s Trigonomet y the g with ubdivi i n minute, tenth i s s e s a a r. i of minute , econd , with which the tud nt f mili It s
u se a r un a a ra dia n often convenient to nothe it ngle c lled the .
1 sure of An les “ ra 6 . g fi A dia n is a positive a ngle eh its vertex is pla ced a t the center of a
r r a rc is ua ra u s ci cle the inte cepted eq l in length to the di . This un it is thu s a little le ss tha n one of the a ngles o f a n eq uila te ra l tria ngle ; in fa ct it fo llows from the geometry o f the
r s n a semicircu m ci cle , i ce the length of
is 7 7 a ference 3 th t
1 1T ra dia n s where 1T ( ) EM 81 .
° 1 ra a 57 1 7 o r 5 whence di n 7 a pproxim a tely. It is ea sy to cha nge from degree s to ra dia n s a n d vice versa b a s a o on y me n of rel tion w hich sh uld be rem embered . C
f r — s a s o s u s a re r a s . 9 1 93 ver ion t ble thi p rpo e p inted in T ble , pp .
2 se of Ra a n M ea sure i 6 U . s s n r . di It how in geomet y tha t two a ngle s a t the center of a circle a re to ea ch a s their inter cepted a rcs the refore if a n a n gle a t the cen ter is mea su red in ra dia n s a n d if the ra diu s a n d the intercepted a re a re m ea su red in rms sa a r u m r a su r s te of the me line r unit, thei n e ic l mea e sa tisfy the simple rela tion
2 a re a n le ra d ( ) g x iu s .
So metimes also called circu lar mea 86 PLANE R G N ME R 64 T I O O T Y [IX ,
r r s r a r un s in a rc is In othe wo d , the nu mbe of line it the eq ua l to the product of the nu mber o f ra dia n s in the a ngle by
h e r in r s r t nu mbe of l ea unit in the a diu s .
f w EXA M P LE 1 . Find the difference in latitude o t o places o n the same
di 200 . of 4000 . meri an mi apart , taking the radius the earth as mi ° a rc ra d u s ’ Angle / i in radians 2 51 appro ximately .
An ula r eed In n P is 63 . S . a a a g p rot ti g body point , which a t a s a r r a s r a s r a dis di t nce f om the xi of ot tion , move th ough ta nce 2 during ea ch revolution or through a dista nce 7' while
u r s ro a n a ra a r r v the body t n th ugh ngle of one di n . The efo e if is n a r a a s P in n a r n s er un the li e ( ctu l) peed of ( li e u it p time it,
i n r r e. . er s o a n d m s a ul a s o g feet p ec nd) , if the g peed of the
i r a h a a s e e. ra a s er s t ting body ( r di n p time unit, g. di n p econd) , then their n u merica l mea s ure s sa ti sfy the rela tion
(3) v = r . w;
hence the a ngu la r sp eed of a ro ta ting bo dy is n umerica lly equ a l
the a u l s ee a i n i o h xi o o t i n to ct a p d of p o nt o e u n t fr m t e a s f r a t o . Engineers u s ua lly expre s s the a ngula r speed of the rota tin g
r s o f r r r P M o r pa t m a chine y in evolution s p e minute (R . . . )
s er s n R P s a re a s r to revolution p eco d ( . . The e e ily educed ra dia n s per min u te ( o r per s econd) by remembering tha t one
r o a s 2 r ev lution eq u l 7r a dia n s . f EXA M P L 1 . fl h f f u o E A yw eel o radius 2 t. ro tates at an ang lar Speed
P . S . o f o o n . R . Find the linear speed a p int the rim
I n o to 2 7r 5 7r fo r o 2 ft . o n radia s per sec nd , z x , and a p int fr m
of o o 17 ft o . the axis r tati n 2 x 5 1r . per sec nd
f 4- o n o E XA M P LE 2 . Find the angular speed o a 3 inch wheel an aut
o o er o r. m bile g ing 20 mi . p h u ' Every time the wheel turns thro ugh a radian the car go es fo rward 1 7
o 352 m. o in o f 20 . , . (the length the radius) , and mi per h ur i per sec nd
theref o re the wheel turns thro ugh 352 17 radians per seco nd .
ota ti n a s a s ra a a s r 64 . N o . In me uring ngle in di n me u e w e sha ll a dopt the pra ctice universa l in a dva nced work a n d w rite on ly the n u merica l mea sure of the a ngle in term s of the unit
PLANE R G NOME R 4 T I O T Y [IX , 6
6 . If i of hi mi o the l near speed a ve cle is 30 . per h ur , what is the of o n f h f angular speed e o its wheels w ich is 4 t. in diameter A 2 o ns . 2 radians per sec nd .
f . ft 7 . A wheel 5 t in diameter is co nnected by a belt 40 . in length 4 ft o o with a wheel . in diameter . If the large wheel makes 30 rev luti ns
h o w o f i per minute , ften do es the seam o the belt pass th s wheel What is the angular speed o f the smaller wheel ? 1 Ans . 5 sec. 3 s o . 1 T , 44 radian per sec nd
wo o s o 8 . Find the angular distance o n the earth between t p int wh se
i o o o n of is 800 m . d stance fr m each ther , the arc a great circle , iles f 1 1° 2 ’ [ Take the radius o the earth to be 4000 miles ] A ns . 7
’ 9 . Find the distance in miles between two po ints o n the earth s sur ° face whose angular distance is 1 between two po ints who se angular
i . d stance is radians . A ns . 1000
10 f . Find the length o the subtended arc of an angle of radians f f o o i An s . at the center a circle rad us 5 .
1 1 ° . Find the length of the subtended arc of an angle o f 55 at the
o f o f center a circle radius 3 . Ans .
12 . Find the angle at the center which subtends an arc of 3 ft . o n a
i o f 4 ft. E c rcle radius xpress the angle in radians and in degrees , and o o o two A n c mpare the w rk d ne in the cases s . radian
1 to f I R o . 9 1 3 . educe radian measure by means Tables V , p
° ° ’ ° ( a ) 23 ( b) 68 45 ( c) 138
Ans .
1 to 92—93 4 . R educe degree meas ure by means o f the Tables pp . ( a ) ( b) ( c) ( d) ° ' 1' Ans . 1 98 14 3 s ,
1 5 o to . Reduce the f ollo wing angular speeds to degrees per sec nd ; revo lutio ns per second to revo lutions per minute
FUNCTIO NS O F ANY AN G LE
ol ion o orces P ro ection s 65 es ut f . . In 26 4 . 3 R F j , p . , sa w fin d n s a r o r a we how to the compone t of fo ce , velocity, on a n n a s r n r o n a in a n d y li e , the p oj ectio of the fo ce th t l e ; we sa w tha t the componen ts of a for F o n a of o er Y ce e ch tw p P: (my) en dicu la r a s n n p xe , eve whe F , a a is u s a re the ngle obt e ,
1 F P ro F : F co s a ( ) , j, ,
’ 82 . 7 : FIG . 1 P I F F S (1 , 21 i in
s ra r s u r sa ro s m If eve l fo ce occ in the m e p blem , o e of them
° ma y m a ke a n a ngle a g rea ter tha n 1 80 with the po sitive
' r is de n e co s a a n d sin a fo r di ection 0 X . It convenient to fi
° 1 r r a n gle s grea ter tha n 180 so tha t the eq ua tion s ( ) em a in t ue .
so r a s o f a n r If we do , the p oj ection on the two xe y di ected segmen t of length r j oining the origin 0 to a p oin t P a re
2 x Pro r r co s a P ro r r sin a ( ) ja, , y jy , where a is the a ngle between the positive direction 0 X a n d the
s r n P a n d ma a n a a n Size s po itive di ectio O , y be ngle of y , po i
r H n s r defin ition s a re : tive o nega tive . e ce the de i ed
x y
3 co s a . ( ) ’ r
s d fini i r n s s s a r a n The e e t on s a e co i tent with tho e l e dy give ,
°
1 1 35 fo r s a n d o s t. e. a s O § a g . , , the ine the c ine in c e
r r d efin itio n they determine the sa m e va lu es a s the ea lie s.
r o nometr c unctio ns of An enera l De ni ions . 66. G fi t T ig i F y
n le in nd in A . d fin i io n s s a a co s a 65 a g The e t of given h ve, 89 90 PLANE R G N M E R X T I O O T Y 1 , 66
rs ssa r u r s . Ea is a of cou e, no nece y dependence pon fo ce ch nu mber which depen ds o nly on the m a gnitu de a n d sign o f the
A r n i a ngle . pu ely geometric defi it on o f these a n d o f the
r n r n o s o f a n a a o s s othe trigo omet ic fu cti n y ngle , c n i tent with
d efin itio n s 1 1 35 a n d the of , , with the fun da menta l rela tion s between P s a s a a sin oc co s a them , uch t n / , 2 2 sin a co s a 1 r r a re , the ecip oc l m m la tion s . a a a s s , etc , y be de follow P la ce the given a ngle on a pa ir P (W) (W) a ul a r a s a n d s a n of rect ng xe , elect y
F I G “ 83 “ P s or a s a re x point who e co din te ( , y)
d s a r 0 r r o n the termin a l side a t a i t nce > f om the o igin . Then <4> if $ 535
a a <5 é fsszzz,
1 6 ta n a pr o vided 3: 0 ( ) 2,
8 (7) ctn a p rovided y qe O 5 23312;
8 see a rov a: 0 ( ) 2; p ided ;
1 (9) csc a provided y 4 0 . 3; 0 312;
Three a dditiona l fun ction s so m etime s u sed a re 1 (10) The ver s ed sine o f a : v er s a co s a .
: v 1 (11 ) The ha ver sine o f a ha a 4( co s a ) . — 2 r a s ca o f a : s a s ee a 1 . (1- ) The exte n l e nt ex ec
z 1 in a n d a l so the cover s ed sine of a s a .
The exceptio n s no ted are based on the general pri nciple that a fractio n al
' expressio n do es n o t represen t a n u mber if its d en o min ato r is zero .
92 PLANE R G N ME X 7 T I O O TRY [ , 6
—a s is s s a r s o f a n a y xi po itive ; imil ly, the co ine y ngle in the s o r r a ra is a econd thi d qu d nt neg tiv e .
r ma a s s ta n a a sec at csc a . Simil ly, the ign of , ctn , , , etc , y be deter mined directly from a figu re ; they a re a s fo llow s
Q U ADR AN T co s a . ta n a . ctn 0. s ec (1. cs c a
NO 1 a o a co s TE . ( ) tan is p sitive (negative ) when sin and a have like
( unlike) signs ( 2) recipro cals have the same sign .
EXE ISES — FUN O N O F THE G ENER L NG LE RC XXIV . CTI S A A
1 B o n o o d efinitio ns . y placing the angles the axes , sh w fr m the that
° ° ( a ) sin 225 co s 225 v2/2 . ° ° ( b ) sin 1 50 co s 1 50 v3/2 . ° ° v ( 0 ) sin 330 co s 330 3/2 .
d v2 2 cos V2 2 . ( ) sin / , / ( 6 ) sin cos ° 1 80 0 n 0 for n 1 2 3 ( f ) sin , sin ( ; z 4 , z}; , ziz , ° co s 90 0 co s 2 n 0 fo r n : 1 2 3 ( g ) , [ ( ; 4 , 4 , 4 ,
9 2 . of h e o o i o 72 Which t f ll w ng are p sitive and which negative sin , sin sin 850° tan sec sin cos sin cos ctn 280 cos csc cos tan co s
2 2 2 z P o fo r a n n a a co s a 1 . 3 . r ve y a gle that sin [ Use y
Pro ve each o f the o ther Pytha go re a n relatio ns for a ny a n gle a 2 2 2 2 1 ta n a = s ec a o a 0 1 a csc a if sin a fi 0 . + , if c s $ ; + ctn , ?
f a cos a 4 . P o a s ee a a o o r ve that ctn , , csc are the recipr cals tan , ,
a fo r o f a fo r o defined . sin , respectively , all values which b th are
5 a P o i of a n in first o r o . ( ) r ve that the s ne y angle the sec nd quad b P o o of rant is between 0 and 1 . ( ) r ve that the c sine any angle in the l st o r 4th qua drant is between 0 and 1 .
n o t o dd of its 6 . Pro ve that if an angle is an multiple a right angle F o r is a sine is between 1 and 1 ; and conversely . what angles sin — = + l ; sin a z l ; cos a = + 1 ? X 8 F NC N OF ANY ANG E , § 6 ] U TIO S L
7 . o cc ot cos a fo r o f a cos a 0 . Sh w that tan sin / all values , if i
8 . Sho w that tan a and ctn a may have any values whatever °
o o r o . 9 . Show that vers a and hav a are always p sitive zer
° 10 I n 05 0 a s to o . f an a gle starts at and gradually incre es sh w that the behavio r of sin a and cos a will be as indicated in this table
1 1 . By placing the angles indicated o n rectangular axes determine the numbers to fill the blanks in the f o llo wing table
° ° ° ° ° ° ° ° ° ° a . 3 0 45 60 1 20 1 3 5 1 50 225 240 3 1 5 3 3 0
fi fi fl — fi
1 di o 2 . Assuming that the sun passes rectly verhead, trace the change of o o f o o to in the length the shad w an bject fr m dawn sunset . Which trigo no metric functio n do you think of in this pro blem
13 . A o f Ex s . 10 1 1 o ssuming the results and , derive fr m them the ° ° variatio n of the tangent fro m 0 to 360 and its values at each o f the
o Ex . 1 1 . Do fo r angles menti ned in the same ctn a , sec 06, csc 0L.
— 8 . R a d o f a bl s n a d 6 e in g T e . Si e n Co sine of 9 a n d In order to fin d theva lu e of a n y one of the trigon o metric fun c
s a n a o s tion of give ngle w e c n ult the ta ble s . In the ta bles the va lu es of the different function s a re prin ted only u p to
° To fin d the s ine of a n a cute a ngle g rea ter tha n 45 we ma ke ° u se a sin a co s 90 of the rel tion ( a ) . The ta ble s a re a r
° ra nged to fa cilita te thi s by ha ving the a n gles a bove 45 printed a t o f a a n d a s a the bottom the p g e , the colu mn he ding ch nged
r m s s . Ta bles . f o ine to co ine , etc (See , p If we Wish to fin d the sine of a n a ngle g rea ter tha n we mu st fin d a wa y to exp re ss the sine in term s of some function of 9 4 PLANE R G N ME R X T I O O T Y [ , 68
r n a n a cu te a ngle . We p oceed to fi d expre ssion s for the va lu es
° ° s a nd s n a n s 90 9 1 80 of the ine the co i e of the gle j: , j : 9,
° ° 27 0 a n d 3 60 9. 1 9,
To con s truct these a ngles we dra w a circle of ra diu s r with
its n r a t r n a nd ra a rs HN ma k ce te the o igi d w the di mete , KS ing the a ngle 9 with the
—a s to a n d l y xi the right eft,
a nd a so a rs P ill l the di mete , TL m a king the a ngle 9 with
w- s a n d the a xi bov e a below .
a s X OH X K The ngle , O ; X L X M X N X O , O ; O , OS ;
a ‘ b - a - ( - l . b) a nd X O T a re a n s , the gle
m a a n d X P entioned bove , O i D s the a ngle 9 . enote the P coordin a tes o f. the point
b a b ca u s y ( , ) ; then be e the
4 . FIG . 8 tria ngle OAH is congru ent
r a O OP r n a s H a re I) a to the t i ngle the coo di te of the point ( , ) ,
a n d sa wa the o n a s s If L Ill in the m e y co rdi te of the point , , ,
N S T a re a s s to a s a fi u re. , , e ily een be indic ted in the g We a re n o w a ble to rea d o ff the v a lues of the trigonometric fun c
s o f a r us a n s r m fi u re in r s a b tion the v io gle f o the g , te m of , , a n d
thu s sin 9 b
° co s 9 b r co s 90 b r ) / , ( / , ° sin 1 80 9 sin 9 — b r ( ) ) / ,
° ° co s 27 0 — b r co s 27 0 b r ( / , ( / , ° sin (360 sin b/r. Hen ce we ha v e
° ° co s (90 sin 9 co s (90 9) ° ° sin (180 sin 9 sin (1 80 9) ° co s sin 9 co s (27 0 9) ° sin (3 60 sin ( 9) sin 9
96 PLANE R G N ME R X T I O O T Y [ , 70
° 3 60 r o f s s multiple of to eithe the olution j u st found . The
° ° solution s which lie between 0 a n d 360 ca n be fo u nd from the
a s b a s r u a s t ble y me n of the fo m l given a bove .
Illustra t ve xa m les on om o s 70 . i E p C p ition a nd Resolution
f Forces .
EX LE 1 . o o E E of of two AMP the c mp nents, x , ”, the resultant . Find ° o first f 12 lb . of 30 o o f rces, the o acting at an angle with the h riz ntal , the ° r o of 2 l a ctin 60 o o . sec nd 0 b . g, at an angle with the h riz ntal
L TI T o o o f ro ec SO U ON . s lve this we make use the principle that the p j tio n o n a ny lin e o f the resultant of any number o f fo rces is the algebraic
sum of the projectio ns o f the co mpo nent f o rces . 5 o o o o of first By equations § 6 , the h riz ntal c mp nent the is ° f o 20 cos 60 12 cos and o the sec nd, hence
° ° z 12 cos 30 20 co s 60 E ,
In a similar manner we fin d
° ° R 12 sin 30 20 60 y sin
We can easily find the magnitude of the resultant fro m the equatio n R 2 R 2 R 2 , y Hence
The directio n of the resultant is given by the equatio n
9 R R tan y , Hence 0 48°
n o o f of EXAMPLE 2 . Find the mag itude and the directi n the resultant
two o F G 24 the f rces ( ,
rm o o 24 213 o o f 24 [ N o The n tati n ( , means a f rce magnitude acting ° o az- at an angle o f 213 with the p sitive axis . ] The metho d of so lutio n is the same as in Example 1 we find
° ° F 17 c o s 128 1 7 38 b , sin ( y ° ° 24 co s 213 24 co s 33 b G , ( y Hence ° ° 17 38 24 co s 33 Rx sin
Similarly we o btain
° E 17 COS 38 24 33 . 3 25 u sin 2 R R : 1) 1 2 1 00 9 arctan arctan . 0 06 ) 8 x 70 F NC N OF ANY ANG E , ] U TIO S L
X — R AD N O F TAB E — RED C O TO E ERCISES XXV . E I G L S U TI N FUNCTIO NS O F ACUTE ANG LES
1 E o o o o f n o t . xpress the f ll wing as functi ns acute angles greater than M ake use of co ngruent angles whenever advan tageo us
° ° ° ( a ) sin 150 ( 5) co s 125 ( 0 ) tan 283 ° ° ( d) ctn 3 6 ( 6 ) sec ( f ) csc 210 ° ° ° ( y) sin 9 43 ( h) co s 55 1 ( t ) tan 546
o find of o o o s 2 . Fr m the tables the values the f ll wing l garithm
° ° ( a ) log co s 16 1 log sin 16 1 ° ° ( 6 ) lo g sin 21 7 ( d) lo g co s 252
N o a c d o [ te that the numbers in parentheses in ( ) , ( ) , and ( ) are p si
if s o f o . tive ; the minu sign were absent , each them w uld be negative Negative numbers have no real lo garithms ]
Co of o o o o 3 . mpute the values the f ll wing expressi ns by l garithms
° ° ° ( a ) sin 148 ( b) cos 1 60 ( c) co s 320
o o o o o 4 . S lve the f ll wing trig n metric equations
o 2 2 ( a ) c s t sin t sin t.
I o co s? t SOLUTION . n this equati n may be replaced by its equal 1
2 t o o d z 2 sin ; the equati n then bec mes a qua ratic in sin t, vi . : 2 sin t +
t 1 0 . o e u iva len t to o n e e sin This equati n is q the given ; t. . every o f f so luti n o either is a so lutio n o the o ther . The so lutions may n ow be fo und by facto ring 2 1 ( sin t ) (sin t 0 .
H t 1 0 t 1 ° ence we have either sin , whence sin , and t 270 o r ° ° t 270 k o r 2 t z 1 else sin , whence sin t and t 30 ° ° k 360 o r t 1 50 k n o o o There are ther s lutio ns .
2 — 2 b 2 s in cc co s cc : 1 . o w ta n x z ( ) ( g ) se + 3 . 2 2 2 c cos w= sin az. h 4 sec w ta n = ( ) ( ) + a 7 . = d cos 2 cc 5 sin a3 3 . i mm ctn cc 2 ( ) + ( ) ta + : . ( e ) co s 2 a; ( j ) 2 i s : ( f ) 5 s n w+ 2 co az 5 . ( k)
5 . R 9 of o 100 150 Find the resultant ( , ) three f rces ( , ( , 200 F a force ' of ( , where ( , ) indicates a magnitude F and direc o a ti n .
6 . o o o n of a o o f 4 1b Find the c mp nents the axes f rce magnitude 5 . 7 . ° ’ a n of 215 20 i o o f - which makes angle w th the p sitive end the maxis .
7 . Find the magnitude and the directio n of a fo rce whose components on two i a x F F perpend cular es are , y H CHAP T E R XI
THE ADDITION FO RM ULAS
7 1 The Add o n ormul a s r o f . iti F . In the eduction certa in trigonometric expre ssion s to s impler o r m o re convenient form s it is s ometimes de s ira ble to express a trigonom etric fun ction of the su m o r difi eren ce of two a ngles in term s of function s of
s a ra a s u m r i r the ep te ngle forming the s o d ffe ence . Without refl ection the s tudent might think tha t sin ( a + B) would be equa l to sin a + sin B by a n a logy with the formu la % (a b)
a b a r a or s a a s s s , but t i l of one two peci l c e will how
s is a wa s s sin is a to thi not l y true ; thu , eq u l one ,
° ° sin 60 sin 30 is a 3 is a r ha n but equ l to 5, which gre te t o n r r fin d r a s fo r sin a a n d e. In o de to the co rect form ul ( B) co s ( a B) we m a ke u se of the theo ry of directed q ua ntities a s
a i 26 55 57 5 8 a n d 65 . expl ned in , , , , S u ppos e a force of m a gnitude A m a ke s a n a ngle a with the
ma a kes s wa xis e a rc o f. B a n po itive , whil nother fo e gnitude m
° a ngle 04 + 90 with this a xi s ; then the resul ta nt R of A a n d B is repre sented by the dia gona l OP of the recta ngle of which
- A a n d B a re two s s . o o R o f s ide The y c mp nent, , thi su lta n t is
1 R = A s n a B sin a ( ) , i (
A sin a + B co s a .
a x- R is Simil rly, the component of
2 R A co s a B co s a ( ) , (
— s A cos a B in a .
N ow b 65 . 85 . y , FIG
3 R R co s a R R sin a ( ) ( B) , , ( B) ,
n d where Bis the a ngle between A a the res ulta nt R . 98
100 PLANE R G ONOME R T I T Y [XI , 73
u m cosine) of the s of two a ngles . The method depen d s on formula s (6) a n d (7 a n d upon the fa ct tha t a ny two numbers a re r r a s a nd s s a n p opo tion l to the ine the co ine of ome gle .
E XA M P LE 1 . E 3 cos a 4 a o k a xpress sin in the f rm sin ( B) . To so lve this we first find an angle who se sine and co sine are pro po r
tio n a l to 3 4 . We o o so and may clearly ch se an angle B that sin B g, and cos B g; hence we may write 3 co s a 4 sin a = co s a + § sin a ) 5 ( sin B cos a cos B sin Hence by fo rmula ( 6) we have
o a 4 a 3 c s sin a 5 sin (B ) . ° From the tables B 36
T EXERCISES XXVI . ADDI IO N FO RMULAS
1 a find a . . Given sin sin B sin ( B) a o o o ( a ) When and B are b th acute ( b) when a and B are b th btus e . ° ° ° ° 45 cos 45 30 cos 30 (c in 2 . Find sin ( ( sin ( ( ) c terms of sin a: and os c .
a; i ce 3 . Given that and y are bo th o btus e angles and that s n
fin d c sin g sin ( a: y) and o s ( 9: y) .
° o o to a cos 90 —x 4 . Use the additi n f rmulas express sin ) and ( M ) f in terms o sin a and co s a .
° ° P o sin 6 cc co s 30 93 c . 5 . r ve that ( 0 ) ( ) sin f 6 . E a 9 of o o a xpress sin ( B ) in terms sines and c sines , B,
and 9 . [ HI N L' Let a B a n d o btain sin ( 95 then replace by its
a . value , B]
f a 7 . E cos a 9 o f o o xpress ( B ) in terms sines and c sines , B,
and 9 .
o o o f two o o o n 5 o t 8 . R educe the c mbinati n simple harm nic m ti s c s 12 sin t to the f o rm r cos ( t
9 R 3 t 4 cos t to o r t . educe sin the f rm sin (
1 R of o o w to o o f 0 . educe each the f ll ing the pr duct a number and the sine o r the cosine of a single angle — / — 2 co s m. ( a ) sin z ( e) x 3 co s cc sin x. — ( b) 3 cos y 4 sin y. — o s 9 12 9 . . ( c ) 5 c + sin ( g ) 7 cos 9 sin 9 .
t 3 0 68 t. h . in co . ( d) 3 sin ( ) 55667 s c . 5 s c 75 ADD N F RM A 1 1 XI , ] ITIO O UL S 0
° 1 1 two o o f n 2 3 o f . Given f rces inte sities and that make angles 30
i o fc- find o o and respectively , w th the p sitive axis ; the h riz ntal and the vertical compo nents of their resultant witho ut fin ding the resultant itself fin d the same quantities by using the resultant .
12 . . 56 . 5 co s .34 fin d 9 Given sin c c , an angle , and a number r,
a . 56 c . 5 co s r sin e o f 70 such th t sin c ( by means . Then ,
io r c 9 .3 4 fin d c o o f m sin ( ) , sin ( and theref re ( fr m the
find fin d 0 . Tables) 0 9 . Hence
4 Double An les rm n d 1 r 7 . . a s 6 a 7 7 a e g Sin ce fo ul ( ) ( ,
r a ll a s a a a n a a r t ue for ngle , they hold when , y ngle wh teve , a n d a sa a n B , the me gle ; hence ,
sin a a sin a co s a co s a sin a ( ) ,
in in a n d co s (a a ) co s a co s a s a s a .
Therefore the following form ula s hold fo r a n y a ngle wha t ever : (10) sin 2 a = 2 5 in a co s a °
2 (1 1) co s 2 a co s2 a sin a ; o r s n sin 2 a co s2 a 1 , i ce ,
2 2 Z = 1 — 2 : 2 — cos a 8in a co s a 1 .
Ta n en of a um or o a eren ce t S f Difi . in rm 75 . g S ce fo ula s
6 a n d 7 fo r a ll a u s a a n d r u a ( ) ( hold v l e of B, the fo m l
sin (a + B) sin a co s B+ co s a sin B co s ( a B) co s 0: co s B sin a sin B holds good fo r a ll va lue s o f a a n d B except tho se which ma ke
: e. a o r o r n co s ( a B) 0 t. except when B a a ngle tha t diffe rs from one of the s e by a n integ ra l number of
° s F o r a s n o t for a 47 43 time ex mple , it doe hold B
D n u ra r a n d a r b co s a co s ividing both me to denomin to y B, we obta in the form ula
ta n a ta n 3 13 + | ( ) , 1 ta n a ta n B
s for a ll a n s a a n d su a a a n d a which hold gle B ch th t , B, + B
n ha ve ta gents . 102 ANE R G N ME R XI 7 PL T I O O T Y [ , 5
Simila rly fro m formula s (8) a n d we obta in
o s fo r a ll a s a a n d s a a a n d a which h ld ngle B uch th t , B, B ha ve ta ngents . F rom formula s (10) a n d ( 11) we fin d
2 ta n a 15 ta n 2 a ’ 1 — ta n2 a
Which hold s fo r every a ngle a s uch tha t a a n d 2 a ha ve ta n
e s sa o r a ma o a i 1 3 g nt . The me f m ul y be bt ned directly fro m ( ) by puttin g a in pla ce of B.
li ion s r fr A ca t . a s s a a e e 76 . pp The for m ul of thi ch pter q uently u sed for reducing exp ressio n s who se va lue s a re to l b e a u a a o a s ma s . c lc ted , to for m in which l g rithm y be u ed
o o f o CD to EXAMPLE . Supp se the height an bject is be determined and that it is n ot co nven ient to measure a base line bearing directly
o o l o o t ward the base 0 . The f l wing meth d is then f o at s ometimes emplo yed . The angle o elevati n is mea sured fro m so me co nvenient po int A ; a line AB d is then me as ured at right angles to the
A fin a ll o f o o h line O y the angle elevati n, B, is o m B A served fr . The height h can then be de
termin ed by s o lving a successio n o f triangles .
’ With the aid o f the fo rmul a s o f this chapter it is frequen tly po ssible in such cases to reduce the . 86 FIG . o calculatio n to a single lo garithmic co mputati n . In the case jus t mentio ned we have
B O z h ctn A C : h ctn a B , 2 2 d? 137 7 21 6 h? ( 0 mm ctn2 a ) h2 ( ctn B ctn a ) ( ctn B ctn a ) (sin a co s cos a sin B) (sin a co s B+ co s a sin B) h? B sin2 a sin? B
o l 6 and hence , using f rmu as ( ) we have
(1 a h sin sin B \/sin ( a B) sin ( a +B)
L o o o et the student sh w, by Opening a b k and studying the dihedral wo angle f o rmed by t leaves, that a B.
104 PLANE R G N M E R XI 7 T I O O T Y [ , 7
un ions O l n l s c Ha e . Th 77 . F t f f A g e fo rmul a s
z 2 éo s a + sin a = 1
2 2 co s a sin a co s 2 a a re a ll a s true for v lues of a . If we ubtra ct o n e o f the se fr o m
er a n d a s a dd a n o r s the oth , if we l o them , we obt i the f m ula
2 1 6 2 si a = 1 — s 2 a ( ) n co , 2 1 7 2 co s a 1 co s 2 ( ) a .
Thes e fo rmu la s a re true fo r a ll v a lues o f a ; fo r a they beco m e 2 ’ 2 s in 1 co s a a n d 2 ’ 2 co s 1 co s a ,
' o r s s a re a ll a s o f a ma ince the e true for v lue , we y write
1
° -c
fo r a ll a s o f sa a s ma which hold good v lue a . The m e for mul y
’ b e obta ined from (1 2) by s o lving fo r sin o r fo r ’ co s a r a 2 fo r a fte putting / . F rom (1 8) a n d (19) we g et by div is io n
/1 co s a sin a (20) ta n oi/z : 1 + cos a which ho ld for a ll va lues of a except when a denomina tor
a s ra a is r va nishe s . The mbiguity of ign of the dic l dete mined in a given ca s e by the fa ct tha t ta n ( a /2) is po sitive or nega tive a ccording a s a /2 is o r is n o t in the first o r second
r q ua d a nt . The rela tion s between a n a ngl e a n d its ha lf a re frequ ently u s eful in problem s tha t rela te to a chord of a circle a n d the
s n s a t r s s fo r a a ngle which it ubte d the cente ; thi occur , ex mple, XI 77 ADD N FORM ULAS 105 , 1 ITIO in la ying out ra il ro a d curves whe re it is co n venien t to ma ke
T s is m ea s u rements a long cho rds of the cu rve . hi illu stra ted
r s r in some of the ex ercise s below . The ela tion a e a l so u seful in simplifying trigonometric exp re s sion s a n d in a da ptin g fo r
r o mula s to loga ithmic computa ti n .
— — EXERCISES XXVIII . HALF ANGLE FO RM ULAS
° ’ 1 si o o f 22 30 o . Find the ne , the c sine , and the tangent fr m the kno wn values o f sin cos tan
° o f 2 . Find the sine , c sine , and tangent o 1 5
05 05 3 . Given that sin and that is an acute angle find sin and ta n
° fi ° ’ 4 . Given tan 26 n d tan 13 17
° ’ ° 5 n 6 52 find o . Give tan 3 sine , c sine , and tangent of 18
r o 6 . If den tes the radius o f the circle in the a cco m fi u re o panying g , 0 a ch rd , and 9 the angle which 0 sub tends at the center sh o w that sin : c/( 2 r)
fi u r B 7 . I e D n the g , draw the line tangent to the
AD to B D o o o circle , and perpendicular fr m the pp site o f o BA o a [ ABD end the ch rd . Sh w that ( ) ( 0) BD : A B co s 2 r sin co s 87 . FIG . r Sin 9 .
° 8 P o 4 05 . 5 05 05 . r ve that tan ( sec tan , if tan exists
° ° ° P o 5 45 2 45 if 9 . r ve that tan ( 4 tan ( tan tan (1 exists .
2 1 05 05 if 05 . 0 . P ro ve that tan 2 sin ctn sin , sin 0
1 1 o 2 csc 05 if 05 . . P r ve that tan ctn , sin 0
1 o co s 1 05 fo r o f 05. 2 . P r ve that [ sin sin all values
1 o cos 3 . P r ve that [ sin f 05 1 sin 05 for all values o .
I fi ure G OA of 14 . n the g , is a diameter a circle o f radius r' A OP 05 is any acute angle O CP by geometry ; and P B is perpendicular to C
o GA . Sh w that
B r co s 05 B P : r sin 05 BA O , ,
r y 05 G B r 1 co s ers , (
2
03 co s Em . 88 . 106 PLANE TRIG ONOM ETRY
1 5 . o Ex . 14 o o ns of Fr m , sh w that the functi can be fro m the figure in the fo rm
( Sin 0‘ sin
l cos 05 cos
tan
f 16 . If n o o of 05 a umerical value any functi n is given , all the o the of 05 of o o o f o Ex . 14 . functi ns and can be und ge metrically fr m Thus ,
06 Off O P : 5 BP : 4 B 2 2 : if sin is given, lay , ; then O 5 4 3 .
B = BA = 2 a n d Hence , C 8, ; It f o llo ws that sin 05 co s 05 tan 05
2 4 x/5 5 sin / , — co s 0 tan
1 ni o of 05 o o f o f 7 . Find the remai ng functi ns and th se by means
Ex . 1 6 it cos 05 05 , if tan
1 8 o s o f o of 05 o . The remaining functi n and th se can be f und
o of is o fi u re of Ex . 14 o when any functi n given fr m the g , by dr pping D o s if a perpendicular fro m 0 to UP . thi tan — 2 x 4 B P CB BA ho fi r o f E . 1 o 19 . u e Since , in the g , by ge metry , s w
05 05 2 that ( 1 co s ) vers sin 05.
0 D o o o as o o 2 . erive trig n metric f rmul fr m the ge metric identities x 14 ( E . ) 2 B > 3 B P P A [ Bi P 0 1 0 .
a c r ormu a s I n a r r a s 78 . F to F l . a d pting trigonomet ic fo mul to loga rithmic co mputa tion it is often desira ble to exp re ss the su m ( or difference) o f two S ines ( o r cosines) a s the p roduct of
n s other fu ction .
° ° ° ° A M P LE R 35 15 to o 2 25 co s 10 E X 1 . educe sin sin the f rm sin
d o a; 11: T o this , set y y a nd so lve fo r a: and y a: y a: 3: cos co s a: Then sin ( y) sin y sin y, — sin ( x y) : sin zc cos y cos msin y ; di a: 11; 2 9: co s whence , ad ng, sin ( y) sin ( y) sin y ° ° ° substituting a: y we get sin 35 sin 15 2 sin 25 cos
PLANE TRIG ONOMETRY [XL 78
cos 3 y) + cos 3 y
° ° 1 45 w 4 cos 2 . sin ( ) sin ( 5 (c.
1 3 . sin 3 a: sin 5 a; 2 sin 4 as cos az.
14 . ( a ) cos ( b c) co s a 2 sin ( s 0) sin ( 3 — ( b) co s a co s ( s a ) ;
— sin 3 + sin ( s c) tan g; ( a + b)
0 100 s in(3 — — 8 ta n x tan y sin ( x y)
” so — “ o o f o for i 1 6 . The called meth d ffsets lay ng fi e o ut a circul ar track is illustrated in the adj o ining gur . OA B 0 to O B ’ dis The track is tangent at , and the ’ ’ OA ’ A B A 'A B o to tances , , , C , are easily sh wn be fi ure A OA ’ f as marked in the g , where A is hal
0 - the angle at the center subtended by a 1 0 foot chord . ’ ’ I OA’ B run A B’ n practice, the line is , and and o B ’B to marked . Sh w that , the distance actually be o ff o B’ is laid fr m ,
’ B’ B A A CB 200 05 co s 89 . sin FIG . CHAP T E R XII
G RAP HS O F TRIG O N O M ETRIC F UN CTIO N S
les nd ni s ra si S ca a U t . g: is a 7 9 . The g ph of the function n
e a ss a ll s o s o o a s cc curv p ing through point wh e c rdin te ( , y) ,
in a a n r r sa ti sfy the eq ua tion y s x . The g r ph of y othe t igo
i m r r n x . s s i n o m etric a s co s x ta a . function , , etc , i il ly dete m ned The ra dia n is the unit a n gle commonl y u s ed in plottin g the g ra ph s a n d in the fu rther study of the trigonometric fun ction s
n d o r a a a a su in the Ca lcu lu s a in the dv a nced m them tic l bj ects .
s s s s ecified a sin a: is un rs U nle otherwi e p , the equ tion y de tood
is s o f x a a s a s a in 64 to mea n tha t y the ine r di n expl ined . In plotting cu rve s it is of a dv a nta ge in m a ny w a ys to m a ke
o r zo n a a n d r a s a uni s sa a n d s the h i t l ve tic l c le t the me , thi shoul d be done if n o t too in convenient ]
l ttin P oints In a r 80 P o . a e a u s o f . g T ble V given the v l e the s s a n d a a a s a s r ra a n s ine , co ine , t ngent of cute ngle m e u ed in di which a re very convenient fo r plotting the gra phs of the se
r - s o r function s on c oss ecti n pa pe .
h o sin Dr in n d 81 G ra f x. a a a r r a a s a . p w p i of coo d te xe choo se the s ca le unit 1 0 sma ll division s of the cros s- s ection
r a k r a s s a s in first p a pe . T e f om T ble V the ine of the ngle the q ua dra nt fo r ea ch tenth ra dia n a n d ta bula te
In a n sin a: m n i n f x n o f n . y case , y ea s that y s the si e o u its a gle The ° ° n 60 n 45 n d o r a n o n m right a gle , the a gle , the a gle , the egree , y ther a gle ight
n u n o n n n . be cho se as the it , if it were c ve ie t TIf we were to take the two scale units the same in plo tti n g the curve y
u n n d n o f o l 180 sin 2: where the it a gle is the egree , o e arch the curve w u d be un its lo ng a n d o n ly 1 un it high . , 1 10 PLANE R G N ME R 1 T I O O T Y [XII , 8
Plot the se poin ts a n d d ra w a s mooth cu rve throu gh them a s
fi u r 0 A in the g e.
FIG . 90 .
It is rea dily s een by the prin ciples of 68 tha t the exten s r s r a n d r a ion of the curve th ough the econd , thi d , fou th qu d ra s is a s s b AB B a n d CD a n d a the r nt hown y , C, ; th t cu ve extends to the left a n d to the right of the origin in a succes
s o a r s s a s OAB B OD . i n of che uch , , etc The gr a ph o f sin a: ca n be dra wn witho ut the a id o f T a ble V a s fo llow s : Choos e a co nvenient s ca le unit a n d la y o ff on the
7? x—a s P a o a a n d s s x O pr xim tel , divide thi egment i 5 p y n a n n n u r u a a r s 15 sa n s i to co ve ie t n mbe of eq l p t , y ; the poi t of
s n o rr s to 95 0 ° a k ro divi io c e pond ’ T e f m 30 30 2
a a s s s a s r . 21 fo r a t ble of ine , uch the one p inted on p ex mple ,
° the sines o f the a ngle s in the first q ua dra n t fo r ea ch 6 a n d ta bula te
P l o t thes e points a n d dra w a s m o o th cu rve through them .
1 12 PLANE R G N M E R T I O O T Y [XII , 82
Then the length of AP is a pp roxim a tely equa l to the length of the a rc
u s e s fo r s ru n ra s sin 93 To thi m ethod con t cti g the g ph of ,
co s a w r a s F i . 93 a dr the unit ci cle, in g , t ngent to the y- a xi s a t the origin a n d divide the ra dia n a re into a convenient
r a a s sa 5 b s S s . 2 nu mbe of equ l p rt , y , y line from to the point ,
- h 4 . 6 . 0 11 a s t e a s s o a ra . , , etc , the y xi ( l t divi i n of the q u d nt will + Ma rk s 0 2 4 . 6 o rs . 1 7 . . . . 8 of c u e be only long) the point , , , , ,
FIG . 93 .
1 :v-a s a n d r r e a s e a , on the xi e ect pe p ndicul r qu l
s e to the ordina tes o f the corre sp o nding p o ints o n the a rc . The
' give p o ints o n the g ra ph o f 3
By erecting perpendicula rs to the x -a xis equa l to the ho ri
' zon ta l di sta n ces fro m CQ of the co rres p o nding points on the
o s cc a rc we sha ll g et points o n the g ra ph of c . By dra wing ra dia ting lines fr o m the center 0 of the unit circle thro ugh the points o f divis ion o f the a re w e ca n la y o ff the ta ngents of the s e a rcs o n the y- a xi s a n d con struct the gra ph
9“ The pro o f o f this can n o t b e given u n til the student has stud i ed Calcu l us . 3 n 55 b u t o e n . 017 £5 . The The distance AP is greater tha , the err r is l ss tha ° AB a n f u 4 r o u . 017 o u n o o 7 o greatest erro r , ab t , cc rs whe is arc ab t ° f r 4 . n when x appro ximately . The erro r o a 5 arc is 007 a d fo r a q u ad
n . 006 . ra t, XII 2 R G N M E R C G RAPHS 1 13 , 3 ] T I O O T I
n r s F i 9 1 ra of ta n <5 ; a n d in a obviou s m a nne ( ee g . ) the g ph of
n r n s ra s ca n r u sec a: ca be d a w . The e g ph be extended th o gh
o r r a ra s a n d -a s a s in the the th ee q u d nt , to the left of the y xi ,
1 a n n r a s s 2 77' ra a n s a u s § 8 . If the gle i c e e beyond ( di ) the v l e of a ll the trigonom etric fun ction s repea t them s el ve s a n d the g ra ph from x : 2 77 to will be a repetition of thos e fro m to 7r. F unction s which repea t thems elve s a s 95 in crea ses a re ca lled
er d c fun ct on s er is un p io i i . The p iod the s m a lle st a mo t of increa se in a; which produce s the repetition of the va lue of the
s sin x is a r n c a r function . Thu , pe iodic fu tion with pe iod of
2 7r r ta n a: is 7r. , while the pe iod of
— RAP O F R M T O N EXERCISES XXX . G HS T IGO NO ETRIC F UNC I S
1 P o s of o l o o us V . l t the graph the f l wing functi ns ing Table , and
Table VI when necessary . ( a ) co s a; ( b) tan a: ( c ) vers a: ( d) ctn cc ( 0 ) sec s); ( f ) csc a: ( g ) sinz cc ( h) m s“ : ( 73) x/ sin a:
P o o o o i o us e of 2 . l t the graphs of the f ll wing functi ns w th ut the tables : ( a ) co s a: ( b) tan ce ( c ) sec cc
f o o f 3 . P lot the graph o c s a: by dividing the sec nd quadrant o the i of unit circle into fifths of a radian ( see Fig . 9 1) and mak ng use the fact that co s a: sin ( 1r/2 f 4 . P o o n o (5 z 2 55 l t the same axes the graphs sin , sin % , sin , and
2 sin a .
-- 5 . P o o n o f cos x cos az cos a n d l t the same axes the graphs , g ,
3 cos c .
6 D r o f az n nx n sin n . iscuss the g aphs sin / , sin , and a; ( where is a vi o f of Ex natural number) in ew the results . 4 and 5 .
7 P o of a; cos a: o . l t the graph sin by adding the co rrespo nding rdi
of a; co s o nates the curves y sin and y x pl tted o n the same axes .
P o s o f o o 8 . l t the graph the f ll wing functio ns by adding o rdinates — ( a ) simm cos a: ( b ) 2 sin w+ co s w — — ( c) ta n m 2 sin a: ( d ) cos a: ) — ( e) a + sin a> ( f ) az co s a:
9 . P o o n s s o f ac 0: 1r l t the same axe the graph sin , and sin (
1 0 . P o o n o f :5 co s x cos a: 1r 2 . l t the same axes the graphs sin , , and ( / ) 1 14 PLANE R G N ME R 84 T I O O T Y [XII ,
Invers e unction s s n 69 a 83 . a . F We h ve ee in th t the equa tion (1) y sin a: ca n be s olved fo r x if y is a n y nu mber wha tever between 1
in fin i m r u a n d 1 a n d a a re a n te u s s . , th t there n be of ol tion
‘X‘ An y one of the se solution s is denoted by
2 x a rcs n ( ) i 11.
s s a a is a s r in ra a s 2 m a n s If we uppo e th t the ngle me u ed di n , ( ) e tha t x is the n u mber of ra dia ns in a n a ngle (o r a re) who se sin e ” is is r a a rc s n o r a n a s s is y; it e d i e y ngle who e ine y. Likewise a rcco s y den otes a n a ngle who se cosin e is y a ro ta n n who se a n n t y denotes a n a gle t ge is y.
r s s s sin 55 a: a r s a re a s s The exp e ion y , c in y, two pect of
“ r a s a s a re s a s A is u one el tion , j u t the two t tement the ncle
” “ ” of B a n d B is the nephew of A ; either one implie s the o r o in the ; b th mea n the sa m e th g .
As s s a rcsin e unction a n d in a r u a r we wi h to tudy the f , p tic l to
a r s is a n d s a r comp e it with the ine function , it convenient cu tom y
n k a s n sa a r a <5 a n d r to thi of it depe ding on the me v i ble , w ite
3 : a r s n x t. e. a: sin . ( ) y c i , [ y]
a 3 is a r 1 b s s a so lv We note th t ( ) obt ined f om ( ) y two tep , ( ) ing (1) fo r £5 ; a n d (b) in tercha nging m a n d y in Two func tion s so rela ted tha t ea ch ca n be obta ined from the other in this m a nn er a re ca lled invers e functions ; ea ch is the inverse
r of the othe .
sa s n s co s x a nd a r s w ta n a: a n d In the me e e, y y cco ; y y= a rcta n 55 ; y : sec 55 a n d y : a rcs ec 55 ; y vers x a n d y
r r r r m ction s . a cve s 55 . a e s fl ; etc , inve e
ra h ca l Re res en a tion of Inverse unctions . n 84 . G p i p t F Si ce the equa tion s (1 ) y sin 90 a n d (2) a; a rcsin y
— ' The n o tatio n sin 1 y also is used very frequently to den o te a rcsin y ; it is - 1 - 1 necessary to n o tice carefully that sin y do es no t mean (sin y) .
PL E R G N M ETR XI I 34 1 16 AN T I O O Y [ ,
i r o th e a o f co s :5 F i . 9 5 we Sim la ly fr m gr ph , g , derive the
' i 9 7 a n d in sa e wa ra s g ra ph of a rcco s a: in F g . ; the m y the g ph
a c a x a cs ec x a rcctn x a r s w a rcvers 55 ca n a of r t n , r , , cc c , , be dr wn
o s ta n 55 sec 93 w csc w ers 93. from th e of , , ctn , , v
96 . . 9 7 . FIG . FIG
EXE ISES — NVER C IO S RC XXXI . I SE FUN T N
2 “ Draw the graph o f y arcsin a: as in 8 .
f o 82 F Draw the graph o y arcc s cc as in . ° f 5 Draw the graph o y arctan (5 . s 9 Draw the graph o f y arc ec (5 . II 84 IN ERSE F NC I NS X , ] V U T O
98 . FIG .
99 . FIG .
14 DAY USE RETURN TO DESK FRO M WHICH BO RR O WED l O AN DEPT
Th k i u n a e is bo o s d e o the last d te sta mp d belo w , o r
o n the da te to which renewed.
R e ks a r s a enew d bo o e ubject to immedi te reca ll .
LD 2 1 A—5 0 m ( A1 7 2 4 8 1 0 ) 4 7 6 B