DOCSLIB.ORG

Solutions of Triangle

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Solutions of Triangle 19. SOLUTIONS OF TRIANGLE 1. INTRODUCTION A In any triangle ABC, the side BC, opposite to the angle A is denoted by a; the side CA and AB, opposite to the angles B and C respectively are denoted by b and c respectively. The c b semi-perimeter of the triangle is denoted by s and its area by ∆ or S. In this chapter, we shall discuss various relations between the sides a, b, c and the angles A, B, C of ∆ ABC. B a C 2. SINE RULE Figure 19.1 The sides of a triangle (any type of triangle) are proportional to the sines of the angle opposite to them in triangle abc ABC, = = sinA sinB sinC sinA sinB sinC Note: (i) The above rule can also be written as = = abc (ii) The sine rule is a very useful tool to express the sides of a triangle in terms of sines of the angle and vice-versa abc in the following manner: = = = k( Let) ; ⇒=a k sinA, b= k sinB, c= k sinC sinA sinB sinC sinA sinB sinC Similarly, = = = λ(Let) ; ⇒=λsinA a , sinB= λ b , sinC= λ c abc 3. COSINE RULE bca222+− cab222+− abc2+− 22 In any ∆ABC , cosA = ; cosB = ; cosC = 2bc 2ac 2ab Note: In particular ∠=A 60 ⇒ b222+−= c a bc ∠=B 60 ⇒ c222+− a b = ca ∠=C 60 ⇒ a2+−= b 22 c ab Figure 19.2 4. PROJECTION FORMULAE If any ∆ABC :(i) a= bcosC + ccosB (ii) b= ccosA + acosC (iii) c= acosB + bcosA i.e. any side of a triangle is equal to the sum of the projection of the other two sides on it. 19.2 | Solutions of Triangle Case I: When ∆ABC is an acute angled triangle, BD cosB=⇒= BD AB.cosB ⇒= BD c.cosB and AB CD cosC=⇒= CD AC.cosC ⇒= CD bcosC AC then, BD+DC=BC Figure 19.3 ∴ a = c cos B + b cos C Case II: When ∆ABC is an obtuse angled triangle, CD cosC=⇒= CD AC.cosC AC BD CD= b.cosC and cos( 180−= B) ⇒ BD =− c.cosB then, AB a = BC and CD – BD ⇒ a= bcosC + ccosB Figure 19.4 Illustration 1: If A= 75ο , B= 45ο , then what is the value of b+ c2? (JEE MAIN) Sol: Here, c=−= 180ο 120 οο 60 . Therefore by using sine rule, we can solve the above problem. a= k sin 75o abc Use sine rule = = = K ⇒=b k sin 45o sin75οοο sin45 sin60 c= k sin 60o 31++ 31 consider, (b+= c 2) k( sin45ο + 2 sin600 ) =k = 2k =2k sin75ο = 2k sinA = 2a 2 22 Illustration 2: In a ∆ ABC, if B= 30ο and c= 3b , then find the value of A. (JEE MAIN) cab22+− 2 Sol: Here, by using cosine rule cosB = we can easily solve the above problem. 2ca cab222+− 33bab222+− We have cosB = ⇒= ; ⇒a22 − 3ab + 2b =⇒− 0 (a 2b)(a − b) = 0 2ca 2 2×× 3b a ⇒−=ab 0 OR a – 2b = 0 ⇒Eithera =⇒= b A 30ο or aa== 2b 2b ⇒ ⇒ a a22 ⇒ ⇒ 4b 4b 2 2⇒ = = b b 22 22 + + c c ⇒= ⇒= A A 90 90οο . Illustration 3: Prove that asin B−+ C bsin(C −+ A) csin A −= B 0 (JEE MAIN) ( ) ( ) abc Sol: By sine rule i.e. = = = k we can simply prove the above equation. sinA sinB sinC In equation asin( B−+ C) bsin(C −+ A) csin( A − B) = 0, putting a = k sin A, b = k sin B, c = k sin C = k( sinAsin( B−+ C) sinBsin( C −+ A) sinCsin( A − B)) =0 (expanding all terms gets cancelled) (Using sin (α−β ) = sin α . cos β− sin β cos α) bc22− Illustration 4: Prove that sin( B−= C) sinA (JEE MAIN) a2 Mathematics | 19.3 bc22− Sol: Given, sin(B−= C) sinA ⇒ a2 sin(B−= C)( b22 − c) sinA a2 Takin L.H.S., a2sin(B − C) = a2(sinBcosC − cosBsinC) sinA sinB sinC acb22+− 2 abc2+− 22 Now using sine rule, = = = k (say) and cosine rule, cosB = , and cosC = abc 2 ac 2 ab abc2+− 22 acb 222 +− abcacb2+−−+− 22222 =a2 kb − ×=kc ka =sinA ×−= b22 c RHS. ( ) 2 ab 2 ac 2 Illustration 5: The angles of a triangle are in 4:1:1 ratio. Find the ratio between its greater side and perimeter? (JEE ADVANCED) Sol: Here, the angles are 120ο , 30ο , 30ο . Therefore, by using sine rule, we will get the required ratio. Angles are 120ο , 30ο , 30ο . If the sides opposite to these angles are a, b and c respectively, a will be the greatest side. a bc a bc a bc Now from sine formula, = = ; ⇒==; ⇒===k (say) sin120οοο sin30 sin30 3/2 1/2 1/2 3 11 3k 3 then a= 3k , perimeter= (2+ 3k) ; ∴ Required ratio = = . (2+ 3k) 23+ CB Illustration 6: Solve bcos22+ ccos in term of k where k is perimeter of the ∆ABC . (JEE ADVANCED) 22 1+θ cos 2 Sol: We can solve the given problem simply by cos2 θ= 2 CBbc Here, bcos22+ ccos =(1 + cosC) ++ (1 cosB) [using projection formula] 2222 bc+ 1 ab ++ c C BKk = +=a ; ∴+=bcos22 ccos [where k=a+b+c, given] 22 2 2 222 AB− tan ab− 2 Illustration 7: In any triangle ABC, show that = (JEE ADVANCED) ab+ AB+ tan 2 Sol: We can derive the values of a, b and c using sine rule and putting it to L.H.S. we can prove the above problem. abc We know that, = = = k sinA sinB sinC ⇒=a k sinA , b= ksinB , c= k sinC …(i) On putting the values of a and b from (1) on L.H.S., we get AB+− AB 2cos sin a−− b k sinA k sinB sinA − sinB L.H.S. = = = = 22 a++ b k sinA k sinB sinA + sinB AB+− AB 2sin cos 22 AB− tan A+− B AB 2 = cot tan = = R.H.S. 22 AB+ tan 2 19.4 | Solutions of Triangle 5. NAPIER’S ANALOGY (LAW OF TANGENTS) BC−− bc A AB−− ab C CA−− ca B In any ∆ABC, (i) tan = cot (ii) tan = cot (iii) tan = cot 2 bc+ 2 2 ab+ 2 2 ca+ 2 bc− A k sinB− k sinc A sinB− sinc A Proof:(i) R.H.S. = cot= cot [By the sine rule] = cot bc+ 2 k sinB+ k sinc 2 sinB+ sinc 2 BC−+ B C 2sin cos 22 A = .cot[By C & D formulae] BC+− B C 2 2sin cos 22 BC−+ BC A BC− A A = tan .cot ⇒ cot = tan .tan .cot [By condition identities] 22 22 22 BC− = tan= LHS 2 Similarly, (ii) and (iii) can be proved. Illustration 8: In any triangle ABC, if A= 30ο , b=3 and c= 33, then find ∠B and ∠C . (JEE MAIN) CB−− cb A Sol: By using formula, tan = cot , we can easily obtain the values of ∠B and ∠C . 2 cb+ 2 BC+ A Here ∠=A 30ο ∴ =90ο −= 90 − 15οο = 75 … (i) 22 Since c> b ⇒∠ C >∠ B and B + C = 150ο … (ii) CB−− cb A cb − BC + C−− B 33 3 ο tan = cot= tan; ⇒=tan tan75 2 cb++ 2 cb 2 2 3( 31+ ) CB−−π cb cb − A ⇒tan =cot −= A tan 2 cb++ 2 cb 2 CB−+3( 31−−) ( 31) tan45οο tan30 [Using (1) ] ⇒tan = tan 45οο+= 30 ( ) οο 2 3( 31++) ( 31) 1− tan45 tan30 1 31− 1 + ( ) 31−+ 31 CB− ο = 3 = = ⇒=45 tan45ο = 1 1 31+ 1 31+− 31 ; 2 ; ( ) 1 − 3 ⇒−=C B 90ο … (iii) Solving (ii) and (iii), we get ∠=B 30ο and ∠=C 120ο 6. TRIGONOMETRIC RATIOS OF HALF ANGLES Sine, cosine and tangent of half the angles of any triangle are related to their sides as given below. Note that the perimeter of ∆ABC will be denoted by 2s i.e. 2s=a+b+c and the area denoted by ∆ . Mathematics | 19.5 A B C Formulae for sin, sin, sin for any ∆ABC 2 2 2 A (sbsc−−)( ) B (scsa−−)( ) C (sasb−−)( ) (i) sin= (ii) sin= (iii) sin= 2 bc 2 ac 2 ab A B C Formulae for cos, cos, cos for any ∆ABC 2 2 2 A ss( − a) B ss( − b) C ss( − c) (i) cos= (ii) cos= (iii) cos= 2 bc 2 ac 2 ab A B C Formulae for tan, tan, tan for any ∆ABC 2 2 2 A (sbsc−−)( ) B (scsa−−)( ) C (sasb−−)( ) (i) tan= (ii) tan= (iii) tan= 2 ss( − a) 2 ss( − b) 2 ss( − c) sa−−− sb sc Illustration 9: In a triangle ABC, if = = , then find the value of tan2 (A/ 2) . (JEE MAIN) 11 12 13 A (sbsc−−)( ) Sol: As we know, tan= . Therefore, by using this formula, we can solve the above problem. 2 ss( − a) −− sa−−− sb sc 3s−( a ++ b c) s 2 A (sbsc)( ) 12× 13 13 = = = = ; Now tan = = = 11 12 13 11++ 12 13 36 2 ss( − a) 36× 11 33 AB C Illustration 10: In a triangle ABC, prove that (a++ b c) tan + tan = 2ccot . (JEE MAIN) 22 2 A (sbsc−−)( ) B (scsa−−)( ) Sol: Here by using tan= and tan= we can prove the above problem. 2 ss( − a) 2 ss( − b) AB (sbsc−−)( ) ( scsa −−)( ) sc−− sb sa − L.H.S. = (a++ b c) tan + tan =2s +=2s + 22 ss−− a ss b s sa−− sb ( ) ( ) (sc− ) sbsa−+− 2ssc− ss( − c) 2c C =2s = (a++−− b c b a) = 2c = =2ccot = R.H.S. s sasb−− C 2 s−− as b s −− as b ( )( ) tan 2 ∆∆ 1 1 2∆( 2s −− a b) 2c∆2 2cs( s− c) Alternate: L.H.S. = 2s + =∆+=2 = = −− ∆ ssa( −) ssb( −) sa sb (sasb−−)( ) ( sasb −−∆)( ) ∆∆ 1 1 2∆( 2s −− a b) 2c∆2 2cs( s− c) c 2s + =∆+=2 = = = 2c.cot = R.H.S.
Load more

Recommended publications
  • Geodesy. Application of the Theory of Least Squares to the Adjustment Of
    SI No Serial ^) DEPARTME]S'T OF COMMERCE 0. S. COAST AND GEODETIC SURVEY it E. LESTER JONES, Supei^intendent GEODESY APPLICATION OF THE THEORY OF LEAST SQUARES TO THE ADJUSTMENT OF TRIANGULATION RY OSCAR S. ADAM^ COMPUTER UNITED STATES COAST AND GEODETIC SURJ^Y No. 28 Special Publication { WASHINGTON GOVERNMENT I'RINTING OFFICE 1915 Serial No. 9 DEPARTMEISTT OF COMMERCE U. S. COAST AND GEODETIC SURVEY E. LESTER JONES, Superintendent GEODESY APPLICATION OF THE THEORY OF LEAST SQUARES TO THE ADJUSTMENT OF TRIANGULATION BY OSCAR S. ADAMS COMPUTER UNITED STATES COAST AND GEODETIC SURVEY Special Publication No. 28 WASHINGTON GOVERNMENT PRINTING OFFICE 1815 ADDITIONAL COPIES OF THIS PUBLICATION MAY BE PROCURED FROM THE SUPERINTENDENT OF DOCUMENTS GOVERNMENT PRINTING OFFICE WASHINGTON, D. C. AT 25 CENTS PER COPY QB 32/ CONTENTS Page. General statement 7 Station adjustment 7 Observed angles 8 List of directions 8 Condition equations , 9 Formation of normal equations by differentiation 9 Correlate equations 11 Formation of normal equations 11 Normal equations 12 Discussion of method of solution of normal equations 12 Solution of normal equations 13 Back solution 13 Computation of corrections 13 Adjustment of a quadrilateral 14 General statement 14 Lists of directions 16 Figure 16 Angle equations 17 Side equation 17 Formation of normal equations by differentiation 17 Correlate equations 18 Normal equations 18 Solution of normal equations 19 Back solution 19 Computation of corrections 19 Adjustment of a quadrilateral by the use of two angle and two
    [Show full text]
  • Hyperbolic Trigonometry in Two-Dimensional Space-Time Geometry
    Hyperbolic trigonometry in two-dimensional space-time geometry F. Catoni, R. Cannata, V. Catoni, P. Zampetti ENEA; Centro Ricerche Casaccia; Via Anguillarese, 301; 00060 S.Maria di Galeria; Roma; Italy January 22, 2003 Summary.- By analogy with complex numbers, a system of hyperbolic numbers can be intro- duced in the same way: z = x + hy; h2 = 1 x, y R . As complex numbers are linked to the { ∈ } Euclidean geometry, so this system of numbers is linked to the pseudo-Euclidean plane geometry (space-time geometry). In this paper we will show how this system of numbers allows, by means of a Cartesian representa- tion, an operative definition of hyperbolic functions using the invariance respect to special relativity Lorentz group. From this definition, by using elementary mathematics and an Euclidean approach, it is straightforward to formalise the pseudo-Euclidean trigonometry in the Cartesian plane with the same coherence as the Euclidean trigonometry. PACS 03 30 - Special Relativity PACS 02.20. Hj - Classical groups and Geometries 1 Introduction Complex numbers are strictly related to the Euclidean geometry: indeed their invariant (the module) arXiv:math-ph/0508011v1 3 Aug 2005 is the same as the Pythagoric distance (Euclidean invariant) and their unimodular multiplicative group is the Euclidean rotation group. It is well known that these properties allow to use complex numbers for representing plane vectors. In the same way hyperbolic numbers, an extension of complex numbers [1, 2] defined as z = x + hy; h2 =1 x, y R , { ∈ } are strictly related to space-time geometry [2, 3, 4]. Indeed their square module given by1 z 2 = 2 2 | | zz˜ x y is the Lorentz invariant of two dimensional special relativity, and their unimodular multiplicative≡ − group is the special relativity Lorentz group [2].
    [Show full text]
  • 7 TRIGONOMETRY Objectives in This Chapter You Will Learn About • Revision of Solution of Triangles
    7 TRIGONOMETRY Objectives In this chapter you will learn about • Revision of solution of triangles. • Solving problems in 2–dimensions and in 3-dimensions by constructing and interpreting models. 1 | P a g e 7.1 Revision In previous grades, we dealt with the types of triangles named according to side and angles. In trigonometry, we are able to find the unknown sides and angles when given the magnitudes of the minimum required number of angles and sides. We looked at two main types of triangles: • Right angled triangles. • Non-right angled triangles Right angled triangles When a right angled triangle is given, the following may be useful: • Pythagoras theorem • Definitions of Trigonometric ratios Non-right angled triangles • Sine rule • Cosine rule • Area rule NOTE: • Not all triangles that you need to solve are right-angled triangles. In grade 11 you learned about rules which can be used to solve non-right-angled triangles. • Theorems/axioms involving lines, triangles, quadrilateral and circles are also useful. 7.1.1 The Sine Rule The sine rule expresses the relationship between the two of the sides of the triangles and two of the angles. In any triangle, a is the side opposite angle A, b is the side opposite angle B and c is the side opposite angle C, then the lengthof any side the length of any other side = the sine of the angle opposite that side the sine of the angle opposite that side That is: 2 | P a g e In ABC, a b c = = sin A sinB sinC This can also be written as: sin A sinB sinC = = a b c The sine rule can be used when the following information about a triangle is given: • Two sides and an angle opposite to one of the two sides • One side and any two angles Worked Example 1 Solve the triangle given below.
    [Show full text]
  • 4 Trigonometry and Complex Numbers
    4 Trigonometry and Complex Numbers Trigonometry developed from the study of triangles, particularly right triangles, and the relations between the lengths of their sides and the sizes of their angles. The trigono- metric functions that measure the relationships between the sides of similar triangles have far-reaching applications that extend far beyond their use in the study of triangles. Complex numbers were developed, in part, because they complete, in a useful and ele- gant fashion, the study of the solutions of polynomial equations. Complex numbers are useful not only in mathematics, but in the other sciences as well. Trigonometry Most of the trigonometric computations in this chapter use six basic trigonometric func- tions The two fundamental trigonometric functions, sine and cosine, can be de¿ned in terms of the unit circle—the set of points in the Euclidean plane of distance one from the origin. A point on this circle has coordinates +frv w> vlq w,,wherew is a measure (in radians) of the angle at the origin between the positive {-axis and the ray from the ori- gin through the point measured in the counterclockwise direction. The other four basic trigonometric functions can be de¿ned in terms of these two—namely, vlq { 4 wdq { @ vhf { @ frv { frv { frv { 4 frw { @ fvf { @ vlq { vlq { 3 ?w? For 5 , these functions can be found as a ratio of certain sides of a right triangle that has one angle of radian measure w. Trigonometric Functions The symbols used for the six basic trigonometric functions—vlq, frv, wdq, frw, vhf, fvf—are abbreviations for the words cosine, sine, tangent, cotangent, secant, and cose- cant, respectively.You can enter these trigonometric functions and many other functions either from the keyboard in mathematics mode or from the dialog box that drops down when you click or choose Insert + Math Name.
    [Show full text]
  • A Treatise on Spherical Trigonometry, and Its Application
    A T REATISE SPHEEICAL T RIGONOMETKY. WORKSY B JOHN G ASEY, ESQ., LLD., F. R.8., FELLOWF O THE ROYAL UNIVERSITY OF IRELAND. Second E dition, Price 3s. A T REATISE ON ELEMENTARY TRIGONOMETRY, With n umerous Examples AND E&uesttfms f or Exammatttm. OKEY T THE EXERCISES IN THE TREATISE ON ELEMENTARY TRIGONOMETRY. Fifth E dition, Revised and Enlarged, Price 3s. 6d., Cloth. A S EQ,UEL TO THE FIRST SIX BOOKS OF THE ELEMENTS OF EUCLID, Containing- a n Easy Introduction to Modern Geometry . With a umerstts Examples. Seventh E dition, Price 4s. 6d. ; or in two parts, each 2s. 6d. THE E LEMENTS OE EUCLID, BOOKS I.-YL, AND PROPOSITIONS I.-XXI. OP BOOK XI. ; Together ' with an Appendix on the Cylinder, Sphere, Cone, &>c. @opdatt$ J tettotatiotts & aumeraus Exercises. Second E dition, Price 6s. AEY K TO THE EXERCISES IN THE FIRST SIX BOOKS OF CASEY'S "ELEMENTS OF EUCLID." Price 7 s. 6d. A T REATISE ON THE ANALYTICAL GEOMETRY OF THE POINT, LINE, CIRCLE, & CONIC SECTIONS, Containing a n Account of its most recent Extensions, Wxih t tttmeratts Examples. Price 7 s. 6d. A T REATISE ON PLANE TRIGONOMETRY, including THE THEORY OF HYPERBOLIC FUNCTIONS. LONDON : L ONGMANS? CO. DUBLIN: HODGES, FTGGIS & CO. A T REATISE ON- SPHERICAL T RIGONOMETRY, ANDTS I APPLICATION- TO GEODESYND A ASTRONOMY, ■WITH BY JOHN C ASEY, LL.D., F.E.S., Fellowf o the Royal University of Ireland; Memberf o the Council of the Royal Irish Academy ; Memberf o the Mathematical Societies of London and France ; Corresponding M ember of the Royal Society of Sciences of Liege; and Professorf o the Higher Mathematics and Mathematical Physics in t he Catholic University of Ireland.
    [Show full text]
  • Trigonometry for the Solution of Triangles
    Trigonometry for the Solution of Triangles _________________________ name Trigonometry was originally a branch of geometry. Its name means triangle measure (from trigos + metry). If two angles of a triangle have measures which are known, then the third angle can be calculated by knowing that the three angles have a sum of 180o. This does not require any knowledge of trigonometry. mCA B = 38.09 m ACB = 75.23 mFDE = 33.75 F mGEF = 102.25 C A D E B G So m < B = ___________ and m < F = ______________ If any two sides of a right triangle have lengths which are known, then the length of the third side requires the use of the Pythagorean Theorem. This also does not require any knowledge of trigonometry. J K M 3 12.4 45.1 H L 8 I h2 + j2 = i2 m2 + l2 = k2 32 + 82 = i2 12.42 + l2 = 45.12 so i2 = 73 so l2 = 45.12 – 12.42 i = 73 8.544 l = 1880.25 43.362 When there is a combination of information about angles and sides, then there is a need for additional knowledge. This is where trigonometry comes in. Trigonometry and Triangles page 2 To see how the ideas of trigonometry are developed, we use the ideas of the ratios of the three sides of right triangle. (Later, an extension of these ideas allows trigonometry to be used on triangles which do not have right angles.) Using the ideas about similar triangles (which are triangles which have the same shapes but different sizes), F G H I E D C B A all of the right triangles are similar.
    [Show full text]
  • MASTER COURSE OUTLINE Prepared By: Date: November 14, 2018
    MASTER COURSE OUTLINE Prepared By: Date: November 14, 2018 COURSE TITLE Precalculus II GENERAL COURSE INFORMATION Dept.: MATH& Course Num: 142 (Formerly: MATH 152) CIP Code: 27.0102 Intent Code: 11 Program Code: Credits: 5 Total Contact Hrs Per Qtr.: 55 Lecture Hrs: 55 Lab Hrs: 0 Other Hrs: 0 Distribution Designation: Math Science MS, Symbolic or Quantitative Reasoning SQR COURSE DESCRIPTION (as it will appear in the catalog) In preparation for calculus this is a comprehensive study of trigonometry, circular functions, right triangle trigonometry, analytical trigonometry. Sequences, series and induction are also covered. PREREQUISITES MATH& 141 or concurrent enrollment in MATH& 141 TEXTBOOK GUIDELINES College level text or worksheets at the discretion of the instructor COURSE LEARNING OUTCOMES Upon successful completion of the course, students should be able to demonstrate the following knowledge or skills: 1. Apply trigonometric functions to the solution of triangles and application problems 2. Manipulate trigonometric functions to prove identities and solve equations 3. Employ summations, sequences, and series in mathematical induction INSTITUTIONAL OUTCOMES IO2 Quantitative Reasoning: Students will be able to reason mathematically. COURSE CONTENT OUTLINE 1. Trigonometry a. Angles and their measure b. Right triangles and trigonometric functions c. Trigonometric functions of real numbers d. Graphs of sine and cosine e. Graphs of other trigonometric functions f. Additional graphing techniques g. Applications of trigonometry 2. Analytic Trigonometry a. Applications of the fundamental identities b. Verifying trigonometric identities c. Solving trigonometric equations d. Sum and difference formulas e. Multiple-angle formulas and product-sum formulas 3. Additional applications of trigonometry a. Law of sines and cosines b.
    [Show full text]
  • MAT 119 Trigonometry
    MISSOURI WESTERN STATE UNIVERSITY COLLEGE OF LIBERAL ARTS AND SCIENCES DEPARTMENT OF COMPUTER SCIENCE, MATHEMATICS, AND PHYSICS COURSE NUMBER: MAT 119 COURSE NAME: Trigonometry COURSE DESCRIPTION: The Trigonometry course includes the degree and radian measure of angles; the definition of the six trigonometric functions; the solution of triangles and appropriate application problems; the graphing of the circular functions; the solution of conditional trigonometric equations; the verification and usefulness of identities; the inverse trigonometric functions; the trigonometric form of a complex number and polar curves. Offered Fall and Spring semesters. PREREQUISITE: ACT math score of at least 20 or the equivalent. (Not open to the student with credit in MAT 177) TEXTBOOK: TECHNOLOGY: The use of a graphing calculator will be required throughout the course and each student must have his own calculator. The calculator must have at least the capability of the TI 83. (The TI 83 or TI 83+ is recommended.) While graphing calculators other than the TI 83 may be used, classroom instruction will be given only for the TI 83. COURSE OBJECTIVES: The student will learn to: 1. Measure angles in degrees and radians and convert from one type of measure to the other. 2. Use trigonometry to solve right and oblique triangles. 3. Use a calculator to evaluate trigonometric functions of angles. Last updated 9-2016 4. Find trigonometric functions of real numbers. 5. Graph a trigonometric equation of the form y = af[b(x – c)] + d. 6. Graph algebraic/trigonometric functions by addition of ordinates. 7. Solve conditional trigonometric equations. 8. Recognize and verify trigonometric identities.
    [Show full text]
  • Chapter 3 CNC Math 53 54 CNC Machining N60G01X3.25F.002 N70G04X0.5 N80X3.35F.05 N90G00X5.0Z0T0101 U Bisect
    This sample chapter is for review purposes only. Copyright © The Goodheart-Willcox Co., Inc. All rights reserved. N10G20G99G40 N20G96S800M3 N30G50S4000 N40T0100M8 N50G00X3.35Z1.25T0101 Chapter 3 CNC Math 53 54 CNC Machining N60G01X3.25F.002 N70G04X0.5 N80X3.35F.05 N90G00X5.0Z0T0101 u Bisect. To divide into two equal parts. 01111 N10G20G99G40 u Congruent. Having the same size and shape. N20G96S800M3 N30G50S4000 u Diagonal. Running from one corner of a four-sided figure to the N40T0100M8 N50G00X3.35Z1.25T0101 opposite corner. N60G01X3.25F.002 N70G04X0.5 u Parallel. Lying in the same direction but always the same distance apart. N80X3.35F.05 Chapter 3 u Perpendicular. At a right angle to a line or surface. u Segment. That part of a straight line included between two points. CNC Math u Tangent. A line contacting a circle at one point. u Transversal. A line that intersects two or more lines. Angles An angle () is the figure formed by the meeting of two lines at the Objectives same point or origin called the vertex. See Figure 3-1. Angles are measured Information in this chapter will enable you to: in degrees (n), minutes (`), and seconds (p). A degree is equal to 1/360 of a circle, a minute is equal to 1/60 of 1n, and a second is equal to 1/60 of 1`. u Identify various geometric shapes. There are many types of angles, Figure 3-2. An acute angle is greater u Apply various geometric principles to solve problems. than 0n and less than 90n. An obtuse angle is greater than 90n and less u Solve right triangle unknowns.
    [Show full text]
  • Chapter 5 Solution of Triangles
    Available online @ https://mathbaba.com Chapter 5 114 Solution of triangle Chapter 5 Solution of Triangles 5.1 Solution of Triangles: A triangle has six parts in which three angles usually denoted by ,, and the three sides opposite to denoted by a, b, c respectively. These are called the elements of Triangle. If any three out of six elements at least one side are given them the remaining three elements can be determined by the use of trigonometric functions and their tables. This process of finding the elements of triangle is called the solution of the triangle. First we discuss the solution of right angled triangles i.e. triangles which have one angle given equal to a right angle. In solving right angled triangle denotes the right angle. We shall use the following cases B Case-I: When the hypotenuse and one Side is given. c β Let a & c be the given side and a hypotenuse respectively. Then angle can be found by the relation. γ Sin = A b C Fig.5.1 Also angle β and side “b” can be obtained by the relations β= 90o and Cos = B Case-II: β c β When the two sides a and b are a given. Here we use the following relations to find , β & c. λ λγ Tan = , β = 90o , A b C Fig.7.25.2 c = ab22 Case-III: When an angle and one of the sides b , is given. The sides a , c and β are found Available online @ https://mathbaba.com Available online @ https://mathbaba.com Chapter 5 115 Solution of triangle from the following relations.
    [Show full text]
  • A History of Trigonometry Education in the United States: 1776-1900
    A History of Trigonometry Education in the United States: 1776-1900 Jenna Van Sickle Submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy under the Executive Committee of the Graduate School of Arts and Sciences COLUMBIA UNIVERSITY 2011 ©2011 Jenna Van Sickle All Rights Reserved ABSTRACT A History of Trigonometry Education in the United States: 1776-1900 Jenna Van Sickle This dissertation traces the history of the teaching of elementary trigonometry in United States colleges and universities from 1776 to 1900. This study analyzes textbooks from the eighteenth and nineteenth centuries, reviews in contemporary periodicals, course catalogs, and secondary sources. Elementary trigonometry was a topic of study in colleges throughout this time period, but the way in which trigonometry was taught and defined changed drastically, as did the scope and focus of the subject. Because of advances in analytic trigonometry by Leonhard Euler and others in the seventeenth and eighteenth centuries, the trigonometric functions came to be defined as ratios, rather than as line segments. This change came to elementary trigonometry textbooks beginning in antebellum America and the ratios came to define trigonometric functions in elementary trigonometry textbooks by the end of the nineteenth century. During this time period, elementary trigonometry textbooks grew to have a much more comprehensive treatment of the subject and considered trigonometric functions in many different ways. In the late eighteenth century, trigonometry was taught as a topic in a larger mathematics course and was used only to solve triangles for applications in surveying and navigation. Textbooks contained few pedagogical tools and only the most basic of trigonometric formulas.
    [Show full text]
  • ED053927.Pdf
    DOCUMENT RESUME ED 053 927 SE 011 332 AUTHOR Brant, Vincent TITLE Trigonometry, A Tentative Guide Prepared for Use with the Text Plane Trigonometry with Tables. INSTITUTION Baltimore County Public Schools, Towson, Md. PUB DATE 65 NOTE 99p. AVAILABLE FROM Baltimore County Public Schools, Office of Curriculum Development, Towson, Maryland 21204 ($2.00) EDRS PRICE EDRS Price MF-$0.65 HC-$3.29 DESCRIPTORS Curriculum, *Curriculum Guides, Instruction, Mathematics, *Secondary School Mathematics, *Teaching Guides, *Trigonometry ABSTRACT This teacher's guide for a semester course in trigonometry is prepared for use with the text "Plane Trigonometry with Tables" by E. R. Heineman. Included is a daily schedule of topics for discussion and homework assignments. The scope of each lesson and teaching suggestions are provided. The content for the course includes trigonometric functions, solution of right triangles, trigonometric equations and identities, oblique triangles, and inverse trigonometric functions. Also included are two supplementary units on special right triangles and set theory. (Author/CT) U.S. DEPARTMENT OF HEALTH. EDUCATION & WELFARE OFFICE OF EDUCATION THIS DOCUMENT HAS BEEN REPRO. DUCED EXACTLY AS RECEIVED FROM THE PERSON OR ORGANIZATION ORIG . INATING IT POINTS OF VIEW OR OPIN IONS STATED DO NOT NECESSARILY REPRESENT OFFICIAL OFFICE OF EDU CATION POSITION OR POLICY CURRICULUM GUIDE TRIGONOMETRY A Tentative Guide Prepared for Use with the Text Plane Trigonometry with Tables by E. R. Heineman Guide Produced by Vincent Brant Supervisor of Senior High School Mathematics William S. Sartorius, Superintendent Tow son, Maryland 1965 2 BOARD OF EDUCATION OF BALTIMORE COUNTY Aigbu.rth Manor Towson, Maryland 21204 T. Bayard Williams, Jr.
    [Show full text]