UNIT 6 SPHERICAL TRIGONOMETRY Trigonometry

Spherical UNIT 6 SPHERICAL TRIGONOMETRY Trigonometry

Structure 6.1 Introduction Objectives 6.2 Definitions 6.3 Formulae for a Spherical Triangle 6.3.1 Cosine Formula 6.3.2 Sine Formula 6.3.3 Supplemental Cosine Formula 6.4 Latitudes and Longitudes 6.5 The Haversine Formula 6.6 Cotangent Formula or Four Consecutive Parts Formula 6.7 Right-angled Spherical Triangles 6.7.1 Formulae for a Right-angled Triangle 6.7.2 Napier’s Rules 6.8 Quadrantal Spherical Triangles 6.9 Polar Triangle 6.10 Summary 6.11 Answers to SAQs 6.1 INTRODUCTION

Originally trigonometry was conceived only to deal with the measurement of triangles but this branch of mathematics later included the treatment of all angular measurements and calculations. Plane trigonometry deals with such relationships when all the angles and the lines lie on a plane. The spherical trigonometry on the other hand, deals with these relationships when the angles and the lines lie on the surface of the sphere. Spherical trigonometry is thus concerned with the sphere and with the spherical triangles. We have already seen in the earlier units that a straight line drawn from the centre of the sphere to its surface is called its radius and the straight line drawn through the centre and terminated both ways on the surface is called its diameter. Objectives After studying this unit, you should be able to • distinguish between plane trigonometry and spherical trigonometry, • define and explain, great and small circles, axis and poles of great circles, • define spherical angle and also the angle of intersection between two great circles, • discuss and derive cosine, sine and supplemental cosine formulae and the limitations of their application, • explain the Haversine formula, discuss its advantage over sine and cosine formulae as also its application to problems in spherical triangles, • define right-angled spherical triangle and discuss its properties,

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Mathematics-I • define quadrantal spherical triangles and their use in the solution of problems, • explain Nepier’s rule for right-angled triangles and quadrantal triangles and its working rule, and • explain polar triangles and their relationship with the primitive spherical triangles and their use in the solution of problem in spherical trigonometry. 6.2 DEFINITIONS

Great and Small Circles When a plane cuts a sphere the section thus formed is called a circle and when the plane passes through the centre of the sphere the circle is defined as a great circle. If the plane does not pass through the centre of the sphere, the circle of the section is known as a small circle. A great circles divides the sphere into two identical parts, each called a hemisphere. Axis and Poles The extremities of the diameter perpendicular to a section of the sphere are called the poles of that section while the diameter itself is known as the axis of that section. The pole of a circle is a point on the surface of the sphere which is equidistant from all points on the circle. P is also the pole of the great circle LL′, and all great circles as is clear, have two poles. The arc PA of the great circle PAP′ is called the spherical radius of the circle QQ′ (Figure 6.1). P

Q r O Q’

L C L’ B

P’ Figure 6.1 The Diameter and the Lune The intersection of the planes of two great circles is a diameter of the sphere; and these two great circles cut off on the surface of a sphere two pairs of congruent areas, which are known as Lunes. Thus a lune is a portion of the surface of a sphere enclosed by two great circles. For instance in Figure 6.1, PQLP′AP is a lune and ∠ QPA is the angle of the lune. Angle between Two Great Circles (a) The angle between the two great circles is the angle between the tangents

drawn to the two great circles at their point of intersection. Thus ∠ TAT1 is the angle between the two great circles at the point A, which is the angle of

their intersection. Clearly ∠ TAT1 is the angle between the planes, i.e. ∠ EOB (Figure 6.2). Also ∠ EOB = ∠ LOD. This angle is defined as spherical angle.

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D L Spherical Trigonometry

O E θ

B T A T Figure 6.2 : Spherical Angle 1 (b) The angle between any two great circles is measured by the arc they subtend on the great circle to which they are perpendicular, i.e. the angle between the planes QPQ′ and QAQ′ is equal to the arc AP of the great circle PAP′ to which QOQ′ is perpendicular (Figure 6.3). Q

P O P ’

Q ’ Figure 6.3 (c) If the two great circles have their poles at the points L and L′, the angle of intersection of the two great circles will be equal to the angle subtended at the centre of the sphere by the arc joining the poles L and L′ of the great circles. In other words this is the angle subtended at O by the arc of a great circle joining L and L′. This is also equal to the inclination of the planes of the two great circles AA′ and BB′ (Figure 6.4). L L’

A’ A O B’

Figure 6.4 Length of Small Circle Arc Consider a small circle arc BD of the circle BDB′ with its centre at O and pole L. Let the great circle arcs LD and LB meet the great circle AEA′, having its pole at L, in the points E and A (Figure 6.5). Then the arc BD will be given by arc BD = arc AE cos AB meaning thereby that the length of the arc BD is equal to the arc AE multiplied by the cosine of the angular distance of arc BD from the arc AE.

129 L

B’ O B Mathematics-I D

A’ C A

Figure 6.5 Spherical Triangle (Definition and Properties) A spherical triangle is defined as a triangle on the surface of a sphere contained by the arcs of three great circles. In the Figure 6.6 there are three great circles ABA′B′, ACA′C′, BCB′C′ of the sphere with centre O and BAC is one of the spherical triangles formed by them. It must be noted that the sides of the spherical triangles are formed only by the arcs of great circles and never by small circles. The arcs BC, CA, AB of the spherical triangle ABC are known as the sides of the triangle and are denoted by the a, b, c. Since they are respectively proportional to the angles BOC, COA and AOB that they subtend at O, they are measured by these angles in degrees and minutes of arc.

C ’ O A ’ A C

B ’

Figure 6.6 The angles of the spherical triangle ABC are the angles between the planes of the three great circles and are denoted by A, B, C where A is the angle between the planes ABA′B′ and ACA′C′; B is the angle between the planes, ABA′B′ and BCB′C′ and C is the angle between the planes ACA′C′ and BCB′C′. The three sides and the three angles together are called the six elements of the spherical triangle. Polar Triangle Let ABC, A′B′C′ be two spherical triangles on the surface of a sphere with centre at the point O such that A′ is the pole of the great circle BC and lies on the same side of it as the vertex A, B′ is the pole of the great circle AC on the same side of it as the vertex B and C′ is the pole of the great circle AB on the same side of the circle as the vertex C. Then the triangle A′B′C′ is known as the polar triangle of the spherical triangle ABC. The original triangle ABC is called the primitive triangle of the polar triangle. It can be readily seen that if A′B′C′ be the polar triangle of the triangle ABC, then ABC will be the polar triangle of A′B′C′. Now produce B′O to the point R, and since OR is perpendicular to the plane COA, and OA′ is perpendicular to the plane BOC, the angle ROA′ is equal to the angle 130 between the planes COA and BOC and is therefore equal to the angle C.

Spherical R Trigonometry

A′

b ’ b

C ’ C ’

C C

a ’ a

B ’ B

Figure 6.7 Let a′, b′, c′ be the sides of the spherical triangle A′B′C′, then cBOAAOR′′′=∠ =π−∠ ′ =π−C (from above) Similarly it can be shown that bBa′′=π−and =π−A

As ABC is the polar triangle of A′B′C′, it follows that aAbBc=π−′;; =π−′′ =π−C By transposing we get A′′′=π−aB;; =π− bC =π− c The above relations connecting the sides and angles of a polar triangle with those of the sides and angles of its primitive can be put in the form of the following statements. (a) The sides of the polar triangle A′B′C′ are the supplements of the corresponding angles of its primitive triangle ABC. (b) The angles of the polar triangle A′B′C′ are the supplements of the corresponding sides of its primitive triangle ABC. In a Spherical Triangle ABC (a) a + b + c < 360o (b) A + B + C > 180o, but < 540o Proof (a) Consider the spherical triangle ABC. Produce AB and AC to meet at the point A′ (Figure 6.8), then since the sum of two sides in the triangle is greater than the third, we have BA′ + CA′ > BC . . . (i) Also ABA′ = ACA′ = 180o . . . (ii) 131

Mathematics-I A

B C

A ’

Figure 6.8 Adding AB + AC to both sides in Eq. (i), we get (AB + BA′) + (AC + CA′) > AB + AC + BC i.e. 180o + 180o > a + b + c or a + b + c < 360o (b) In the polar triangle, the sides a′, b′, c′ are respectively given by 180o − A, 180o − B, 180o − C ∴ a′ + b′ + c′ = (180o − A + 180o − B + 180o − C) < 360o or A + B + C > 180o Also since each angle is less than 180o, therefore A + B + C < 540o Note : The quantity A + B + C − 180o is called the spherical excess and is denoted by E. Thus, E = A + B + C − 180o

6.3 FORMULAE FOR A SPHERICAL TRIANGLE

There are various formulae connected with the spherical triangle corresponding to those connecting the sides and angles of a plane triangle and these include the cosine, the sine and the cotangent formulae, which we shall discuss in the following. 6.3.1 Cosine Formula In any spherical triangle ABC cos a = cos b cos c + sin b sin c cos A cos b = cos c cos a + sin c sin a cos B cos c = cos a cos b + sin a sin b cos C Let ABC be the spherical triangle on the surface of a sphere with the centre at O. The sides a, b, c of the triangle are measured by the angles BOC, COA; and AOB at the centre of the sphere as shown in the Figure 6.9.

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Spherical Trigonometry

c b b D a B O a C E

Figure 6.9 At the point A, draw AE and AD perpendiculars to OA. Then angle ∠ EAD=∠ A . In the plane triangle AED, we have by cosine formula ED2 = AE2 + AD2 − 2AE· AD cos A . . . (6.1) Also in the triangle OED ED2 = OE2 + OD2 − 2OE· OD cos a . . . (6.2) Again from the right angle triangles OAE and OAD, we get OE2 = OA2 + AE2 and OD2 = OA2 + AD2 . . . (6.3) Combining Eq. (6.2) and Eq. (6.3), we have ED2 = 2OA2 + AE2 + AD2 − 2OE · OD cos a . . . (6.4) Eq. (6.1) and Eq. (6.4) together give 2OA2 + AE2 + AD2 − 2OE · OD cos a = AE2 + AD2 − 2AE · AD cos A or 2OE · OD cos a = 2OA2 + 2AE · AD cos A . . . (6.5) Dividing both sides by OE, OD, we get OA OA AE AD cosa=+ . . cos A . . . (6.6) OE OD OE OD or cos a = cos b cos c + sin b sin c cos A This relation proves the first formula. Similarly we can prove the other two relations of the formula which are as follows : cos b = cos c cos a + sin c sin a cos B . . . (6.7) cos c = cos a cos b + sin a sin b cos C . . . (6.8) These three formulas are defined as cosine formulae. They are known as fundamental formulae because it is with the help of these formulae that we can derive many other formulae. Further more by transposing, the cosine formulae can also be written in the form as cosab− cos cos c cos A = . . . (6.9) sinbc sin cosbca− cos cos cos B = . . . (6.10) sinca sin cosca− cos cos b and cos C = . . . (6.11) sinab sin Important (a) The first three formulae are used when the two sides and the included angle of a spherical triangle are given and we have to calculate the third side. 133

Mathematics-I (b) The second set of three formulae are used when all the three sides of a triangle are given and we have to find out the cosine of any angle. 6.3.2 Sine Formula In any spherical triangle ABC sinA sinBC sin == . . . (6.12) sinab sin sin c i.e. the sines of the angle of a spherical triangle are proportional to the sines of the opposite sides. This relation is known as sine-rule. We have seen from cosine formula that cosab− cos cos c cos A = sinbc sin

2 22⎛⎞cosabc− cos cos Now sinAA=− 1 cos =− 1 ⎜⎟ ⎝⎠sinbc sin sin22b sin c−+ (cos 2 a cos 2 b cos 2 c − 2 cos abc cos cos ) = sin22bc sin

(1−−−− cos22222b ) (1 cos c ) cos a cos b cos c + 2 cos abc cos cos = sin22bc sin

1−−−+ cos222bca cos cos 2 cos ab cos cosc ) = sin22bc sin Dividing both sides by sin2 a, we get sin2222A 1−−−+ cosa cos b cos c 2 cos abc cos cos ) = sin22a sinabc sin2 sin2

sin A [1−−−+ cos222a cos b cos c 2 cos abc cos cos ] ∴ = sin a sinabc sin sin 2λ = sinab sin sin c The radical above is taken with a positive sign because sin b, sin c and sin A are all positive. sinB sin C By symmetry, it can also be shown that and are each equal to sinbc sin 2λ . sinabc sin sin

where 2λ= [1 − cos222abc − cos − cos + 2 cos ab cos cosc ] Important This formula gives the relation between two angles and the two sides opposite to them. If three of the four quantities are known, the fourth one can be calculated with the help of this formula. 6.3.3 Supplemental Cosine Formula In the spherical triangle ABC, 134

cosA + cosBC cos Spherical cos a = sinBC sin Trigonometry

cosB + cosCA cos cos b = sinCA sin

cosCA+ cos cos B cos c = sinAB sin

Let A′B′C′ be the polar triangle of the spherical triangle ABC. While defining polar triangle, we have shown that aA′′′=π−,,bBcC =π− =π− and A′′′=π−aB,, =π− bC =π− c

Applying cosine formula to triangle A′B′C′, we have cos a′ = cos b′ cos c′ + sin b′ sin c′ cos A′ Using the above relation in this, we get cos (π − A) = cos (π − B) cos (π − C) + sin (π − B) sin (π − C) cos (π − a) or − cos A = cos B cos C − sin B sin C cos a i.e. cos A = − cos B cos C + sin B sin C cos a Rewriting the above relation, we obtain cosA + cosBC cos cos a = sinBC sin

cosB + cosCA cos Similarly, cos b = sinCA sin

cosCA+ cos cos B and cos c = sinAB sin

6.4 LATITUDES AND LONGITUDES

Let S′ be the sphere representing the earth with its centre at O and let NS be its polar axis. Then N is called the North pole and S the South pole of the earth. The great circle LL′ of which N and S are the poles and which is at right angle to the line NS is defined as the terrestrial equator. (Figure 6.10) N S ’

L ’ O L

A B

S Figure 6.10 If P is any point on the circle NBSN, the angle POB subtended by the arc PB at the centre O of the circle is termed the latitude of P. The latitude is measured from 0 to 90o north or the south of the equator. NP, the compliment of PB is called the colatitude. 135

Mathematics-I All great circles which pass through N and S are called the meridians. The meridian that passes through the Greenwich observatory, by International Standards is known as the Prime Meridian. In the Figure 6.10, NAS is the prime meridian while NBS is any other meridian. If O is the centre of the earth, the angle AOB between the planes NAS and NBS is known as the longitude of the point P on the great circle NBSN. Longitude is measured from 0 to 180o east or the west of the prime meridian. Example 6.1 Show that in any spherical triangle ABC. sin (A + Ba ) cos+ cos b = sinCc 1+ cos If a and b are complementary, then deduce that c (cosbbC+= sin ) sin 2 cos2 sin ( A+B ) 2 Solution sin (AB++ ) sin A cos B cos A sin B sin A sin B ==cosB + cos A sin CCCsin sin sin C sinA sinaB sin sin b Using sine formula, i.e. ==; sinCc sin sin C sin c sin (AB+ ) sin a sin b We get =+cosB cos A sinCc sin sin c Using cosine formula for cos B and cos A, the right hand side can be written as sinab⎛⎞⎛⎞ cos−− cos ca cos sin ba cos cos b cos c ⎜⎟⎜⎟+ sinccacbc⎝⎠⎝⎠ sin sin sin sin sin 1 =+−−(cosbacab cos cos cos cos cosc ) sin2 c (cosab+− cos ) (1 cos c ) cos a + cos b == 1cos− 2 c 1cos+ c sin (A + Ba ) cos+ cos b or = sinCc 1+ cos This proves the result. π π Again as a and b are complementary ab+ ==or a− b. Substituting in the 22 above result we get ⎛⎞π cos⎜⎟−+bb cos sin (AB++ ) 2 sinb cos b ==⎝⎠ c sin Cc1+ cos 2cos2 2 c or (cosbbC+= sin ) sin 2 cos2 sin ( A+B ) 2 Example 6.2 In a spherical triangle ABC, show that (a) sin 2B + sin 2C = 0 when b + c = π 136

A π Spherical (b) sin 2ca= cos sec2 when bc+ = 2 2 Trigonometry Solution (a) sin 2B + sin 2C = 2 sin B cos B + 2 sin C cos C

⎡sin B ⎤ =+2sinCB⎢ cos cosC⎥ . . . (6.13) ⎣sin C ⎦ Using both sine and cosine formula in Eq. (6.13), the right hand side (RHS), can be written as

⎡⎤sinbb⎛⎞ cos−− cos ca cos cos c cos a cos b 2sinC ⎢⎥⎜⎟+ ⎣⎦sincca⎝⎠ sin sin sin ab sin Since b + c = π, sin c = sin (π − b) = sin b sin b ∴ = 1, or sin b = sin c sin c

and we get,

2 sinCb⎛⎞ cos−− cos cac cos cos cos a cos b ⎜⎟+ sin ac⎝⎠sin sin b 2sinC =+(cosbc cos ) (1− cosa ) sinab sin

Again c = π − b ; gives cos c = cos (π − b) = − cos b Substituting cos c = − cos b, we prove the result. (b) Starting with the RHS, we have A cosaa 2 cos cosa sec2 == A 21cos2 + cosA 2 2cosa (by cosine formula) cosab− cos cos c 1 + sinbc sin

2cosabc sin sin = sinbc sin+− cos a cos b cos c

ππ Since bc+=, b = −∴ csin b = cos c , cos b = sin c 22 Using these relations, we get Aacc2cos cos sin cosa sec2 = 2 sinbb cos+− cos a sin b cos b

2cosacc cos sin ==2 sincc cos= sin 2c cos a

This proves the result. Example 6.3

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Mathematics-I a ABC is a spherical triangle in which b = c. Prove that cosBb= cot tan , 2 hence find B if a = 60o 25′, b = 55o 18′ Solution By Cosine formula cosbc− cos cos a cos B = sinca sin Since b = c, we get cosbcab− cos cos cos (1− cosa ) cos B == sinca sin sin ba sin a cotb 2 sin2 a ==2 cotb tan aa 2sin cos 2 22 This proves the result. a Given a = 60o 25′, = 30o 12.5′, b = 55o 18′ 2 log cot b = log cot 55o 18′ = 184038. a log tan = log tan 30o 12.5′ = 176508. 2 a log cos B = log cot b + log tan = 160546. 2 From the log table, B = 66o 13.5′. Example 6.4 If arcs be drawn from the angles of a spherical triangle to meet the mid-points of the opposite sides, and if α, β be the parts of arc which bisects the side a, show that sin α a = 2cos sinβ 2 A

b α 2

θ b D 2 G β θ b φ B 2

a 2 E a C 2 Figure 6.11 Solution Let ABC be the spherical triangle and let D and E be the mid-points of the sides AC and BC respectively. Let the arcs AE and BD intersect in the point G. Then AG = α, GE = β 138

Also let ∠=θ∠=φ∠=AGD, GBEand ADG ψ Spherical Trigonometry Applying sine-rule to triangle AGD, BCD and BEG, we get bb sin sin sin α sin a ==22; sinψθ sin sin φ sin ( π−ψ ) a sin sin β and 2 = sinθφ sin Multiplying these results, we get a sin α sin = sin a sin β 2 sin α a Hence = 2cos sinβ 2 Example 6.5 Two ports are in the same parallel of latitude, their common latitude being l and their difference of longitude 2λ; show that the saving of distance in sailing from one to the other on the great circle, instead of sailing along the parallel of latitude is 2cossin(sincos)rl⎡⎤λ−−1 λl ⎣⎦ λ being expressed in circular measure and r being the radius of the earth. Solution Let TR be the equator, P the pole and A, B the ports in the same parallel of latitudes (Figure 6.12).

A B l l R T

D C 2 λ

Figure 6.12 The difference of longitude between the ports A and B is the great circle CD or the ∠=APB 2λ.

∴ Arc CD = 2λ radians = 2λ r (linear units). Now the arc AB of the small circle = Arc CD cos AC = 2λ r cos l (linear units) . . . (6.14) Join AB by the great circle arc. Then from Δ PAB, we have cos AB = cos AP cos BP + sin AP sin BP cos APB = cos (90 − l) cos (90 − l) + sin (90 − l) sin (90 − l) cos APB = sin2 l + cos2 l cos 2λ

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Mathematics-I 2222AB 22 or 12sin−=+−λ=−λ sincos(12sin)12sincosll l 2 AB AB ∴ sin222=λ sin cosll or sin=λ sin cos 22

∴ AB =λ2 sin−1 [sin cosl ] radians

=λ2rl sin−1 [sin cos ] linear units . . . (6.15) Hence the saving of the distance is Eq. (6.14) − Eq. (6.15) = 2λ r cos l − 2 r sin− 1 [sinλ cos l] = 2r [λ cos l − sin− 1 (sin λ cos l)] 6.5 THE HAVERSINE FORMULA

Definition The ratio (1 − cosine) is defined as Versine and half of the versine is called the Haversine. 1cos− θ Thus, Hav. θ= 2 1 The Hav. 0o is 0. Hav. 90o is and Hav. 180o is 1 so that Haversine has the 2 advantage that it is always positive, i.e. positive in the first two quadrants where its value increases from 0 to 1 as the angle increases from 0o to 180o. To Derive the Haversine Formula By cosine formula, we have cos a = cos b cos c + sin b sin c cos A . . . (6.16) But cos A by the above definition is cos A = 1− 2 Hav. A Substituting in (6.16), we get 1 − 2 Hav. a = cos b cos c + sin b sin c (1 − 2 Hav. A) or 1 − 2 Hav. a = cos b cos c + sin b sin c − 2 sin b sin c Hav. A or 1 − 2 Hav. a = cos (b ∼ c) − 2 sin b sin c Hav. A and since cos (b ∼ c) = 1 – 2 Hav. (b ∼ c) ∴ we get 1 − 2 Hav. a = 1 − 2 Hav (b ∼ c) − 2 sin b sin c Hav. A or Hav. a = Hav (b ∼ c) + sin b sin c Hav. A This is the Haversine formula and sometimes called natural Haversine formula. This is used in variety of problems in navigation. In this form it is used to solve the problem. “Given two sides and the included angle, to find the third side.” The above formula on transposition gives Hav.ab− Hav. (: c ) Hav. A = sinbc⋅ sin In other words Hav.sideopposite− Hav. (difference sides of adjct) Hav. angle = product of sinesof adjacentsides 140

In the later form, it is used when all the three sides of the spherical triangle ABC Spherical are given and we have to find any angle. Trigonometry Note : The above forms of the formula may be used in preference to cosine formula. Example 6.6 In a spherical triangle ABC. Given AB = 50o 10′, AC = 64o 17′, BC = 27o 37′. Find C. Solution Hav. C = [Hav. c − Hav. (b − a)] cosec b cosec a c = 50o 10′ b = 64o 17′ a = 27o 37′ b − a = 36o 40′ Nat. Hav. = 0.09894 Nat. Hav. c = 0.17972 . . . (A) Nat. Hav (b – a) = 0.09894 . . . (B) A – B = 0.08078 log A – B = 8.90732 log cosec b = log cosec 64o 17′ = 10.04530 log cosec a = log cosec 27o 37′ = 10.33390 ∴ log Hav. C = log A – B + log cosec b + log cosec a = 9.28652 ∴ C = 52o 11′ Example 6.7 In a spherical triangle ABC. Given c = 20o, a = 36o 28′, b = 109o 15′, solve the triangle cos a (a) cos A = sin b

(b) cosCa=− cot cot b

Solution

(a) log cos a = 190537.

− log sin b = 197501. ______log cos A = 193036.

∴ A = 31o 35′′′ 15 . (b) log cot a = 0.13132 − log cot b = 154311. ______log cos C = 167443.

∴ C = 61o 48′′ 05′. 141

Mathematics-I Example 6.8

In a spherical triangle ABC, A = 88o 24.5′, a = 87o 01′, c = 100o 09′, b = 98o 10′, Find B. Solution Hav. formula for the calculation of B is given by Hav. B = [Hav. b − Hav. (a − c)] cosec a cosec c Nat Hav. b = Hav. 98o 10.0′ = 0.57103 . . . (A) Hav (a – c) = Hav. 13o 08′ = 0.01308 . . . (B) A – B = 0.55795 log (A – B) = − 1 + 0.74661 log cosec a = 0.00059+ log cosec c = 0.00685 log Hav. B = log A – B + log cosec A + log b cosec c = − 1 + 0.75405 ∴ B = 97o 46′ Remark We discuss the solution of the above problem using sine-rule sinA sin BAsin sin b ==or sin B sinab sin sin a

log sin A = log sin (88o 24′.5) = + 99983.01

log sin b = log sin (98o 10′) = + 99557.01 + 99540.01

log sin a = log sin (87o 1′) = + 99940.01

log sin B = + 99600.01

B = 97o 46′ or 82o 14′ To remove the anamoly between the two values, we know that the greater side must have the greater angle opposite to it. Now Opposite side Opposite angle a = 87o 1′ A = 88o 24.5′ b = 98o 10′ B > 88o 24.5′ Hence B should be 97o 46′ and not 82o 14′. Therefore, it is advisible to use Haversine formula wherever feasible. SAQ 1 (a) In a spherical triangle ABC, a = 49o 08′, b = 58o 23′ and C = 71o 20′. Calculate A and B. (b) In a spherical triangle PZX, z = 70o 45′, x = 62o 10′ and P = 50o 00′ find p. 142

(c) Solve the following spherical triangles : Spherical Trigonometry (i) PQR, given P = 53o 05′, PQ = 70o 20′ and PR = 110o 14′. Find Q and R. (ii) ABC, given a = 49o 08′, b = 58o 23′, C = 71o 20′. Find A and B. (iii) PAB, given A = 33o 14′, a = 80o 05′, b = 70o 12′. Find B. (d) In a triangle PAB in which P is the pole and A and B two places in the northern hemisphere, given A = 68o 0′, AB = 60o 30′, P = 80o 16′, find the latitude B. (e) In an equilateral spherical triangle ABC, prove that aa1 (i) cos sin = 222 (ii) sec A = 1 + sec a a (iii) 1−= 2 cosA tan2 2 a (iv) 12coscot+=a 2 2 (f) In a spherical triangle ABC, prove that sin C cos b = sin A cos B + cos A sin B cos C cot a sin c = cot A sin B + cos C cos B (g) In a spherical triangle ABC, if θ, φ, ψ be the arcs of great circles drawn from A, B, C perpendicular to opposite sides, then prove that sin a sin θ = sin b sin φ = sin c sin ψ

=−(1 cos222abc − cos − cos + 2cos ab cos cosc )

sina⎛⎞ 1− cos abc cos cos (h) In a spherical triangle ABC, prove that = ⎜⎟. sinA ⎝⎠ 1− cosABC cos cos (i) If one side of a spherical triangle is divided into four equal parts and α, β, γ, δ be the angles subtended at the opposite corner by the parts taken in order, show that sin (α + β) sin β sin δ = sin (γ + δ) and sin α sin γ (j) If the three sides of a spherical triangle ABC be halved and a new triangle c b formed, show that the angle θ between new sides and is given by 2 2 1 bc cosθ= cosA + tan tan sin2 θ 222 (k) Given a spherical triangle ABC with (i) c = 50o 10′, b = 64o 17′, a = 27o 37′, find A, B, C by means of cosine formula. (ii) b = 64o 17′, a = 27o 37′, A = 28o 29′, find B using sine formula.

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Mathematics-I 6.6 COTANGENT FORMULA OR FOUR CONSECUTIVE PARTS FORMULA

In a spherical triangle ABC cot a sin b = cot A sin C + cos b cos C From cosine formula, we know that cos a = cos b cos c + sin b sin c cos A . . . (6.17) cos c = cos a cos b + sin a sin b cos C . . . (6.18) Again from sine-formula sinAC sin sin C =∴=sinca sin . . . (6.19) sinac sin sin A

Substituting for cos c and sin c from (6.18) and (6.19) in (6.17), we get sin C cos a = cos b [cos a cos b + sin a sin b cos C] + sin b sin c cos A sin A

sin C =+cosab cos2 sin abbC sin cos cos + sin ab sin cos A sin A

or cos a (1 − cos2 b) = sin a sin b [cos b cos C + sin C cot A]

cosab sin2 or =+cosbC cos sin C cot A sinab sin

or cot a sin b = cos b cos C + sin C cot A . . . (6.20) which is the result to be derived. Note : As we go round the triangle ABC, the four parts used are next to one another. It should be borne in mind that while arranging these parts, we should begin with a side and end with an angle. These two may be referred as outer parts and other two as inner parts.

A Other Angle Inner Side

c Inner Angle C

C a B Other Side

Figure 6.13 As an aid to memory, the formula may be remembered as cosine of the (inner side) × cosine of the (inner angle) = sine of the (inner side) × cotangent of the (other side) − sine of the (inner angle) × cotangent of the (other angle).

144

Spherical 6.7 RIGHT-ANGLED SPHERICAL TRIANGLES Trigonometry

Definition If one of the angles of a spherical triangle is 90o, it is called a right-angled triangle; and if one side of a triangle is 90o that is to say, a quadrant it is called a quadrantal triangle. Note (a) In any right-angled triangle, an angle and its opposite side are of the same affection. That is both are less than 90o or both are greater than 90o. (b) In a right-angled triangle the magnitudes of the sides are either (a) all three less than 90o or (b) two greater than 90o and one less than 90o. (c) The solution of a right-angled triangle can equally well be obtained by the use of Sine, Cosine and Haversine formulae where the value of sine 90o will be taken as unity and cos 90o as zero. 6.7.1 Formulae for a Right-angled Triangle If one of the angles say ‘C’ of a spherical triangle ABC is taken as 90o, then all the formulae derived in the earlier sections get simplified by putting sin C = 1 and cos C = 0, as will be seen in the following : (a) cosine formula; cos c = cos a cos b + sin a sin b cos C We get cos c = cos a cos b sinA sinBC sin (b) sine formula; ==, we get sinab sin sin c sin a = sin A sin c; sin b = sin B sin c (c) From cos A = − cos B cos C + sin B sin C cos a cos B = − cos C cos A + sin C sin A cos b We get cos A = sin B cos a cos B = sin A cos b (d) From cos C = − cos A cos B + sin A sin B cos c, we get cos c = cot A cot B (e) From cos b cos C = sin b cot a − sin C cot A We get cos a cos C = sin a cot b − sin C cot B sin b = tan a cot A; sin a = tan b cot B 6.7.2 Napier’s Rules The formulae derived above can be easily written by a rule known as Napier’s Rules. Consider a right-angled spherical triangle ABC with a right angle at A = 90o. To use these mnemonic rules divide a circle into five sectors by means of five radii and place b, c, πππ −−−B,,aC, in that order, one in each sector. Taking one of the sectors to be 222 known as the middle part, the sectors immediately adjoining it will be known as “adjacent parts”, and the other two sectors will be the “opposite parts.” It will be found that if we select any three of the five parts, one part will be in the middle (in the sense that it is equidistant from the other two) and the other two will be either both be π “adjacent” or both “opposite”. For instance, if we choose − a as the middle part, then 2 π π − B and − C will be called adjacent parts and b and c designated as opposite parts. 2 2

145 A

b b π 2 c c Mathematics-I π − C π B 2 − B 2 a π − a 2

Figure 6.14 Napier’s Rules State Rule (1) The sine of the middle part = product of the tangents of the adjacent parts. Rule (2) The sine of the middle part = product of the cosines of the opposite parts. ⎛π ⎞ Thus, if ⎜− a⎟ be taken as the middle part, from rule (1) we have ⎝⎠2 ⎛⎞⎛πππ ⎞⎛ ⎞ sin⎜⎟⎜−=aB tan − ⎟⎜ tan −C⎟ ⎝⎠⎝222 ⎠⎝ ⎠ i.e. cos a = cot B cot C and by rule (2) we have ⎛⎞π sin⎜⎟−=abc cos cos ⎝⎠2 i.e. cos a = cos b cos c. Example 6.9 ABC is a spherical triangle in which A = 90o, B = 120o, c = 60o, find a, b, C. Solution Using the formula given above, we have ⎛⎞ππ ⎛⎞ sin⎜⎟−=B tanca tan ⎜ −⎟ ⎝⎠22 ⎝⎠ cosB cos 120o or cot a == tan c tan 60o 11 1 =−. =− 2 323 ∴ a = 106o 6′ 33 Also tan b = tan B sin c = tan 120o sin 60o =−31 × =− =− .5 22 ∴ b =−180oo 56 18.5′ = 123 o 41.5′

133 Now cosCcB== cos sin cos 60oo sin 60 =×= 22 4 ∴ C = 64o 20′ Example 6.10 146

In a spherical triangle ABC, Angle C = 90o. Angle B = 30o. Side AB = 70o. Find Spherical side AC and the angle A. Trigonometry

AC BC

Com p A B

Figure 6.15 Solution (a) sin AC = cos (90 − B) cos (90 − c) sin AC = sin B sin c

log sine 30o = 169897.

log sine 70o = 97299.1

log sine AC = 167196. ∴=AC 28o 1.5′

(b) sin (90o − A) = tan (90o − c) tan 28o 1.5′ = tan 20o tan 28o 1.5′

log tan 20o = 156107.

log tan 28o 1.5′ = 72612.1

log cos A = 128719. ∴=A 79o 49′ 48′′

Example 6.11 In a spherical triangle ABC, α, β be the arcs drawn from right angle C respectively perpendicular to and bisecting the hypotenuse C. Show that c sin22 (1+α=β sin ) sin2 2 C

a b

β α

A B 0 90 c/2 c/2 L γ M

Figure 6.16 Solution 147

Mathematics-I ABC is the given spherical triangle, right-angled at C. Let CM be perpendicular on AB and CL be arc bisecting the side AB. Let LM = γ. In the right-angled triangle CMB, ⎛⎞c We have cosa = cosα− cos ⎜⎟γ ⎝⎠2 Again in the triangle CMA we have ⎛⎞c cosb = cosα+ cos ⎜⎟γ ⎝⎠2

2 ⎛⎞⎛cc⎞ ∴ cosab cos= cosα−γ+ cos⎜⎟⎜ cos γ⎟ ⎝⎠⎝22⎠ 2 cos α 222⎛⎞c =+γ[]cosc cos 2=αγ cos⎜⎟ cos− sin 22⎝⎠ Again in the triangle ABC cos c = cos a cos b

222⎛⎞c ∴ cosab cos==αγ− cos c cos⎜⎟ cos sin ⎝⎠2

2222222cc⎛⎞ 2c or 1−=αγ−=αγ−α 2sin cos⎜⎟ cos sin cos cos cos sin 22⎝⎠ 2 Also cos β = cos α cos γ cc ∴ 1−=αγ−α 2 sin22222 cos cos cos sin 22 c =β−αcos22 cos sin2 2 c =−1sin22 β−− (1sin)sin α 2 2 c or sin22β =+ sin (1 sin 2α ) 2 which proves the result. Example 6.12 In a spherical triangle ABC, if C is a right angle and D is the mid-point of AB, show that c 4cos22 sinCD=+ sin 2 a sin 2 b 2 C

b a

A c/2 D c/2 B Figure 6.17 Solution 148

Since the triangle is right angle at C, we have by cosine formula Spherical Trigonometry cos c = cos a cos b . . . (6.21) D is the mid-point of AB c ∴ AD== BD 2 In the triangle CAD, we have cc cosCD =+ cos cosb sin sin b cos A . . . (6.22) 22 Again in the triangle CBD cc cosCD =+ cos cosa sin sin a cos B . . . (6.23) 22 Multiplying (6.22) and (6.23), we get

2 ⎛⎞cc⎛ cc⎞ cosCD =+⎜⎟ cos cosb sin sin b cos A ×+⎜ cos cosa sin sina cos B⎟ ⎝⎠22⎝ 22⎠ ccc =+cos2 cosab cos cos sin (cosabA sin cos + cos baB sin cos ) 222 c + sin2 sinab sin cos A cos B 2 Using (6.21), we get c 1 =+cos2 cosc [cos abc sin sin cos A+ cos bac sin sin cos B ] 22 c + sin2 sinab sin cos A cos B 2 1cos+ c 1 =+−+−coscaabcbba [cos (cos cos cos ) cos (cos cos cosc )] 22

2 cab⎡⎤⎛⎞cos−− cos coscbc⎛⎞ cos cos cos a + sin sinab sin ⎢⎥⎜⎟⎜⎟ 2 ⎣⎦⎝⎠sinbc sin ⎝⎠sin ca sin 1cos+ c 1 =++−coscabab (cos22 cos 2cos cos cosc ) 22 c sin2 +−++2 [cosab cos cos c (cos22 b cos a ) cos 2 cab cos cos ] sin2 c 1cos+ c 1 =+coscca (−++ 2cos222 cos cosb ) 22 c sin2 +−2 coscba (1 cos222− cos+ cosc ) cc 4sin22 cos 22 1cos+ c 1 =+−−coscca (2sin222 sin sinb ) 22 cos c ++−(sin222ab sin sinc ) c 4cos2 2 Now cos2 CD = 1 − sin2 CD 149

Mathematics-I 221cos+ c 2 ∴ sinCD=− 1 cos CD =− 1 cosc− sin c 2 ⎛⎞ ⎜⎟1cossincoscc2 c ++(sin22ab sin ) − + ⎜⎟cc ⎜⎟2 4cos22 4cos ⎝⎠22 cos22cc++ cos (sin a sin2 b ) ⎛⎞ csin2 c cos c =−1s−in22cc + 2cos−cos − cc⎜⎟ 224cos22⎝⎠4cos 22 cos22cc cos 1 sinc cos c =−cos22ca −+(sin + sin2b ) − cc 224cos224cos 22

2 22cccccos− cos ⎛⎞2 2 2 2 or 4cos sin CD =−⎜⎟4cos sinc cos c+ sin a+ sin b 222⎝⎠ = (1 + cos c) cos c (cos c − 1) – cos c (1 − cos2 c) + sin2 a + sin2 b = cos c (cos2 c – 1 – 1 + cos2 c) + sin2 a + sin2 b = sin2 a + sin2 b which proves the result. Example 6.13 In a spherical triangle ABC the great circle arc AN cuts BC at right angles at N; establish the following results : cosBNA cos B (a) (i) = cosCN cos AC (ii) cos C = cot AC tan NC

C 900

N B Figure 6.18 Solution (a) (i) Since the angle at N = 90o, using cosine formula we have cos AB = cos AN ·cos BN and cos AC = cos AN ·cos CN By division, we get cosABB cos N = cosACC cos N

(ii) Also from spherical triangle ACN, using Napier’s rule We have 150

⎛⎞ππ ⎛ ⎞ Spherical sin⎜⎟−=CCNA tan tan ⎜ −C⎟ Trigonometry ⎝⎠22 ⎝ ⎠ or cos C = tan CN cot AC which proves the result. (b) Evaluate NA, NB, NC where BC = 54o; AC = 65o and C = 78o From (ii) above tan CN = cos C tan AC = cos 78o tan 65o

log tan CN = log cos 78o + log tan 65o = 1 =+ 64921.133133.031788.

∴ CN = 24o 2′; BN = BC − CN = 54o − 24o 2′ = 29o 58′

cosAC cos65o cos AN == cos CN cos 24o 2′

∴ log cosAN =− log cos 65oo log cos 24 2′

= 1 − = 66533.196062.162595.

i.e. log cos AN = 66533.1 ∴ AN = 62 o 26′ Example 6.14 An aeroplane flies in a great circle course from a point A (lat 30oN; long 10oE) to a point B on the equator, the initial direction of departure being 20oE of S. Find the longitude of B, and the length of the journey, taking the earth to be a sphere of radius 4000 miles. Solution In the (Figure 6.19) N and S are the north and the south poles on the earth’s surface with its centre at the point O. The meridian through A meets the equator in C. In the spherical triangle ABC, we have from the data in question A = 20o, C = 90o, b = 30o N

O C B

S Figure 6.19 Using Napier’s rules, we have

⎛⎞π sinba=−= tan tan⎜⎟ A tan a cot A ⎝⎠2 or tan a = sin b tan A

o o i.e. tan a = sin 30 tan 20 151

Mathematics-I 1 =×0.36397 = 0.18199 2 ∴ a = 10o 19′ (nearly) Hence the longitude of B is 10o + 10o 19′ = 20o 19′ E Again from Napier’s rule ⎛⎞π sin⎜⎟−=ca cos cos b ⎝⎠2 or cos c = cos a cos b ∴ log cos c = log cos a + log cos b = log cos (10o 19′) + log cos (30o) = 1 + = 93045.193753.199292.

∴ c = 31o 33′ = 0.55065 radians Hence the distance AB = 4000 × c = 4000 × 0.55065 = 2203 miles. 6.8 QUADRANTAL SPHERICAL TRIANGLES

A quadrantal spherical triangle is one in which a side is 90o, i.e. a quadrant. If the side c = 90o, then all the formulae derived earlier will get simplified by putting cos 90o = 0 and sin 90o = sin c = 1. As in the case of right angled triangle, here also if the values of any two other quantities (sides or angles) are given, the values of the remaining three can be calculated. The solution of quadrantal spherical triangles can be obtained with ease by the use of Napier’s Rules for circular parts. Divide a circle into five sectors and label them with numbers as shown below. A B C b 1 2 3 5 0 90 − b c 900 − c

A − 900 4

B C

Figure 6.20 In the triangle ABC, let BC be taken as the quadrant, i.e. 90o. Then B and C are next to the quadrant, the other three are written down as complements. For examples in parts (1) and (2) are inserted the angles containing the quadrant. In part (3) insert the complement of the side opposite to the angle in (1). In part (5) insert the complement of the side opposite the angle in (2). Finally in part (4) insert (Angle opposite the quadrant minus 90o) Rule for Writing the Formula 152

If a = 90o, then arrange the circular parts as shown above. Spherical Trigonometry (a) Sine of the middle part = product of the tangents of the adjacent parts. (b) Sine of the middle part = product of the cosines of the opposite parts. For illustration, if part number (5) is taken as the middle part, (1) and (4) are its adjacent parts while (2) and (3) are its opposite parts. Example 6.15 In a quadrantal triangle ABC, b = 78o 14′, c = 49o 08′ and a = 90o. Determine the angles A and B.

B 0 90

C Figure 6.21 Solution To Find A Since b and c are given and A is the middle part, we have by Napier’s Rule

o ⎛⎞⎛ππ⎞ sin (A −= 90 ) tan⎜⎟⎜ −cb tan −⎟ ⎝⎠⎝22⎠

or − cos A = cot c cot b

log cot c = log cot 49o 08′ = 9.93712

log cot b = log cot 78o 14′ = 9.31870 log cos A = 9.25582

∴ cos A = 79o 37′ But cos A is negative

∴ cos A = 90o + 10o 23′ = 100o 23′ To Find B sine B and c are together and they are both opposite of b which is the middle part, we have

⎛⎞ππ ⎛⎞ sin⎜⎟−=bB cos cos ⎜⎟ −= cB cos sin c ⎝⎠22 ⎝⎠

or cos b = cos B sin c or cos B = cos b cosec c 153

Mathematics-I log cos B = log cos b + log cosec c

= log cos 78o 14′ + log cosec 49o 08′ log cos 78o 14′ = 9.30947 log cosec 49o 08′ = 10.12134 log cos B = 9.43081 ∴ B = 74o 21′ (As B and b must be of the same affection). Example 6.16 In a quadrantal triangle ABC, a = 90o, C = 60o 20′, A = 115o 40′, Find AB. Solution sine middle = product of cosine opposites

⎛⎞⎛⎞ππ sinCcAc=− cos⎜⎟⎜⎟ cos −= sin sin A ⎝⎠⎝⎠22

sinC sin 60o 20′ ∴ sin c == sin A sin 115o 40′

log sine 60′ 20′ = 9.93898 log sine 115o 40′ = 9.95488 ∴ log sine c = 9.98410 ∴ c = 74o 35.5′ Example 6.17 In a quadrantal triangle ABC, c = 52o11′, B = 69o 47′ and a = 90o. Determine A, C and b. Solution To Find A Draw a diagram and start at the top with the 90o side. sine middle = tan, adjacents sin (90o − c) = tan B tan (A − 90o) tan (A − 90o) = cos c cot B tan (A − 90o) = cos 52o 11′ cot 69o 47′ log cos 52o 11′ = − 1 + 0.78756 log cot 69o 47′ = − 1 + 0. 56615 + log tan (A − 90o) = − 1+ 0.35371 A − 90o = 12o 43.4′ ∴ A = 12o 43.4′ + 90o = 102o 43.4′

154

B c Spherical c b Trigonometry 90 − c 90 − b

A − 90 B C a

Figure 6.22 To Find C sin B = tan C tan (90o − c) tan C = sin B tan c = sin 69o 47′ tan 52o 11′ log sin 69o 47′ = − 1 + 0.97239 log Tan 52o 11′ = + 0.11006 + log Tan C + 0.08245 ∴ C = 50o 24.3′ To Find b sine middle = cosines opposites sin (90o − b) = cos B cos (90o − c) cos b = cos B sin c = cos 69o 47′ sin 52o 11′ log cos 69o 47′ = − 1 + 0.53854 log sin 52o 11′ = − 1 + 0.89761 + log cos b = − 1 + 0.43615 ∴ b = 74o 09.5′ Important Note In the segment “Angle opposite the 90o side” we have used (the angle − 90o) which is basically not the same as the complement of the angle. To overcome this anamoly, when the three parts are written down, the two adjacents or the two opposites are both sides or both angles we must prefix a negative sign to the product. We illustrate this by studying the following example. Example 6.18 In a spherical triangle ABC, c = 90o, a = 110o 11′, B = 14o 20′. Find the angle C and the side b = AC. Solution To Find C sine middle = tan·adjacents ⎛⎞ππ ⎛ ⎞ sin⎜⎟−=aB tan tan ⎜ −C⎟ ⎝⎠22 ⎝ ⎠ cos a = tan B cot C cosa cos 110o 11′ cot C == tan B tan 14o 20′

log cos 110o 11′ = 9.53785 − o o log Tan 14 20 = 9.40742 + 155

Mathematics-I log cot C = 10.13043

1′ ∴ C = 36o 31 2 Note Extra sign was introduced because we used the angle C opposite the 90o side or because the two adjacents were both angles. C

A b B a

o π π 90 − b − a 2 2 π − C c 2 A B Figure 6.23 To Find b sine middle = cosines opposites

⎛⎞ππ ⎛⎞ sin⎜⎟−=ba cos ⎜⎟ − cos B ⎝⎠22 ⎝⎠ or cos b = sin a cos B = sin 110o 11′ cos 14o 20′ log sin 110o 11′ = 9.97248 log cos 14o 20′ = 9.98627 log cos b = 9.95875 ′ 3 ∴ AC = b = o 3424 4 Note There is no extra sign in this case because we have not used the angle opposite the 90o side or in other words one opposite was a side and the other opposite the angle. Example 6.19 In a spherical triangle, if θ, φ, ψ be the arcs bisecting the angles A, B, C respectively and terminated by the opposite sides. AB C Show that cot θ+φ+ψ=++cot cot cot cot cot cotabc cot cot . 22 2 Solution Let the bisector of the angle A meet the side BC in D. Suppose angle ADB is α. Then angle ADC is π − α.

156

A/2 A/2 Spherical E θ ψ F Trigonometry φ C/2 B/2 α C/2 B/2 C B D ∠ ADC = π − α ∠ ADB = α

Figure 6.24 Using cotangent formula for the triangles ADB and ADC, we have AA cosθ=θ− cos sin cotc sin cot α 22 AA cosθ cos = sin θ cotb − sin cot ( π−α ) 22 Adding the above two relations, we get A 2cosθ=θ+ cos sin (cotbc cot ) 2 A or 2cotθ=+ cos cotbc cot . . . (6.24) 2 Similarly from the other two relations, we get B 2cotθ=+ cos cotca cot . . . (6.25) 2 C 2cotθ=+ cos cotab cot . . . (6.26) 2 Addition of (6.24), (6.25) and (6.26) gives the desired result. Example 6.20 In a spherical triangle ABC, given AC = 32o, BC = 116o Angle C = 90o, Find A. C b 900 1160 320 a

A B Figure 6.25 Solution By Napier’s Rules sin b = tan a tan (90o − A) = tan a cot A ∴ cot A = sin b cot a = sin 32o cot 116o

log sin 32o = 172421. 157

Mathematics-I log cot 116o = 68818.1

log cot A = 141239.

A = 75o 30′ approximately, i.e. 75o 30′ 30′′. As the cotangent of 116o in negative, the answer cot A = 75o 30.5′ is not the correct answer as A must be greater than 90o.

11′ ′ ∴ A =+90oo 14 29 = 104 o 29 . 22 6.9 POLAR TRIANGLE

In the earlier sections we have seen that in a spherical triangle when two sides and the included angle or when the three sides are given, the triangle can be solved by the application of cosine formula or the haversine formula. Sine formula can also be used when opposite pair sides and angles are given; but the formula also has limits in its application. In problemes where two angles and the included side or where three angles are given and it is required to solve the spherical triangle, the above formulae do not provide any assistance in getting the solution of the problem. In such situations the solution can possibly be obtained by the use of the polar triangle the definition of which has already been given in the earlier section (ref. Definition 6.2). We have also discussed Supplemental Formula relating to polar triangles and use it to solve the following problems. Example 6.21 In the spherical triangle ABC, the angles A and B are given as A = 54o 01′, B = 121o 25′ and the side c is c = 55o 14′. Determine the angle C. Solution In polar triangle A′B′C′ C′ = π − c = 180o − 55o 14′ = 124o 46′; a′ = π − A = 125o 59′ and b′ = π − B = 58o 35′. Again a′− b′ = 67o 24′. A ′

A b′ c′ c b 550 14′

B a C

B′ a′ C′ Figure 6.26 [Note : The problem has been transformed from the two angles and the included side to one with two sides and the included angle by the use of supplemental formula]. So we have

158 hav c′ = hav (a′ ∼ b′) + hav C′ sin a′ sin b′

log hav C′ = 9.89493 Spherical Trigonometry log sin b′ = 9.93115 log sin a′ = 9.90805 log 9.73413 nat 0.54217 nat hav (a′ − b′) = 0.30785 nat hav c′ = 0.85002 ∴ c′ = 134o 26′ and therefore C = π − c′ = 180o − 134o 26′ log = 45o 34′ Example 6.22 In a spherical triangle ABC, given c = 100o 09′, A = 88o 24.5′, B = 97o 46′ Calculate the third angle C. Solution The data gives two angles and the included side. So in the polar triangle C′ = 180o − c = 180o − 100o.09′ = 79o 51′ a′ = 180o − A = 180o − 88o 24.5′ = 91o 35.5′ b′ = 180o − B = 180o − 97o 46′ = 82o 14′ A

b c

a B

Figure 6.27 Using cosine formula for the polar triangle hav c′ = hav C′ sin a′ sin b′ + hav (a′ ∼ b′) where a′ ∼ b′ = 9o 21′.5 log hav C′ = log hav 79o 51′ = 9.61477 log sin a′ = log sin 91o 35.5′ = 9.99982 log sin b′ = log sin 82o 14 ′ = 9.99599 log = 9.61058 anlilog = 0.40793 nat hav (a′ − b′) = 0.00664 nat hav c′ = 0.41457 ∴ c′ = 80o 10′

o o and C = π − c′ = 180 − 80 10′ = 99 50′ 159

Mathematics-I = 99o 50′

SAQ 2 Solve the following spherical triangles : (a) ABC; given AC = 47o.00′, AB = 67o 00′, A = 66o 30′; Find B. (b) PAB; given a = 80o 05′, A = 33o 14′, b = 70o 12′; Find B. (c) PZX; given P = 50o 00′, x = 62o 10′, z = 70o 45′; Find p. ′ 1 (d) PZX; given P = 47o 28.3′, z = 104o 10′, and X = 41o 03 , Find p, Z, x. 2 (e) ABC; A = 81o 24.3′, B 61o 31.7′, C = 102o 58′, calculate a, b, c. (f) ABC; given B = 84o 38.3′, C = 93o 36′, and a = 64o 18′, find A, b, c. (g) XYZ; given X = 73o 01′, y = 47o 47′, x = 90o. Solve the quadrantal triangle. (h) PXY; given p = 53o 20′, X = 92o 5′, Y = 90o. Calculate P, x, y. (i) In a right angle spherical triangle ABC, in which C = 90o, and a = 26o 27′ 24′′, c = 46o 40′ 12′′. Find b, A, B. (j) In a spherical triangle ABC if CX and CY are the internal and external bisectors of angle C, which is a right angle, show that cot2 CX + cot2 CY = cot2 a + cot2 b (k) If the side c be a quadrant and δ, the length of the arc perpendicular to it from C, show that (i) cos2 δ = cos2 a + cos2 b (ii) cos2 δ = cot2 A + cot2 B (l) The sides of a spherical triangle ABC are all quadrants and x, y, z are the arcs joining any point within the triangle to the angular points, prove that cos2 x + cos2 y + cos2 z = 1 (m) If P is taken in AB, a side of any triangle ABC such that AP = AC, show that sin c cos CP = cos a sin b + cos b sin (c − b) π π (n) In a spherical triangle ABC, Abc= , == and D is the point of the side 42 1 AC. If BDC = α, show that cos α = . Also show that in the spherical 3 triangle ABC, when the arcs a, b, c are very small and is equivalent to a cosine formula for a plane triangle.

160

Spherical 6.10 SUMMARY Trigonometry

We summarise, what we have learnt in this unit : • A section of a sphere by a plane is a circle and if the plane goes through the centre of the sphere, the section is a great circle. • The poles are the extremities of the diameter which is perpendicular to the section of the sphere. • The cosine and sine formulae are the relation connecting the sides and angles of a spherical triangle and are given as cos a = cos b cos c + sin b sin c cos A sinA sinBC sin and == sinab sin sin c These formulae are derived from those connecting the sides and angles of a plane triangle. • Supplemental cosine formula helps in the evolution of a side of a triangle when all its three angles are known cosA + cosBC cos cos a = sinBC sin

• Haversine formula was defined as 1cos−θ Hav. θ =θor cos= 1− 2 Hav. θ 2 and twice of Hav. θ was defined as Versin θ. This formula has the advantage, that it is always positive in the range 1 0 < θ < 180o. Its value is 0 when θ = 0. Its value is where θ is 90o and its 2 value is 1 when θ = 180o. ∴ Its value is positive, lying between 0 and 1 when θ lies between 0 and 180o. The cosine formula connecting two sides and the excluded angle is cosabcbc=+ cos cos sin sin cos A Its analogue in terms of haversine formula is Hav. A = Hav. (b – c) + sin b sin c Hav. A • Cotangent formula or four consecutive parts formula was derived by combining sine and cosine formulae. • Polar Triangle Problems involving two angles and the included side or where three angles are given, it is advisable to solve them by using polar triangle. The polar triangle is connected with the primitive triangle for purpose of getting the solution. The relation connecting the sides and angles of a polar triangle with those if the sides and angles of its primitive can be written in the following statements : (a) The sides of the polar triangle A′ B′ C′ are the supplements of the corresponding angles of its primitive triangle ABC, i.e. aAbBc′′′=π−;; =π− =π−C 161

Mathematics-I (b) The angles of the polar triangle A′ B′ C′ are the supplements of the corresponding sides of its primitive triangle ABC, i.e. A′ = π−aB;;′′ =π− bC =π− c and was derived as cotab sin= cot AC sin+ cos b cos C

• Right-angled Spherical Triangle If one angle of a spherical triangle is 90o, the triangle is defined as a right-angled triangle. In a right-angled triangle, an angle and its opposite side are said to be of the same affection if both are less than 90o or both are greater than 90o. In sine, cosine and other formula derived in the earlier sections get simplified when we replace sin 90o by 1 and cos 90o by zero. • Quadrantal Spherical Triangle If in a triangle, the length of one side is 90o, i.e. a quadrant the triangle defined as quadrantal spherical triangle. As in the case of a right-angled triangle, if the values of any two other quantities (side or angles) are given, the values of remaining three quantities can be calculated. • Napier’s Rules The formulae obtained under the right-angled triangle or the quadrantal spherical triangle can easily be written by a rule known as Napier’s rule. Napier’s Rules Statement Rule 1 Sine of the middle part = Product of the tangents of the adjacent of parts. Rule 2 Sine of the middle part = Product of the cosines of the opposite parts. 6.11 ANSWERS TO SAQs

SAQ 1 (a) A = 59o 0.7′; B = 74o 52.3′; c = 56o 41.6′. (b) p = 46o 19.5′. (c) (i) Q = 124o 29.3′, R = 55o 48.8′ (ii) A = 59o 01′, B = 74o 52′ (iii) B = 31o 34′ (d) a is 54o 58′ and latitude of B is 35o 02′ N. ′ 1 (k) (i) A = 28o 29′, B = 112o 03 , C = 52o 11′ 2 ′ 1 (ii) B = 67o 57 2 162

SAQ 2 Spherical Trigonometry ′ 1 (a) B = 52o 32 2 (b) B = 31o 34′ ′ 1 (c) p = 46o 19 2 (d) p = 66o 41′, Z = 128o 55′, x = 54o 56′ 1 (e) a = o 987 , b = 62o 37′, c = 100o 09′ 2

1′ (f) A = 64o 06′, b = 81o 47′, c = 91o 24 2 (g) z = 107o 51′, Y = 45o 06′, Z = 114o 26.5′ (h) P = 53o 21.5′, x = 92o 36′, y = 91o 33′ (i) b = 39o 57′ 48′′, A = 37o 44′ 52′′, B = 62o 00′ 7′′

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Mathematics-I FURTHER READINGS

Grewal, B. S., (2007), Higher Engineering Mathematics, Khanna Publishers. Kreyszig Erwin, Advanced Engineering Mathematics, John Wiley and Sons. Bali, N. P., Iyenger, N. N., Engineering Mathematics, Laxmi Publications (P) Ltd. Stood, KA., Advanced Engineering Mathematics, Palgrave MacMillan. Shanti Narayan, Analytical Solid Geometry, S. Chand and Sons.

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Spherical APPLIED MATHEMATICS Trigonometry

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