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the angle opposite this side are given, then through the point C there may generally be drawn two arcs of great circles CB, CB', making the same angle with AB, and each of the triangles ABC, A

B AB'C will satisfy the conditions of the problem. There is the same ambiguity in the numerical computation, since the side BC is found by means of its sine (Art. 27). In the preceding example, however, there is no ambiguity, because the angle A is less than B, and, therefore, the side a must be less than b, that is, less than a quadrant.

Ex. 2. In the oblique-angled spherical triangle ABC, the angle A is 128° 45', the angle C=30° 35', and BC=68° 50'. Required the remaining parts.

It will be observed that in this case the perpendicular BD, drawn from the angle B, falls without the triangle ABC, and therefore the side AC is the difference between the segments CD and AD.

AB=37° 28/ 20 Ans. AC=40° 9' 4".

B = 32° 37' 58".

CASE III.

(221.) Given two sides and the included angle, to find the remaining parts.

In the triangle ABC let there be given two sides, as AB, AC, and the included angle A. Let fall the perpendicular CD on the side AB; then, by Napier's rule,

R cos. A=tan. AD cot. AC. Having found the segment AD, the segment BD becomes known; then,

A by Theorem III., Cor. 3,

sin. BD : sin. AD :: tan. A: tan. B. The remaining parts may now be found by Theorem III.

Ex. 1. In the spherical triangle ABC, the side AB=73° 20, AC=41° 45', and the angle A=30° 30'. Required the remain. ing parts.

D

B

cot. AC: cos. A::R: tan. AD=37° 33' 41". Hence

BD=35° 46' 19''. sin. BD : sin. AD :: tan. A. : tan. B=31° 33' 43''. Also, by Theorem III., Cor. 1,

COS. AD : cos. BD :: cos. AC : cos. BC=40° 13' 0". Then, by Theorem III.,

sin. BC : sin. AB :: sin. A: sin. ACB=131° 8' 47'. Ex. 2. In the spherical triangle ABC, the side AB=78° 15', AC=56° 20', and the angle A=120°. Required the other parts.

B = 48° 57' 29". Ans. c 62° 31' 40".

BC=107° 7' 45'

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Case IV.

1

(222.) Given two angles and the included side, to find the remaining parts.

In the triangle ABC let there be given two angles, as A and ACB, and the side AC included be

C tween them. From C let fall the perpendicular CD on the side AB. Then, by Napier's rule, R cos. AC=cot. A cot. ACD.

D Having found the angle ACD, the

А

B angle BCD becomes known; then, by Theorem III., Cor. 4,

cos. ACD : cos. BCD :: cot. AC : cot. BC. The remaining parts may now be found by Theorem III.

Ex. 1. In the spherical triangle ABC, the angle A=32° 10', the angle ACB=133° 20', and the side AC=39° 15'. Required the other parts.

cot. A : cos. AC :: R: cot. ACD=64° 1' 57". Ilence

BCD=69° 18' 3''. Then cos. ACD : cos. BCD :: cot. AC : cot. BC=45° 20' 43''. Also, by Theorem III., Cor. 2,

sin. ACD : sin. BCD :: cos. A : cos. B=28° 15' 47". Then, by Theorem III.,

sin. B : sin. ACB :: sin. AC : sin. AB=76° 23' 5'.

Ex. 2. In the spherical triangle ABC, the angle A=125° 20', the angle C=48° 30', and the side AC=83° 13'. Required the remaining parts.

AB= 56° 39' 9". Ans. BC=114° 30' 24".

B 62° 54' 38''.

CASE V.

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A

(223.) Given the three sides of a spherical triangle, to find the angles.

In the triangle ABC let there be given the three sides. From one of the angles, as C, draw CD perpendicular to AB. Then, by Theorem IV., tan. JAB : tan. }(AC+BC): tan. }(AC-BC): tan. }(AD-BD). Hence AD and BD become known;

D then, by Napier's rule,

B R cos. A=tan. AD cot. AC. The other angles may now be easily found.

It is generally most convenient to let fall the perpendicular upon the longest side of the triangle.

Ex. 1. In the spherical triangle ABC, the side AB=112° 25', AC=60° 20', and BC=81° 10'. Required the angles. tan. 56° 121 : tan. 70° 45' :: tan. 10° 25' : tan. 19° 24' 26". Hence AD=36° 48' 4", and BD=75° 36' 56'. Then R: tan. AD:: cot. AC : cos. A=64° 46' 36''.

" Also, R : tan. BD :: cot. BC : cos. B=52° 42' 12''. Then sin. AC: sin. AB :: sin. B : sin. ACB=122° 11' 6".

Ex. 2. In the spherical triangle ABC, the side AB=40° 35', AC=39° 10', and BC=71° 15. Required the angles.

A=130° 35' 55". Ans. B= 30° 25' 34".

C= 31° 26' 32".

CASE VI.

(224.) Given the three angles of a spherical triangle, to find the sides.

If A, B, C are the angles of the given triangle, and a, b, c its sides, then 180°-A, 180°-B, and 180°-C are the sides of its polar triangle, whose angles may be found by Case V. Then the supplements of those angles will be the sides a, b, c of the proposed triangle.

Ex. 1. In the spherical triangle ABC, the angle A=125° 34', B=98° 44', and C=61° 53'. Required the sides. The sides of the polar triangle are

54° 26', 81° 16', and 118° 7'. From which, by Case V., the angles are found to be

134° 6' 21", 41° 28' 17", and 53° 34' 47". Hence the sides of the proposed triangle are AB=45° 53' 39', BC=138° 31' 43', and AC=126° 25' 13".

Ex. 2. In the spherical triangle ABC, the angle A=109° 55', B=116° 38', and C=120° 43'. Required the sides.

a= 98° 21' 20". Ans. {b=109° 50' 10".

c=115° 13' 7"

TRIGONOMETRICAL FORMULÆ.

B

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D

A

(225.) Let ABC be any spherical triangle, and from the angle B draw the arc BD perpendicular to the base AC. Represent the sides of the triangle by a, b, c, and the segment AD by 2; then will CD be equal to b-x. By Theorem III., Cor., 1, cos. C : cos. a :: cos. X : cos. (6-2)

cos. b cos. x+sin. b sin. x

R (Trig., Art. 72), formula (4).

Whence

R cos. a cos. x=cos. b cos. C cos. X+sin. b cos. c sin. X; or, dividing each term by cos. x, and substituting the value of

C

:: COS. X:

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R' cos. a=R cos. b cos. ctsin. 6 cos. c tan. X.

But by Theorem II., Cor. 2, we have

R cos. A

cos. A sin. c tan. x=

(Art. 28). cot. C

COS. C

(3)

COS. A

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Hence R' cos. a=R cos. b.cos. ctsin. b sin. c cos. A,

=(1) from which all the formulæ necessary for the solution of spherical triangles may be deduced. In a similar manner we obtain

R’ cos. b=R cos. a cos. ctsin. a sin. c cos. B, (2)

R’ cos. C=R cos. a cos. b+sin. a sin. 6 cos. C. These equations express the following Theorem:

The square of radius multiplied by the cosine of either side of a spherical triangle, is equal to radius into the product of the cosines of the two other sides, plus the product of the sines of those sides into the cosine of their included angle. (226.) From equation (1) we obtain, by transposition,

R' cos. a-R cos. E cos. C

sin, b sin. C a formula which furnishes an angle of a triangle when the three sides are known. If we add R to each member of this equation, we shall have

R’ cos. a+R sin. b sin. C-R cos. b cos. C
R+cos. A=

sin. b sin. c

2 cos. ?LA But, by Art. 74, R+cos. A

R
And, by Art. 72, formula (2), by transposition,

R sin. b sin. c-R cos. b cos. c=-Rocos. (b+c).
Hence, by substitution, we obtain

cos. "LA_R®(cos. a-cos. (b+c))
R

sin. b sin. C
2R sin. i(a+b+c) sin. 3(b+c-a)

sin. 6 sin. C by Art. 75, formula (4).

If, then, we put s=i(a+b+c), that is, half the sum of the sides, we shall find

sin. s sin. (s-a) cos. JA=RV

(4)

sin. b sin. C By subtracting cos. A from R instead of adding, we shall obtain, in a similar manner,

-

(5) sin. b sin. c

sin. BA=RV sin. (s—b) sin. (s—c)

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