5. In the right-angled spherical triangle ABC, given the base AB=12° 30', and the opposite angle C=24° 45'; required the other parts of the triangle

Answer, AB=28° 44' 37", BC=31° 7' 48", C=68° 27' 53. 6. In the right-angled spherical triangle ABC, given AC=45° 15', and the angle C=63° 20°; required the rest ?

Answer, AB=54° 44', B=51° 0'50", BC=66° Ở 56". 7. In the right-angled spherical triangle ABC, given the leg AB=54° 30', and the adjacent angle B=44° 50'? required the rest?

Answer, LC=650 49' 53", AC=38° 59' 11", BC=63° 10' 4'. 8. In the right-angled spherical triangle ABC, given the sides about the right-angle respectively 55° 28', and 63° 15'; required the hypothenuse and angles ?

Answer, BC=75° 13' 2", C=67° 27' 1", B=58° 25' 46". 9. Given the two legs of a right-angled spherical triangle respectively 15° 20', and 31° 57 ; required the hypothenuse BC, and the angles B and C.

Answer, C=67° 1' 22", B=27° 23' 27", BC=35° 5' 3". 10. In the right-angled spherical triangle ABC, given the angle B=64° 40', and the angle C=46° 15; required the sides ?

Answer, AC=40° 5'6", AB=53° 40' 36", BC=630 3' 5". Jl. In the spherical triangle ABC, right-angled at A, given the angles B and C respectively 72° 20', and 24° 50' ; required the sides?

Answer, BC=46° 30' 31", AC=17° 44' 21", AB=13° 43' 48". 12. In the spherical triangle ABC, right-angled at A, given AB=10° 39' 40", and the adjacent angle B=23° 27' 42" ; required the remaining angle C, and the sides AC and BC ?

Answer, BC=11° 35' 40", AC=4° 35' 26", C=66° 58' 1".


98. Here there are as many varieties, as there are different combinations among the six parts of a triangle, taken four by four. But there are fifteen of these combinations (Page 111), and each admits of three different cases. Consequently there are 45 problems to solve relating to oblique-angled triangles. But as the resolution of one frequently includes that of several others, they may be reduced to the twelve following cases :

I. 99. In an oblique-angled spherical triangle ABC, given the angles B and A, and the side opposite to the angle A, to find the side AC opposite to the other angle,

Let A=61° 25', B-82° 36', BC=59° 40'. C
By (82) we have the proportion

sin A : sin B::sin BC : sin AC.
But arith. co. log sin A (610 25) = 10.056445

log sin B (82° 36') = 9.996368
log sin BC (59° 40') = 9.9360628
log sin AC

9.988875 Whence AC=77° 5', or 102, 55'. Nor can we decide between these two results, unless we previously know of what species the side we seek for must be.


C 100. Given the two angles A and B, with the side BC, opposite to the angle A, to A А

А find the third an


B gle.

Let fall the perpendicular arc CD, and in the triangle BCD we shall have (90)

R: cos BC :; tang B : cot BCD and then (86) cos B : cos A :: sin BCD: sin ACD.

Adding these two angles, or subtracting one from the other, according as the given angles A and B are of the same or of different species, we shall have the angle C required. Let A=610 25', B=82° 36', BC=59° 40', and we shall have neg. log radius

log cos BC (59° 40) 9.703317
log tang B (82° 36') = 10.886467
log cot BCD

= 10.589784
Whence the angle BCD=14° 25' (by 78).
co. log cos B (82° 36') = 10.890099

log cos A (61° 25') 9.679824
log sin BCD (140 25') = 9.396150
log sin ACL

9.966073 Consequently the angle ACD=670 39', or 1120 21; which gives 820*4', or 1266 46', for the required angle ACB; and nothing in the question enables us to determine which value is to be preferred.

= 10.


101. Given two angles A and B, with the opposite side BC, as in the preceding case, to find the side contained between these two angles (see preceding figure).


= 10.

The perpendicular arc CD forms the two right-angled triangles ACD and DCB, the last of which gives (87)


: cos B :: tang BC : tang BD Besides we have (89)

tang A : tang B:: sin BD, sin AD. Consequently we shall have AB=AD+DB, according as the anglęs given are of the same or of different species. Preserving the same data as in the preceding case, neg. log radius

log cos B (82° 36') 9.109901
log tang BC (59° 40') = 10.232745
log tang BD

Therefore BD-12° 25'.
arith. co. log tang A (61° 25') =

189.736269* log tang B (82° 36') 10.886467 log sin BD (12° 25') 9.332478 log sin AD

9.955214 Hence the side AD=640 25', or 1150 35'; and as in this example the angles A and B are of the same species, we must add the two segments BD and AD; consequently the side AB must be either 76° 50', or 128o.



102. Given the two angles A and C, with the side on which these angles are formed, to find the third angle B (see preceding figure.)

First, (90) R : cos AC:: tang A : cot ACD
And then, (86) sin ACD: sin BCD:: cos A : cos B.

The angle BCD is found by subtracting ACD from ACB, when the arc falls perpendicularly within the triangle ; but when this arc falls without, we must subtract the angle ACB from the angle ACD, in order to find BCD.


103. Given as before, the two angles A and C, with the contained side AC, to find the two other sides, BC for example.

The log taug. 61° 25'=10.263731, subtracting each figure from 9. except the first on the right, as before explained, we obtain 89.756269; and prefixing the negative unit as usual, the arithmetical equivalent becomes as above, 189.736969. The arithmetical equivalent for negative numbers is of considera. ble use in many arithmetical operations. See Nicholson's Treatise on Invo lution and Evolution.

First, (90) R: cos AC :: tang A : cot ACD
And then, (86) cos BCD: cos ACD :: tang AC: tang BC.


104. Suppose any two sides given, AC and BC for example, with one of the angles opposite to those sides, such as the angle A; to find the angle B, opposite to the other side.

This problem may be resolved by the B well known proportion

sin BC: sin AC :: sin A : sin B If we assign to the given sides and angle A, the same values as in the preceding problems, we shall find the angle B by the following calculation : arith; co. log sin BC (59° 40') = 10.063938

log sin AC (779 5') 9.988869
log sin A (61° 25') 9.943555
log sin B

9.996362 The angle B is therefore 82° 36', as we have before supposed it, or 97° 24', as we might also have supposed it.

This problem falls within the first case ; they differ only by an inversion of the terms in the proportion which resolves both.




105. The same things being given, required the third side AB.

(87) R: cos A:: tang AC: tang AD

(83) cos AC : cos BC :: cos AD: cos BD. By the first of

с these two proportions we find AD, and by the second BD: therefore we A

A shall have AB=


AD + BD,
cording to the position of the perpendicular arc.
If we had referred the known quanti-

ties in the triangle ABC to the supple-
mental triangle, we should merely have
to resolve this problem (99). To find the
third angle when we know the other two
angles, and one of the sides opposite to
them. For this problem once resolved,

D we should find the value of the required G

E side AB, by taking the supplement of • this third angle.



106. On the same suppositions, to find the angle C, cuntained between the given sides AC and BC..

(90) Rios iu :: tang Akot ACD, and, (88' t ng BC :tang AĞ: 645 ACD: BCD.

Whenc deduce / ('BACD+BCD, according, as the sides AC and BC are of the same A

A or of different


D species.

B D The supplemental triangle is equally proper to resolve this problem by reducing it to the third case.


107. Given the two sides AC and AB, and the angle A, contained between them, to find the other side BC. (Preceding figure).

(87) R: cos A :: tang AC : tang AD
(83) cos AD: cos BD :: cos AC: cos BC.


108. The same things being given, to find one of the other two angles, the angle B, for example. (Preceding figure.)

(87) R: cos A :: tang AC : tang AD
(89) sin BD : sin AD::tang A: tang B.


109. If the three sides were given, how should we find one of the three angles, A for example? (Preceding figure). We know (88) that A+B

AC-BC AD-DB lang

::tang. : tang

2 This first proportion will give us the segment AD, and the following one will give the angle A.

(87) tang AC :tang AD::R: cos A. Resuming the values we have before supposed, we shall have

: tang AC+BC

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