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(263.) Problem. Three of the six parts of an oblique-angled spherical triangle being given, to solve the triangle.


Let the three sides, S, S', S", be given, to find the three angles, A, A', A", of the triangle.

cos S"

Then, (Art. 230. I.) cos A" =

cos Scos S' sin $ sin $' sin S sin s

cos S"

- cot S.cot S

sin S sin

(Introd. 17.)

Find, therefore, two numbers m and n, such that

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cos S"

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And, from this last equation, the angle A" is determined. Or, since (Art. 230. I.)

cos Scos S cos A" =

therefore (Introd. 30.) sin S sin S'

sin Ssin S' - cos S" +cos Scos S' 2 sin? {A" = 1 - cos A" =

sin S sin S' cos (S' - S) - cos S"

(Introd. 28.)

sin S sin S'
2 sin } (S” – S'+S) sinį (S”+$' – S)

(Introd. 31.) sin S.sin S'

Hence, if P be put for the semi-sum of the three sides of the triangle,

sin (P-S') sin (P - S) sino į A" =

sin S sin S'

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In the same manner, by adding the value of cos A” to unity, it may be shewn that, cos? į A"

sin (P-S") sin P

sin S sin S' And, if the former of the two equations, last deduced, be divided by the latter, then, (Introd. 17.) sin' I A"

sin (P-S'). sin (P-S)
=tan ? A=
cos' L A"

sin P.sin (P-S")

After having, by one of the above four methods, found the angle A", the other angles may readily be found, by Art. 234. IX: and, since (Art. 129.) the greater side is opposite to the greater angle, the order of the angles, according to their magnitude, will be known; and it will appear, also, whether any one of them be obtuse.

The problem proposed, in this first case, is always possible, if any two of the three given sides be greater than the third, and if each of them be less than 180° : neither is there any ambiguity, in this case, in the answer obtained


Let the three angles, A, A', A", be given, to find the three sides, S, S', S", of the triangle. Then (Art. 231. II.)

cos A" +cos A cos A cos S" =

sin A sin A

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And, if



tan y, then

sin A"

* It is manifest, that to this first Case, of the solution of obliqueangled spherical triangles, may be reduced the following problem : “ The inclinations, to one another, of three straight lines, that are not in the same plane, being given, to find the angle made by the projections, of any two of the straight lines, on a plane that is perpendicular to the third."


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First, therefore, find y, from the equation, tan y= cos A cos A

; and S" will then be known from the final sin A equation.

Or, by finding the sum, and the difference, of unity, and the value of cos S", as it is given in Art. 231. II. and then, by dividing the resulting equations, as in the preceding case, if p be put for the semi-sum of the three given angles,

- cos p cos (p - AM) sin S" =

sin A sin A

i cos” į S" =

cos. (P- A) cos (P-A')

sin A sin A

tan’ į S" =

- cos p.cos (p— A') cos (P- A) cos (p - A)

Which expressions might also have been deduced from the solutions of the first case, by means of the polar triangle.

When S" has been found, the other sides may be computed, by Art. 234. IX.

In this second case, the problem proposed is always possible, if the aggregate of the three given angles be less than six right angles, and greater than two right angles : and the problem admits of only one answer.



any two sides as S and S', and the included angle A", be given, to find the other parts of the triangle. Then (Art. 230. I.)

cos S" =cos Scos S' + sin S sin S' cos A".

Find, therefore, two numbers m and n, such that

m = cos S cos s', and n = sin S sin S'cos A" ; then, cos S" = m + n. Or, since cos S" =cos S cos S' + sin S sin S'cos A"

=cos S(cos S' + cos Á" tan Ssin S') (Introd. 17.) if, therefore,

tan x = cos A" tan S,
cos S" = cos Š (cos S'+tan x sin S')

cos S

(cos S' cos x+sin x sin S) (Introd. 17.)


cos S. cos (S' - x)

.* (Introd. 28.)


* It is evident, from Art. 232. VIII. that x is the base of a rightangled spherical triangle, of which S is the hypotenuse, and A" an angle at the base. So that the problem is solved by drawing an arch, at right angles to one of the known sides, from the opposite angle.

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