Question

1.) Isosceles triangle is similar to the Isosceles spherical triangle? justify your answer.
2.) An isosceles spherical triangle has an angle A=B= 54° and side b = 82°. Find the measure of the third angle.
3.) Determine the value of angle B of an isosceles spherical triangle ABC whose given parts are b=c= 54°28’ and a = 92°30’
4.) Solve for the side b of a right spherical triangle ABC whose parts of an isosceles spherical triangle are a = 46°, b = 75° and C = 90°.

Answer 1

1) Isosceles spherical triangle - A spherical triangle is a figure formed on the surface of a sphere by three great circular arcs intersecting pairwise in three vertices. The spherical triangle is the spherical analog of the planar triangle.

Isosceles triangle - An isosceles triangle is a triangle that has two sides of equal length.

The above two triangle is similar as they have a common property of two sides being equal.

2) To solve an unknown angle C, convert the isosceles triangle to a right spherical triangle by constructing a $90\degree$ at the midpoint of the base.

Using Napier's rule :(for triangle ACD)

$\sin b = \tan x* \tan a$

$\cos b = 1 /(\tan x * \tan a)$

$\tan x = 1/(\cos b * \tan a)$

$\tan x = 1/ (\cos 82\degree * \tan 54\degree)$

$\tan x = 5.22$

so x = $79.156\degree$

Thus solving for C:

C = 2x

= 2 * 79.156

$=$ 158°18’43”

3) Let us consider the below figure the value of angle B of an isosceles spherical triangle ABC

$\sin$ co-B $= \tan$ a/2 * $\tan$ co -C

$\cos B = \tan a/2 * 1/\tan c$

$\cos B = \tan 92°30’/2 * 1/\tan 54°28’$

$\therefore B = 41°45’$

4) Let us consider the below figure to find out the the side b of a right spherical triangle ABC

$\sin$ co-C = $\cos a * \cos b$

$\cos c =$ $\cos a * \cos b$

$\cos b = \cos c/\cos a$

$\cos b = \cos 75\degree/\cos 46\degree$

$\therefore b = 68°07′$