An isosceles spherical triangle has angle A=B = 54° , and side b = 82° find the measure of the third angle

$\dfrac{\sin a}{\sin A} = \dfrac{\sin b}{\sin B}$

$a = \checkmark$

With side "a" already known, you may use Napier's analogy or you may use simultaneous equations of cosine law for sides and cosine law for angles.

By Napier's analogy:

$C = \checkmark$ answer

By simultaneous equations

$\cos C - \sin A ~ \sin B ~ \cos c = -\cos A ~ \cos B$

$x - (\sin A ~ \sin B)y = -\cos A ~ \cos B$ ← Equation (1)

Cosine Law for Sides $\cos c = \cos a ~ \cos b + \sin a ~ \sin b ~ \cos C$

$\sin a ~ \sin b ~ \cos C - \cos c = -\cos a ~ \cos b$

$(\sin a ~ \sin b)x - y = -\cos a ~ \cos b$ ← Equation (2)

Solve for x $\cos C = x$

Great from you sir. I can really learn a lot from here. Mechanical engineering student

Welcome to MATHalino.com sir Boaz. Enjoy your stay.

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$\dfrac{\sin a}{\sin A} = \dfrac{\sin b}{\sin B}$

$a = \checkmark$

With side "a" already known, you may use Napier's analogy or you may use simultaneous equations of cosine law for sides and cosine law for angles.

By Napier's analogy:

$C = \checkmark$

answerBy simultaneous equations

$\cos C = -\cos A ~ \cos B + \sin A ~ \sin B ~ \cos c$

$\cos C - \sin A ~ \sin B ~ \cos c = -\cos A ~ \cos B$

$x - (\sin A ~ \sin B)y = -\cos A ~ \cos B$ ← Equation (1)

Cosine Law for Sides

$\cos c = \cos a ~ \cos b + \sin a ~ \sin b ~ \cos C$

$\sin a ~ \sin b ~ \cos C - \cos c = -\cos a ~ \cos b$

$(\sin a ~ \sin b)x - y = -\cos a ~ \cos b$ ← Equation (2)

Solve for x

$\cos C = x$

$C = \checkmark$

answerGreat from you sir. I can really learn a lot from here. Mechanical engineering student

Welcome to MATHalino.com sir Boaz. Enjoy your stay.

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