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:
$C = \checkmark$ answer
By 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
$\cos C = x$
$C = \checkmark$ answer
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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