if 1 side of a triangle is know and all 3 angles, one could use the law of sine to calculate the other 2 sides.
I've come up with a different (uglier) way. I'm wondering why it can't be converted to the law of sine.

Here's my proof:
a, b, c are the sides.
A, B, C are the angles.
a = b*cos(C) + c*cos(B)
b = a*cos(C) + c*cos(A)
c = a*cos(B) + b*cos(A)
let's calculate b when a, A, B, C are known.
b = a*cos(C) + (a*cos(B) + b*cos(A))*cos(A)
b = a(cos(A) * cos(B) + cos(C)) + b*cos(A)^2
b - b*cos(A)^2 = a(cos(A) * cos(B) + cos(C))
b = a(cos(A) * cos(B) + cos(C)) / (1 - cos(A)^2)

Question

if 1 side of a triangle is know and all 3 angles, one could use the law of sine to calculate the other 2 sides.
I've come up with a different (uglier) way. I'm wondering why it can't be converted to the law of sine.

Here's my proof:
a, b, c are the sides.
A, B, C are the angles.
a = b*cos(C) + c*cos(B)
b = a*cos(C) + c*cos(A)
c = a*cos(B) + b*cos(A)
let's calculate b when a, A, B, C are known.
b = a*cos(C) + (a*cos(B) + b*cos(A))*cos(A)
b = a(cos(A) * cos(B) + cos(C)) + b*cos(A)^2
b - b*cos(A)^2 = a(cos(A) * cos(B) + cos(C))
b = a(cos(A) * cos(B) + cos(C)) / (1 - cos(A)^2)

anonymous · Answer

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nnesha · Answer

well this might be the stupid question 
well for law of sines why are u using cosine ?
just asking