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anonymous
 one year ago
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
 one year ago
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)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443024721462:dw

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0well this might be the stupid question well for law of sines why are u using cosine ? just asking
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