• anonymous
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)
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
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