Question 115264: please help me understand these
in triangle ABC given that sides b=10,c=8, angle A=60degrees, find side a?
in triangle ABC given that sides a=4,b=8 and c=10, find cos B where B is the angle opposite to side b?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Law of Cosines Formula:
a^2 = b^2 + c^2 -2bc(Cos(A))
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in triangle ABC given that sides b=10,c=8, angle A=60degrees, find side a?
a^2 = 10^2 + 8^2 -2*10*8*Cos(60)
a^2 = 164 -160*(1/2)
a^2 = 84
a = 2sqrt(21)
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in triangle ABC given that sides a=4,b=8 and c=10, find cos B where B is the angle opposite to side b?
Cos(B) = [a^2+c^2-b^2]/(2ac)
Cos(B) = [16+100-64]/(2*4*10)
Cos(B) = 52/80 = 13/20
B = cos^-1(13/20)
B = 49.46 degrees
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Cheers,
Stan H.
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