Question 535168: solve triangle with A=70 degrees, b= 30, c= 51
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A=70 degrees, b= 30, c= 51
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Use Law of Cosines:
a^2 = b^2 + c^2 -2bc*cos(A)
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a^2 = 30^2 + 51^2 - 2*30*51*cos(70)
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a^2 = 2454.42
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a = 49.54
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Now use the Law of sines to find angle B.
sin(B)/b = sin(A)/a
sin(B)/30 = sin(70)/49.54
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sin(B) = (30/49.54)*sin(70) = 0.5691
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B = arcsin(0.5691) = 34.68 degrees
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Then C = 180-(A+B)
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Cheers,
Stan H.
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