Question 454733
find the smallest angle, to the nearest degree, in the triangle whose sides are a = 5", b =  8" and c =  12"
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Use the Law of Cosines
Cos(A) = [b^2+c^2-a^2]/[2bc]
= [8^2+12^2-5^2]/(2*8*12) 
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= 0.953125
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A = cos^-1(0.953125) = 17.61 degrees
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Use the same pattern to find angle B.
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Then angle C = 180-{ang A + ang B]
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Cheers,
Stan H.