Law of cosines precalc

Question

Given the triangle below, find the measure of angle 

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if side a =20 and the other 2 sides are 16 and 21. keep in mind that angle b and c are opposite of their sides.

Answer 1

b=AC=16, a=BC=20, c=AB = 21

sin A/ 20 = sin B/16 = sin C/21 are the three equations

c^2 = a^2 + b^2 - 2ab cos C

21^2 = 16^2 + 20^2 - 2(16)(20) cos C

441 = 256 + 400 - 640 cos C

441 = 656 - 640 cos C

(441-656)/(-640) = cos C

0.3359375 = cos C

C = 70.37+

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b^2 = a^2 + c^2 - 2ac cos B

20^2 = 16^2 + 21^2 - 2(16)(21) cos B

400 = 256 + 441 - 672 cos B

400 = 697 - 672 cos B

(400 - 697)/(-672) = cos B

0.44196428571428571428571.... = cos B

B = 63.77

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a^2 = b^2 + c^2 - 2bc cos A

16^2 = 20^2 + 21^2 - 2(20)(21) cos A

256 = 400 + 441 - 840 cos A

256 = 881 - 840 cos A

(256 - 881)/(-840) = cos A

0.7440476 = cos A

A=41.92