Find angle B in triangle ABC if a is 10 b 16 and c 12?

We can use the law of cosines here. ( remember, side b is opposite angle B) DEGREE MODE!

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

16^2 = 10^2 + 12^2 - 2(10)(12) cos(B)

256 = 244 - 240(cos B )

12 = -240(cos B )

-0.05 = cosB

arcos(-0.05) = B

B = 93 degrees