posted by MW .
In triangle ABC, b=14, c=15, and cos a= 3/5.
b) Sin B
In this notation, small letters are usually used to represent sides, and capital letters to indicate the angles at the vertex,
so yours : b=14, c=15, cos A = 3/5
direct application of the cosine law
a^2 = 14^2 + 15^2 - 2(14)(15)cos A
once you have a
use the Sine Law to find angle B
ok. I have side A.
How do you find angle B?
sin B/b = sin A/a