In triangle ABC, b=14, c=15, and cos a= 3/5.

Find
a)side a
b) Sin B

Help please?

3 answers

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