The problem above uses a combination of sine and cosine law of triangle to solve for the m∠B.
Given:
m∠A = 60°, b = 9 and c = 6
Cosine Law:
a^2=b^2+c^2-2bccosA
a^2=(9)^2+(6)^2-2(9)(6)cos(60)
a^2=81+36-54
a^2= 63
a=[tex] \sqrt{63} [/tex]
Sine Law
[tex] \frac{sin(A)}{a} = \frac{sin (B)}{9} [/tex]
sin B= 9/ [tex] \sqrt{63} [/tex] sin 60
sin B= 0.98198
B= sin ^-1 (0.98198) = 79.10
Therefore, m∠B is equal to 79.10
