Triangles using law of cosines?

Question

A triangle with, A=60 degrees, b=6, and c=9. Find B and C. I already found that a= 3 square root of 7.

Answer 1

using the law of cosines to find angle B...

b^2=a^2+c^2-2*a*c*cosB

36=63+81-54√7cosB

36=144-54√7cosB

36-144=-54√7cosB

-108=-54√7cosB

-108/(-54√7)=cosB

2/√7=cosB

2√7/7=cosB

(2*2.64575)/7=cosB

5.2915/7=cosB

0.7559286=cosB

using inverse cos finder on your calculator or on the web...

B=40.893425º or about 40.9º

60+40.893425=100.893425

180-100.893425=79.106575º

C=79.106575º or about 79.1º