Precalculus, Chapter 6, 6.2, Section 6.2, Problem 17

The given in the triangle are A=120^o , b=6 and c=7 . To solve for the values of a, B and C, let's apply Cosine Law.
For side a:
a^2=b^2+c^2-2*b*c*cosA
a^2=6^2+7^2-2*6*7cos(120^o)
a^2=36+49-84cos(120^o)
a^2=127
a=sqrt127
a=11.27
For angle B:
b^2=a^2+c^2-2*a*c*cosB
6^2=(sqrt127)^2+7^2-2*sqrt127*7*cosB
36=176-14sqrt127cosB
(36-176)/(-14sqrt127)=cosB
cos^(-1)((36-176)/(-14sqrt127))=B
27.46^o=B
For angle C:
c^2=a^2+b^2-2*a*b*cosC
7^2=(sqrt127)^2+6^2-2*sqrt127*6*cosC
49=127+36-12sqrt127cosC
49=163-12sqrt127cosC
(49-163)/(-12sqrt127)=cosC
cos^(-1)((49-163)/(-12sqrt127)=C
32.54^o=C
Thus, the sides of the triangles are:
a=11.27
b=6
c=7
And its angles are:
A=120^o
B=27.46^o
C=32.54^o