`A = 120^@, b = 6, c = 7` Use the Law of Cosines to solve the triangle. Round your answers to two decimal places.

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`