Question

Math! :)

Answer 1

116.244 I believe but I'm not even sure how I got that... -_- lmao

Answer 2

The law of cosines is what you want to use
\[c = \sqrt(a^2+B^2-2abcos(y))\]

Whole thing is under square root btw

Answer 3

\[c^2=a^2+b^2-2abcos(y)\]
\[c^2-(a^2+b^2)=-2abcos(y)\]
\[(c^2-(a^2+b^2))/-2ab=\cos(y)\]
\[\cos^{-1}((c^2-(a^2+b^2))/-2ab)=y\]
If you plug in values for c, a and b you will get y, which is the measure of angle C