You have to use the Law of Cosines twice here. Let's find angle A first:
![21^(2)= (12)^(2)+(29) ^(2) -2(12)(29)cosA](https://img.qammunity.org/2019/formulas/mathematics/high-school/bt2h7gzo70iy2d5ydv1vz7nujkelqxwbd2.png)
. Doing a bit of simplifying you get
![441=144+841-696cosA](https://img.qammunity.org/2019/formulas/mathematics/high-school/fv9decc7mt7u88lfqoavqy9f92m5rvsjfm.png)
. Simplifying a bit more gives you -544=-696cosA. cosA = .78160919 and that A = 38.6. Actually, now that I think about it we can now move to the Law of Sines to find another angle because the triangle is not a right triangle. Just trust me. Let's find angle B now.
![(sinB)/(12)= (sin38.6)/(21)](https://img.qammunity.org/2019/formulas/mathematics/high-school/e4sabimmv9qsc8pnxuwdkd78f4cwjj7qib.png)
and that
![sinB= (12sin38.6)/(21)](https://img.qammunity.org/2019/formulas/mathematics/high-school/f3wt69lzp7oxle94bca07esgc4bue4vzuw.png)
. sinB = .3565026267 and that angle B is 20.9. Knowing those 2 angles we can find C: 180-38.6-20.9=120.5 and there you have it.