Question

Find angles please please

Answer 1

You have to use the Law of Cosines twice here.  Let's find angle A first: $21^{2}= (12)^{2}+(29) ^{2} -2(12)(29)cosA$.  Doing a bit of simplifying you get $441=144+841-696cosA$.  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. $\frac{sinB}{12}= \frac{sin38.6}{21}$ and that $sinB= \frac{12sin38.6}{21}$.  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.

Answer 2

Law of Cosines is what you need for this problem
To find ∡A: 21² = 12² + 29² - 2(12)(29)cos A
When you solve this equation you get m∡A ≈ 38.6°

To find ∡B: 12² = 21² + 29² -2(21)(29)cos B
When you solve this equation you get m∡B ≈ 20.9°

To find ∡C: 180° - (20.9° + 38.6°) = 120.5°