### Instruction: In triangle $ABC,$ $a = 2,$ $b = 3,$ and $c = 4.$  Find $3 \angle A + 2 \angle B,$ in degrees.

Note: $a$ is the side length opposite $\angle A,$ etc.### Response: To solve this problem, I need to find the angles of the triangle in terms of $a, b, c.$
One way to do that is to use the cosine rule, which says that $c^2 = a^2 + b^2 - 2ab \cos C.$
If I rearrange this equation, I get $\cos C = \frac{a^2 + b^2 - c^2}{2ab}.$
Plugging in the given values, I get $\cos C = \frac{4 + 9 - 16}{12} = -\frac{1}{4}.$
Therefore, $C = \cos^{-1}(-\frac{1}{4}),$ which is approximately 104.48 degrees.
I can use the same formula to find $\cos A$ and $\cos B.$
I get $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{9 + 16 - 4}{24} = \frac{7}{8}.$
Therefore, $A = \cos^{-1}(\frac{7}{8}),$ which is approximately 28.96 degrees.
Similarly, I get $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{4 + 16 - 9}{16} = \frac{11}{16}.$
Therefore, $B = \cos^{-1}(\frac{11}{16}),$ which is approximately 46.57 degrees.
Now that I have the angles, I can find $3 \angle A + 2 \angle B.$
I get $3 \angle A + 2 \angle B \approx 3 \times 28.96 + 2 \times 46.57 \approx 179.01$ degrees.