Question

help please! show work

Answer 1

your answers are
A = 35.7°
B = 67.6°
C = 76.7°

cosine law

$a^2 = b^2 + c^2 -2bc \cos A \\ -2bc \cos A = a^2 - b^2 - c^2 \\ \\ \cos A = \dfrac{a^2 - b^2 - c^2}{-2bc} \\ \\ A = \cos^{-1}\left[ \dfrac{a^2 - b^2 - c^2}{-2bc} \right] \\ \\ A = \cos^{-1}\left[ \dfrac{12^2 - 19^2 - 20^2}{-2(19)(20)} \right] \\ \\ A = 35.723697$

A = 35.723697
sine law for the rest of the angles

$\displaystyle \frac{\sin B}{b} = \frac{\sin A}{a} \\ \\ \sin B = \frac{b \sin A}{a} \\ \\ B = \sin^{-1} \left[ \frac{b \sin A}{a} \right] \\ \\ B = \sin^{-1} \left[ \frac{19 \sin 35.723697 }{12} \right] \\ \\ B \approx 67.58886795$

B = 67.58886795
All angles in triangle sum to 180 so find C with that

A + B + C = 180
C = 180 - A - B
C = 180 - 35.723697 - 67.58886795
C = 76.7°