Question

help please. show work

Answer 1

ans:
A = 93.2°
B = 33.5°
C = 53.4°

cosine law

$a^2 = b^2 + c^2 -2bc \cos A \\ \\ -2bc \cos A = a^2 - b^2 - c^2 \\ \\ \cos A = \dfrac{a^2 - b^2 - c^2}{-2bc} \\ \\ A = \cos^{-1}\left[ \dfrac{a^2 - b^2 - c^2}{-2bc} \right] \\ \\ A = \cos^{-1}\left[ \dfrac{21.9^2 - 12.1^2 - 17.6^2}{-2(12.1)(17.6)} \right] \\ \\ A = 93.15479886$

A = 93.15479886
sine law for the rest of the angles

$\displaystyle \frac{\sin B}{b} = \frac{\sin A}{a} \\ \\ \sin B = \frac{b \sin A}{a} \\ \\ B = \sin^{-1} \left[ \frac{b \sin A}{a} \right] \\ \\ B = \sin^{-1} \left[ \frac{12.1 \sin 93.15479886 }{21.9} \right] \\ \\ B \approx 33.4819378$

B = 33.4819378
All angles in triangle sum to 180 so find C with that

A + B + C = 180
C = 180 - A - B
C = 180 - 93.15479886 - 33.4819378
C = 53.4°