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 = (a^2 - b^2 - c^2)/(-2bc) \\ \\ A = \cos^(-1)\left[ (a^2 - b^2 - c^2)/(-2bc) \right] \\ \\ A = \cos^(-1)\left[ (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 (\sin B)/(b) = (\sin A)/(a) \\ \\ \sin B = (b \sin A)/(a) \\ \\ B = \sin^(-1) \left[ (b \sin A)/(a) \right] \\ \\ B = \sin^(-1) \left[ (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°