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