Question

Solve the triangle PLEASEEEEEE

Answer 1

Part 1

$a=\sqrt{12^2 + 21^2 -2(12)(21) \cos 35^{\circ}} \approx \boxed{13}$

Part 2

$\frac{\sin B}{b}=\frac{\sin A}{a}\\\\\sin B=\frac{b\sin A}{a}\\\\\sin B=\frac{12 \sin 35^{\circ}}{\sqrt{12^2 +21^2 -2(12)(21) \cos 35^{\circ}}}\\\\B=\sin^{-1} \left(\frac{12\sin 35^{\circ}}{\sqrt{12^2 +21^2 -2(12)(21) \cos 35^{\circ}}} \right)\\\\B \approx \boxed{32^{\circ}}$

Part 3

$\angle C=180^{\circ}-\angle A -\angle B=\boxed{113^{\circ}}$