## Precalculus (6th Edition)

$\left| CB\right| \approx 7.032$ $\angle C\approx 21.449$ $\angle B\approx 34.551$
From the laws of cosines $\left| CB\right| ^{2}=5^{2}+3^{2}-2\times 5\times 3\times \cos 121\approx 49.451\Rightarrow \left| CB\right| \approx 7.032$ From the law of sines $\dfrac {\left| CB\right| }{\sin 121}=\dfrac {3}{\sin \angle C}\Rightarrow \sin \angle C=\dfrac {3\sin 121}{\left| CB\right| }\approx 0.365\Rightarrow \angle C\approx 21.449$ Internal angles of a triangle is $180^0$ So $\angle A+\angle B+\angle C=180\Rightarrow \angle B=180-\left( \angle A+\angle C\right) \approx 180-\left( 121+21.449\right) \approx 34.551$