#### Answer

$ \left| CB\right| \approx 7.032$
$ \angle C\approx 21.449$
$\angle B\approx 34.551$

#### Work Step by Step

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
$\angle A+\angle B+\angle C=180\Rightarrow \angle B=180-\left( \angle A+\angle C\right) \approx 180-\left( 121+21.449\right) \approx 34.551$