In triangle $A B C, \angle A$ is $21^{\circ}$ more than $\angle B$, and $\angle C$ is $36^{\circ}$ more than $\angle B$. The size of $\angle B$ is
(A) $20^{\circ}$
(B) $41^{\circ}$
(C) $62^{\circ}$
(D) $46^{\circ}$
(E) $56^{\circ}$
We are told that $\angle A=\angle B+21^{\circ}$ and $\angle C=\angle B+36^{\circ}$.

Since the sum of the angles in a triangle is $180^{\circ}$, then

$$
\begin{aligned}
\angle A+\angle B+\angle C & =180^{\circ} \\
\angle B+21^{\circ}+\angle B+\angle B+36^{\circ} & =180^{\circ} \\
3(\angle B)+57^{\circ} & =180^{\circ} \\
3(\angle B) & =123^{\circ} \\
\angle B & =41^{\circ}
\end{aligned}
$$

ANSWER: (B)

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