In $\triangle A B C, \angle A=3 \angle B$ and $\angle B=2 \angle C$. The measure of $\angle B$ is
(A) $10^{\circ}$
(B) $20^{\circ}$
(C) $30^{\circ}$
(D) $40^{\circ}$
(E) $60^{\circ}$
In $\triangle A B C, \angle A=3 \angle B$ and $\angle B=2 \angle C$. The measure of $\angle B$ is
(A) $10^{\circ}$
(B) $20^{\circ}$
(C) $30^{\circ}$
(D) $40^{\circ}$
(E) $60^{\circ}$

## Solution

Since we have a triangle,

$$
\begin{aligned}
\angle A+\angle B+\angle C & =180^{\circ} \\
3(\angle B)+\angle B+\frac{1}{2}(\angle B) & =180^{\circ} \\
\frac{9}{2}(\angle B) & =180^{\circ} \\
\angle B & =40^{\circ}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-259.jpg?height=241&width=561&top_left_y=1755&top_left_x=1346)

ANSWER: (D)