Question

Say true or false.

In a $\Delta ABC,\angle C=3\angle B=2(\angle A+\angle B)$, then angles are ${20}^{\circ },{40}^{\circ },{100}^{\circ }.$

• A. True
• B. False

Answer 1

The correct option is B False

We know that the sum of the angles of a triangle is ${180}^{\circ }$
$\therefore \frac{\angle B}{2}+\angle B+3\angle B={180}^{\circ }$
$[\because 3\angle B=2(\angle A+\angle B)\Rightarrow \angle B=2\angle A\Rightarrow \angle A=\frac{\angle B}{2}$ and $\angle C=3\angle B$]
$\Rightarrow 4\angle B+\frac{\angle B}{2}={180}^{\circ }$
$\Rightarrow \frac{9\angle B}{2}={180}^{\circ }$
$\Rightarrow 9\angle B={360}^{\circ }$
$\Rightarrow \angle B={40}^{\circ }$
$\therefore \angle A=\angle \frac{B}{2}\Rightarrow \angle A=\frac{40}{2}\Rightarrow \angle A={20}^{\circ }$
Also, $\angle C=3\angle B\Rightarrow \angle C=3\left(40\right)\Rightarrow \angle C={120}^{\circ }$
$\therefore \angle A={20}^{\circ },\angle B={40}^{\circ },\angle C={120}^{\circ }$
The correct answer is B.