In a $\triangle ABC, \angle C = 3 \angle B = 2(\angle A + \angle B)$. Find the three angles.

Given:

In a $\triangle ABC, \angle C = 3 \angle B = 2(\angle A + \angle B)$.

To do:

We have to find the three angles.

Solution:

We know that,

The sum of the angles in a triangle is $180^o$

This implies,

$\angle A+\angle B+\angle C=180$......(i)

$\angle C=3 \angle B=2(\angle A+\angle B$
$3 \angle B=2(\angle A+\angle B)$

$2 \angle A=3 \angle B-2 \angle B$

$2 \angle A=\angle B$

$\angle A=\frac{\angle B}{2}$

Substitute the above in (i),

$\frac{\angle B}{2}+\angle B+3 \angle B=180^o$

$\frac{\angle B}{2}+4 \angle B=180^o$

$\angle B(\frac{1+8}{2})=180^o$

$\angle B \times \frac{9}{2}=180^o$

$\angle B=\frac{180^o\times2}{9}=40^o$

\end{array}
This implies,

$\angle C=3(40^o)=120^o$

$\angle A=\frac{40^o}{2}=20^o$

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Updated on: 10-Oct-2022

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