Question

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

Answer 1

As given that,

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

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

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

$\angle B=2\angle A$

$2\angle A-\angle B=0$ … (i)

As the sum of the measures of all angles of a triangle is 180°. Therefore,

$\angle A+\angle B+\angle C={180}^{\circ }$

$\angle A+\angle B+3\angle B={180}^{\circ }$

$\angle A+4\angle B={180}^{\circ }$ ... (ii)

Multiplying equation (i) by 4, we gets

$\Rightarrow 8\angle A-4\angle B=0$ … (iii)

Adding equations (ii) and (iii), we obtain

$\Rightarrow 9\angle A={180}^{\circ }$

$\Rightarrow \angle A=\frac{{180}^{\circ }}{9}={20}^{\circ }$

From equation (ii), we obtain

$\Rightarrow {20}^{\circ }+4\angle B={180}^{\circ }$

$\Rightarrow 4\angle B={160}^{\circ }$

$\Rightarrow \angle B={40}^{\circ }$

$\Rightarrow \angle C=3\angle B$

$\Rightarrow 3\times {40}^{\circ }={120}^{\circ }$

$\therefore \angle A,\angle B,\angle C$ are ${20}^{\circ },{40}^{\circ },{120}^{\circ }$