Question

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

Answer 1

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)

We know that the sum of the measures of all angles of a triangle is ${180}^{\circ }$. 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 obtain

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

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

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

$\angle A$ = ${20}^{\circ }$

Putting the value of $\angle A$ in equation (i), we get

$\angle B$ = 2 $\angle A$

$\angle B$ = 2 $\times$ ${20}^{\circ }$ = ${40}^{\circ }$

$\angle C$ = 3 $\angle B$ = ${120}^{\circ }$

Hence, required angles of triangle is ${20}^{\circ }$, ${40}^{\circ }$, ${120}^{\circ }$