Question

In a $△ABC$, $\angle C=3\angle B=2(\angle A+\angle B)$. The three angles will be _________.

• A. ${20}^o,{40}^o,{120}^o$
• B. ${30}^o.{60}^o,{90}^o$
• C. ${45}^o,{45}^o,{90}^o$
• D. ${110}^o,{40}^o,{50}^o$

Answer 1

The correct option is A ${20}^o,{40}^o,{120}^o$

We know that in a $△$, the sum of all the interior angles is ${180}^{\circ }$

$\Rightarrow \angle A+\angle B+\angle C={180}^{\circ }........\left(1\right)$

Also given that,

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

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

Substituting values in $\left(1\right)$, we get

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

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

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

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