Question

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

• A. $\angle A={10}^o,\angle B={30}^o,\angle C={110}^o$
• B. $\angle A={18}^o,\angle B={45}^o,\angle C={140}^o$
• C. $\angle A={30}^o,\angle B={80}^o,\angle C={90}^o$
• D. $\angle A={20}^{\circ },\angle B={40}^{\circ },\angle C={120}^{\circ }$

Answer 1

The correct option is D $\angle A={20}^{\circ },\angle B={40}^{\circ },\angle C={120}^{\circ }$

$\Rightarrow \angle A+\angle B=\frac{1}{2}\angle C$
$\because \angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow \frac{1}{2}\angle C+\angle C={180}^{\circ }$
$\Rightarrow \angle C+2\angle C={360}^{\circ }\Rightarrow 3\angle C={360}^{\circ }$
$\Rightarrow \angle C={120}^{\circ }$
$\because \angle C=3\angle B$
$\Rightarrow 3\angle B={120}^{\circ }$
$\Rightarrow \angle B={40}^{\circ }$
$\because \angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow \angle A+{40}^{\circ }+{120}^{\circ }={180}^{\circ }$
$\Rightarrow \angle A={20}^{\circ }$
$Hence,\angle A={20}^{\circ },\angle B={40}^{\circ },\angle C={120}^{\circ }.$