Question

In triangle $ABC$, $2\angle A=3\angle B=6\angle C$. Calculate $\angle A,\angle B$ and $\angle C$.

Answer 1

Given: $2\angle A=3\angle B=6\angle C$.

Let $2\angle A=3\angle B=6\angle C=x$. Therefore,

$\Rightarrow \angle A=\frac{x}{2}$

$\Rightarrow \angle B=\frac{x}{3}$ and

$\Rightarrow \angle C=\frac{x}{6}$

Using, angle sum property in $\Delta ABC$, we get,

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

$\Rightarrow \frac{x}{2}+\frac{x}{3}+\frac{x}{6}={180}^{\circ }$

$\Rightarrow \frac{3x+2x+x}{6}={180}^{\circ }$

$\Rightarrow \frac{6x}{x}={180}^{\circ }$

$\Rightarrow x={180}^{\circ }$

Therefore,

$\Rightarrow \angle A=\frac{x}{2}=\frac{{180}^{\circ }}{2}={90}^{\circ }$

$\Rightarrow \angle B=\frac{x}{3}=\frac{{180}^{\circ }}{3}={60}^{\circ }$

$\Rightarrow \angle C=\frac{x}{6}=\frac{{180}^{\circ }}{6}={30}^{\circ }$