Question

If $2\angle A=3\angle B=6\angle C$ in a $\Delta ABC$, then $\angle A$, $\angle B$, $\angle C$ are

• A. ${30}^{\circ },{60}^{\circ },{90}^{\circ }$
• B. ${90}^{\circ },{60}^{\circ },{30}^{\circ }$
• C. ${30}^{\circ },{90}^{\circ },{60}^{\circ }$
• D. ${45}^{\circ },{90}^{\circ },{45}^{\circ }$

Answer 1

The correct option is B

${90}^{\circ },{60}^{\circ },{30}^{\circ }$

By angle sum property of triangle,

$\Rightarrow$ $\angle A+\angle B+\angle C$ = 180${}^{\circ }$ $\cdot \cdot \cdot$ (i)

From given data,
$2\angle A=6\angle C$
$\Rightarrow$ $\angle A$ = $3\angle C$

$3\angle B=6\angle B$
$\Rightarrow \angle B=2\angle C$

By substituting values of $\angle A$ and $\angle B$ in (i), we get

$\Rightarrow$ $3\angle C+2\angle C+\angle C$=180${}^{\circ }$

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

$\Rightarrow$ $\angle C$=30${}^{\circ }$

Hence,

$\angle A=3\times {30}^{\circ }={90}^{\circ }$

$\angle B=2\times {30}^{\circ }={60}^{\circ }$

$\therefore \angle A,\angle B$ and $\angle C$ of $\Delta ABC$ are ${90}^{\circ },{60}^{\circ },$ and ${30}^{\circ }$ respectively.