Question

In $△$ ABC, $3\angle A=4\angle B=6\angle C$. The angles of a triangle are:

• A. $A={60}^{\circ },B={80}^{\circ }$ and $C={40}^{\circ }$
• B. $A={60}^{\circ },B={65}^{\circ }$ and $C={55}^{\circ }$
• C. $A={70}^{\circ },B={30}^{\circ }$ and $C={80}^{\circ }$
• D. $A={80}^{\circ },B={60}^{\circ }$ and $C={40}^{\circ }$

Answer 1

The correct option is D $A={80}^{\circ },B={60}^{\circ }$ and $C={40}^{\circ }$

Given, $3\angle A=4\angle B=6\angle C$
$\Rightarrow 3\angle A=6\angle C$ and $4\angle B=6\angle C$
$\Rightarrow \angle A=2\angle C$ and $\angle B=\frac{3}{2}\angle C$ $...\left(1\right)$
Since, $\angle A+\angle B+\angle C={180}^o$ [Angle sum property]

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

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

$\Rightarrow \frac{6\angle C+3\angle C}{2}={180}^o$

$\Rightarrow 9\angle C={360}^o$
$\Rightarrow \angle C={40}^o$

By substituting $\angle C={40}^o$ in $\left(1\right).$ we get,
$\angle A={80}^o$ and $\angle B={60}^o.$