Question

In a $\Delta ABC$, if $3\angle A=4\angle B=6\angle C$, find the measure of $\angle A,\angle B$ and $\angle C$.

Answer 1

Given:- $3\angle A=6\angle C\Rightarrow \angle A=2\angle C4\angle B=6\angle C\Rightarrow \angle B=\frac{3}{2}\angle C$

As we know that sum of interior angles in a triangle is $180°$.

Therefore,

$\angle A+\angle B+\angle C=180°$

$2\angle C+\frac{3}{2}\angle C+\angle C=180°$

$\frac{9}{2}\angle C=180°$

$\Rightarrow \angle C=\frac{180°\times 2}{9}=40°$

Therefore,

$\angle A=2\times 40°=80°$

$\angle B=\frac{3}{2}\times 40°=60°$