Question

In $△ABC$, if $3\angle A=4\angle B=6\angle C$, calculate $A,B,C$.

Answer 1

Given:

In $△ABC$

$3\angle A=4\angle B=6\angle C$

Let $3\angle A=4\angle B=6\angle C=x$

$\Rightarrow \angle A=\frac{x}{3},\angle B=\frac{x}{4},\angle C=\frac{x}{6}$

Since, the sum of the angles of triangle is ${180}^{\circ }$.

$\Rightarrow \frac{x}{3}+\frac{x}{4}+\frac{x}{6}={180}^{\circ }$

$\Rightarrow \frac{4x+3x+2x}{12}={180}^{\circ }$

$\Rightarrow \frac{9x}{12}={180}^{\circ }$

$\Rightarrow x={180}^{\circ }\times \frac{12}{9}$

$\Rightarrow x={20}^{\circ }\times 12$

$\Rightarrow x={240}^{\circ }$

Now,

$\angle A=\frac{x}{3}=\frac{{240}^{\circ }}{3}={80}^{\circ }$

$\angle B=\frac{x}{4}=\frac{{240}^{\circ }}{4}={60}^{\circ }$

$\angle C=\frac{x}{6}=\frac{{240}^{\circ }}{6}={40}^{\circ }$