Question

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

Answer 1

Consider $3\angle A=4\angle B=6\angle C=x$
So we can write it as
$3\angle A=x$
$\angle A=\frac{x}{3}$
$4\angle B=x$
$\angle B=\frac{x}{4}$
$6\angle C=x$
$\angle C=\frac{x}{6}$
We know that the sum of all the angles in a triangle is ${180}^{\circ }$.
So we can write it as
$\angle A+\angle B+\angle C={180}^{\circ }$
By substituting the values in the above equation we get
$\left(\frac{x}{3}\right)+\left(\frac{x}{4}\right)+\left(\frac{x}{6}\right)={180}^{\circ }$
$LCM$ of $3,4$ and $6$ is $12$
So we get
$\frac{(4x+3x+2x)}{12}={180}^{\circ }$
By addition
$\frac{9x}{12}={180}^{\circ }$
By cross multiplication
$9x={180}^{\circ }\times 129$
$x={2160}^{\circ }$
By division
$x=\frac{{2160}^{\circ }}{9}$
$x={240}^{\circ }$
By substituting the values of $x$
$\angle A=\frac{x}{3}=\frac{{240}^{\circ }}{3}={80}^{\circ }$
$\angle B=\frac{x}{4}=\frac{{240}^{\circ }}{4}={60}^{\circ }$
$\angle C=\frac{x}{6}=\frac{{240}^{\circ }}{6}={40}^{\circ }$
Therefore, the value of $\angle A,\angle B$ and $\angle C$ is ${80}^{\circ },{60}^{\circ }$ and ${40}^{\circ }$.