Question

In $\Delta ABC,AB=AC$ and $\angle A:\angle B=4:3$. Find the measure of $\angle C$.

Answer 1

As we know that in a triangle sum of all the interior angles is ${180}^{\circ }$.

So,

$\angle A+\angle B+\angle C={180}^{\circ }$ .....(1)

Given that:

Since, $\angle A:\angle B=4:3$

hence,

$\angle A=4x,\angle B=3x$

$Since,AB=AC$

Hence,

$\angle C=\angle B=3x$

Now,

Substituting the values of $\angle A,\angle B,\angle C$ in equation (1), we get-

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

$10x={180}^{\circ }$

$x={18}^{\circ }$

Hence, $\angle C=3\times {18}^{\circ }={57}^{\circ }.$