Question

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.

Answer 1

Given : In $\Delta ABC,AB=AC$

$\angle A=2(\angle B+\angle C)$

To calculate : Base angles,

Let $\angle B=\angle C=x$

$Then\angle A=2(\angle B+\angle C)$

$=2(x+x)=2\times 2x=4x$

$\because \text{Sum of angles of a triangle}={180}^{\circ }$

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

$\Rightarrow x=\frac{{180}^{\circ }}{6}={30}^{\circ }$

$\therefore \angle B=\angle C=30and\angle A=4\times {30}^{\circ }={120}^{\circ }$