Question

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

Answer 1

Let each base angle be $x$ in an isosceles $△ABC$

Then vertex angle be $2(x+x)=4x$

Since sum of angles of a triangle is ${180}^o$ .

Hence $4x+x+x={180}^o$

$\Rightarrow 6x={180}^0$

$\Rightarrow x={30}^0$

$\therefore$ Angles are ${30}^o,{30}^o,{120}^o$