Question

The vertical angle of an isosceles triangle is twice the sum of its base angles. Find each of the angle of the triangle.

Answer 1

Let the base angle $\angle ABC=\angle ACB=x$

Thus, sum of the base angles $x+x=2x$

Therefore, vertical angle $\angle BAC=4x$

By angle sum property in $\Delta ABC$, we get,

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

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

$\Rightarrow x={30}^{\circ }$

Hence, the angles of the isosceles $\Delta ABC$ are ${30}^{\circ },{30}^{\circ },{120}^{\circ }$