Question

The angle of vertex of an isosceles triangle is ${100}^o$. Find its base angle.

Answer 1

Given that vertex of isosceles triangle is ${100}^0$

We know that sum of angles in a triangle$={180}^0$

Since its a isosceles triangle, 2 angles are equal. Let the base angles be $x$

$\Rightarrow {100}^0+x+x={180}^0$

$\Rightarrow {100}^0+2x={180}^0$

$\Rightarrow 2x={180}^0-{100}^0$

$\Rightarrow 2x={80}^0$

$\Rightarrow x={40}^0$

Therefore the base angles of the isosceles triangle are ${40}^{0}and{40}^0$