Question

An exterior angle of a triangle is 100 and its interior opposite angles are equal to each other. Find the measure of each angle of the triangle.

Answer 1

Using the theorem that if the side of a triangle is produced then the interior angle is equal to the sum of two interior opposite angles of the triangle.

Let $\Delta ABC$ be a triangle where the side BC is produced.

By the theorem we have

$\angle A+\angle B={100}^o$ …………..$\left(1\right)$

It is given that the interior opposite angles are equal i.e., $\angle A=\angle B$.

$\therefore$ Equation $\left(1\right)$ becomes

$2\angle A={100}^o$

$\Rightarrow \angle A={50}^o$

$\Rightarrow \angle B={50}^o$

Using angle sum property of a triangle,

Sum of all angles of a triangle $={180}^o$

$\Rightarrow \angle A+\angle B+\angle C={180}^o$

$\Rightarrow 100+\angle C={180}^o$ [from $\left(1\right)$]

$\Rightarrow \angle C={180}^o-{100}^o$

$\Rightarrow \angle C={80}^o$.

Hence $\angle A={50}^o$

$\angle B={50}^o$

$\angle C={80}^o$.