The vertical angle of an isosceles triangle is $100^o$. Find its base angles.

Given:

The vertical angle of an isosceles triangle is $100^o$.

To do:

We have to find its base angles.

Solution:

Let in a $\triangle ABC, AB = AC$ and $\angle A = 100^o$

$AB = AC$   

This implies,

$\angle C = \angle B$                 (Angles opposite to equal sides are equal)

$\angle A + \angle B + \angle C = 180^o$

$100^o + \angle B + \angle B = 180^o$             (Since $\angle C = \angle B$)

$2\angle B = 180^o - 100^o = 80^o$

$\angle B= 40^o$

Therefore,

$\angle C = \angle B = 40^o$

Hence, the base angles are each equal to $40^o$.