# Question #e355e

Jan 4, 2017

$y = {40}^{\circ}$

#### Explanation:

Given that $A B = A C \mathmr{and} \angle C = {40}^{\circ} , \implies \Delta A B C$ is an isosceles triangle.

An isosceles triangle is a triangle that has two sides of equal length.
The unequal side of an isosceles triangle is usually referred to as the 'base' of the triangle. In the figure, $B C$ is the base.
The base angles of an isosceles triangle are always equal. In the figure, the angles $\angle B$ and $\angle C$ are the same.

Given that $\angle C = {40}^{\circ}$
$\implies \angle B = \angle C = {40}^{\circ}$
$\implies \angle A = 180 - \left(\angle B + \angle C\right) = 180 - \left(40 + 40\right) = {100}^{\circ}$
Given $\angle A = 2 y + 20$
$\implies 2 y + 20 = {100}^{\circ}$
$\implies 2 y = {80}^{\circ}$
$\implies y = {40}^{\circ}$