Question

In $△ABC$, if $BC=AB$ and $\angle B=80°$,then $\angle A$ is equal to :

Answer 1

Find the value of $\angle A$

In $△ABC$, we have $BC=AB$

The two sides stemming from vertex $B$ are equal hence, $△ABC$ is an isosceles triangle.

In an isosceles triangle the angles opposite to the equal sides are equal

$\Rightarrow$$\angle BCA=\angle C$ and $\angle BAC=\angle A$ are equal.

Let $\angle A=\angle C=x°$

By applying the angle sum property of a triangle we get

$\Rightarrow$$\angle A+\angle B+\angle C=180°$

Substituting the values we get

$\Rightarrow$$x°+80°+x°=180°$

$\Rightarrow$ $2x°=100°$

$\Rightarrow$ $x°=50°$

Hence, the value of $\angle A$ in $△ABC$ is $50°$