$ \mathrm{ABC} $ is a right angled triangle in which $ \angle \mathrm{A}=90^{\circ} $ and $ \mathrm{AB}=\mathrm{AC} $. Find $ \angle \mathrm{B} $ and $ \angle \mathrm{C} $.
Given:
$ABC$ is a right-angled triangle in which $\angle A=90^o$ and $AB=AC$.
To do:
We have to find $\angle B$ and $\angle C$.
Solution:
Given, that $AB=AC$
We know that,
The angles opposite to the equal sides are also equal.
This implies,
$\angle B = \angle C$
We know that,
The sum of the interior angles of a triangle is always equal to $180^o$
This implies
In $\triangle ABC,$
$\angle A+\angle B+\angle C = 180^o$
Therefore,
$90^o+2\angle B=180^o$ (since $\angle B = \angle C$)
This implies,
$2\angle B=90^o$
$\angle B=\frac{90^o}{2}$
$\angle B=45^o$
Therefore,
$\angle B=\angle C=45^o$.
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