$ABC$ is a right angled triangle in which $\angle A = 90^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.
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:
In $\triangle ABC, \angle A = 90^o$
$AB =AC$
This implies,
$\angle C = \angle B$ (Angles opposite to equal sides are equal)
$\angle A+\angle B+\angle C=180^o$
$90^o+\angle B+\angle C=180^o$
$\angle B + \angle C = 90^o$
Therefore,
$\angle B = \angle C = \frac{90^o}{2} = 45^o$
Hence, $\angle B = \angle C = 45^o$.
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