$ABC$ is a right-angled triangle in which $\angle A = 90^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.
(N/A) In $\triangle ABC$,we have:
$AB = AC$ [Given]
Since the sides are equal,their opposite angles are also equal.
$\Rightarrow \angle ACB = \angle ABC$
We know that the sum of all angles in a triangle is $180^o$.
$\angle A + \angle B + \angle C = 180^o$
$90^o + \angle B + \angle C = 180^o$ [Since $\angle A = 90^o$]
$\angle B + \angle C = 180^o - 90^o = 90^o$
Since $\angle B = \angle C$,we can write:
$2 \angle B = 90^o$
$\angle B = \frac{90^o}{2} = 45^o$
Therefore,$\angle B = 45^o$ and $\angle C = 45^o$.
$AB = AC$ [Given]
Since the sides are equal,their opposite angles are also equal.
$\Rightarrow \angle ACB = \angle ABC$
We know that the sum of all angles in a triangle is $180^o$.
$\angle A + \angle B + \angle C = 180^o$
$90^o + \angle B + \angle C = 180^o$ [Since $\angle A = 90^o$]
$\angle B + \angle C = 180^o - 90^o = 90^o$
Since $\angle B = \angle C$,we can write:
$2 \angle B = 90^o$
$\angle B = \frac{90^o}{2} = 45^o$
Therefore,$\angle B = 45^o$ and $\angle C = 45^o$.
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