In a $\triangle ABC$, if $\angle A = 120^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.
Given:
In a $\triangle ABC$, $\angle A = 120^o$ and $AB = AC$.
To do:
We have to find $\angle B$ and $\angle C$.
Solution:
$AB = AC$
This implies,
$\angle B = \angle C$ (Angles opposite to equal sides are equal)
$\angle A + \angle B + \angle C = 180^o$
$120^o + \angle B + \angle B = 180^o$
$2\angle B = 180^o - 120^o = 60^o$
$\angle B = \frac{60^o}{2} = 30^o$
This implies,
$\angle C = \angle B = 30^o$
Hence, $\angle B = 30^o$ and $\angle C = 30^o$ .
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