In a $\triangle ABC$, if $AB = AC$ and $\angle B = 70^o$, find $\angle A$.
Given:
In a $\triangle ABC$, $AB = AC$ and $\angle B = 70^o$.
To do:
We have to find $\angle A$.
Solution:
$AB = AC$
This implies,
$\angle B = \angle C$ (Angles opposite to equal sides are equal)
$\angle A + \angle B + \angle C = 180^o$
$70^o + \angle A + 70^o = 180^o$
$\angle A = 180^o - 140^o = 40^o$
Hence, $\angle A = 40^o$ .
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