In the figure, $AB = AC$ and $\angle ACD = 105^o$, find $\angle BAC$.
"

Given:

In the given figure, $AB = AC$ and $\angle ACD = 105^o$.

To do:

We have to find $\angle BAC$.

Solution:

In $\triangle ABC, AB = AC$

This implies,

$\angle B = \angle C$             (Angles opposite to equal sides are equal)

$\angle ACB + \angle ACD = 180^o$               (Linear pair)

$\angle ACB + 105^o= 180^o$

$\angle ACB = 180^o-105^o = 75^o$

Therefore,

$\angle ABC = \angle ACB = 75^o$

$\angle A + \angle B + \angle C = 180^o$

$\angle A + 75^o + 75^o = 180^o$

$\angle A + 150^o= 180^o$

$\angle A= 180^o- 150^o = 30^o$

Hence, $\angle BAC = 30^o$.