The triangle ABC is isoceles, this means that sides AB and AC are equal and the base angles
∠ABC and ∠BCA are equal.
The angle ∠BCA and the external angle ∠ACD are a linear pair, this means that they add up to 180º
From this we can calculate ∠BCA as
[tex]\begin{gathered} \angle\text{BCA}+\angle\text{ACD}=180º \\ \angle\text{BCA}+110º=180º \\ \angle\text{BCA}=180º-110º \\ \angle\text{BCA}=70º \end{gathered}[/tex]
Since ∠BCA=∠ACB, then ∠ABC=70º
Now the sum of all interior angles of a triangle is equal to 180º so that
[tex]\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180º[/tex]
Since BCA and ACB are known we can unse this expression to calculate ∠BAC as follows
[tex]\begin{gathered} 70+70+\angle\text{BAC}=180 \\ 140+\angle\text{BAC}=180 \\ \angle\text{BAC}=180-140 \\ \angle\text{BAC}=40 \end{gathered}[/tex]
m∠BAC=40º
