Solution:

According to the given question

\[\angle ACB\text{ }=\text{ }{{90}^{o}}~\]  [Angle in a semi-circle is 90^o]

Also,

\[\angle ABC\text{ }=\text{ }{{180}^{o}}-\angle ADC\] \[=\text{ }{{180}^{o}}-\text{ }{{130}^{o}}~=\text{ }{{50}^{o}}\]

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

By angle sum property of the right triangle ACB, we have

\[\angle BAC\text{ }=\text{ }{{90}^{o}}-\angle ABC\]

\[=\text{ }{{90}^{o}}-\text{ }{{50}^{o}}\]

Hence, \[\angle BAC\text{ }=\text{ }{{40}^{o}}\]