$A$ quadrilateral $ABCD$ is inscribed in a circle such that $AB$ is a diameter and $\angle ADC = 130^{\circ}$. Find $\angle BAC$.

$(40^{\circ})$ Since the opposite angles of a cyclic quadrilateral are supplementary,we have:
$\angle ABC + \angle ADC = 180^{\circ}$
$\angle ABC + 130^{\circ} = 180^{\circ}$
$\angle ABC = 180^{\circ} - 130^{\circ} = 50^{\circ}$
Now,in $\Delta ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semi-circle is $90^{\circ}$).
Using the angle sum property of a triangle in $\Delta ABC$:
$\angle BAC + \angle ABC + \angle ACB = 180^{\circ}$
$\angle BAC + 50^{\circ} + 90^{\circ} = 180^{\circ}$
$\angle BAC + 140^{\circ} = 180^{\circ}$
$\angle BAC = 180^{\circ} - 140^{\circ} = 40^{\circ}$