ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $$\angle ADC = 140^\circ.$$ 
Then $$\angle BAC$$ is equal to:

Given:
ABCD is a cyclic quadrilateral, AB is the diameter of the circle $$, \angle ADC = 140^\circ$$​​
Concept used:
In a cyclic quadrilateral, opposite angles are supplementary: $$\angle ADC + \angle ABC = 180^\circ$$​​
Angle subtended by the diameter in a semicircle is a right angle: $$\angle ACB = 90^\circ$$​​
Solution:
​$$\begin{aligned}\angle ABC &= 180^\circ - \angle ADC = 180^\circ - 140^\circ = 40^\circ \\\angle ACB &= 90^\circ \\\angle BAC &= 180^\circ - \angle ABC - \angle ACB \\&= 180^\circ - 40^\circ - 90^\circ = 50^\circ\end{aligned}$$​​
Correct answer is (b): $$50^\circ$$​​