Question

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:

• A. ${80}^{\circ },$
• B. ${50}^{\circ }$
• C. ${40}^{\circ }$
• D. ${30}^{\circ }$

Answer 1

The correct option is B ${50}^{\circ }$


$\angle ADC+\angle ABC={180}^{\circ }$ [Opposite angles of cyclic quadrilatral]

$\Rightarrow {140}^{\circ }+\angle ABC={180}^{\circ }$

$\therefore \angle ABC={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it.

Now, $\angle C={90}^{\circ }[\because$ Angle in a semi-circle is a right angle]

In $△$ ABC, we have

$\angle BAC={180}^{\circ }-({90}^{\circ }+{40}^{\circ })={50}^{\circ }$

Hence, (B) is the correct answer.