Question

\(\text{In given figure, ABCD is a cyclic} (q)~ 65^\circ\\
~~~\text{quadrilateral whose side AB is a diameter}\\
~~~ \text{of the circle through A, B, C, D.}\\
~~~\text{If} ~\angle ADC = 130^\circ, \text{then find}~ \angle BAC = ___\)

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${100}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is A

${40}^{\circ }$

Since ABCD is a cyclic quadrilateral.
$\therefore \angle ADC+\angle ABC={180}^{\circ }\Rightarrow {130}^{\circ }+\angle ABC={180}^{\circ }\Rightarrow \angle ABC={50}^{\circ }$



Since $\angle ACB$ is the angle in a semicircle.
$\therefore \angle ACB={90}^{\circ }$
Now, in $\Delta ABC$, we have
$\angle BAC+\angle ACB+\angle ABC={180}^{\circ }\Rightarrow \angle BAC={90}^{\circ }+{50}^{\circ }={180}^{\circ }\Rightarrow \angle BAC={40}^{\circ }$