Question

Assertion (A) In the given figure, If O is the centre of the circle and AC is the diameter with $\angle APB={120}^{\circ }$, then $\angle BQC={150}^{\circ }$.

Reason (R) In a cyclic quadrilateral, the sum of opposite angles is ${180}^{\circ }$.

• A. (A) is true and (R) is the correct explanation of (A)
• B. (A) is false and (R) is the correct explanation of (A)
• C. (A) is true and (R) is false
• D. Both (A) and (R) are false

Answer 1

The correct option is A

(A) is true and (R) is the correct explanation of (A)

Given, $\angle APB={120}^{\circ }$

In cyclic quadrilateral ACBP,
$\angle APB+\angle ACB={180}^{\circ }\Rightarrow {120}^{\circ }+\angle ACB={180}^{\circ }\Rightarrow \angle ACB={60}^{\circ }$
Also, $\angle ABC={90}^{\circ }$ [angle in a semi circle]
$\therefore In\Delta ABC,\angle BAC+\angle BCA+\angle ABC={180}^{\circ }\Rightarrow \angle BAC+{60}^{\circ }+{90}^{\circ }={180}^{\circ }\Rightarrow \angle BAC={30}^{\circ }$
Again, in cyclic quadrilateral ACQB,
$\angle CAB+\angle CQB={180}^{\circ }\Rightarrow {30}^{\circ }+\angle CQB={180}^{\circ }\therefore \angle CQB={150}^{\circ }$
Reason (R)True