Question

In the adjoining figure, two circles intersect at $A$ and $B$. The centre of the smaller circle is $O$ and lies on the circumference of the larger circle. If $PAC$ and $PBD$ are straight lines and $\angle APB={75}^o$, find (i) $\angle AOB$, (ii) $\angle ACB$, (ii) $\angle ADB$.

• A. (i) $\angle AOB={150}^o$,
  (ii) $\angle ACB={30}^o$,
  (ii) $\angle ADB={30}^o$.
• B. (i) $\angle AOB={150}^o$,
  (ii) $\angle ACB={30}^o$,
  (ii) $\angle ADB={60}^o$.
• C. (i) $\angle AOB={150}^o$,
  (ii) $\angle ACB={40}^o$,
  (ii) $\angle ADB={30}^o$.
• D. (i) $\angle AOB={120}^o$,
  (ii) $\angle ACB={30}^o$,
  (ii) $\angle ADB={30}^o$.

Answer 1

The correct option is A (i) $\angle AOB={150}^o$,

(ii) $\angle ACB={30}^o$,
(ii) $\angle ADB={30}^o$.
Given that $O$ is the center of the smaller circle and $\angle APB={75}^{\circ }$.

We know, the angle subtended by an arc at the center of the circle has doubled the angle subtended at any point of the circle.

So, $\angle AOB=2\angle APB$

$⟹$ $\angle AOB=2\times {75}^{\circ }={150}^{\circ }$.

Now, since $ACOB$ is a cyclic quadrilateral,

$\angle ACB+\angle AOB={180}^{\circ }$ ...[Opposite angles of a cyclic quadrilateral are supplementary]

$\Rightarrow \angle ACB+{150}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ACB={180}^{\circ }-{150}^{\circ }$

$\Rightarrow \angle ACB={30}^{\circ }$

Also, since AOBD is a cyclic quadrilateral,

$\angle ADB+\angle AOB={180}^{\circ }$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$\Rightarrow \angle ADB+{150}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ADB={180}^{\circ }-{150}^{\circ }$

$\Rightarrow \angle ADB={30}^{\circ }$.

Hence, $\angle AOB={150}^{\circ }$, $\angle ACB={30}^{\circ }$ and $\angle ADB={30}^{\circ }$.

Therefore, option $A$ is correct.