Question

In the below diagram, $O$ is the centre of the circle, $AC$ is the diameter and if $\angle APB={120}^0$, then $\angle BQC$ is

• A. ${30}^0$
• B. ${150}^0$
• C. ${90}^0$
• D. ${120}^0$

Answer 1

The correct option is B ${150}^0$

Quadrilateral $APBC$ is a cyclic quadrilateral.
We know, the opposite angles of a cyclic quadrilateral are supplementary.

Hence, $\angle APB+\angle ACB={180}^0$.

Since, $\angle APB={120}^0$

$\Rightarrow \angle ACB={180}^o-{120}^o={60}^0$.

Here, $AC$ is a diameter and hence $\angle ABC={90}^0$.
Thus, in $\Delta ABC,$

$\angle CAB+\angle ABC+\angle ACB={180}^o$ ...[Angle sum property]

$⟹$ $\angle CAB+{90}^o+{60}^o={180}^o$

$⟹$ $\angle CAB={180}^0-{90}^0-{60}^0={30}^0$.

Now, quadrilateral $ABQC$ is another cyclic quadrilateral.

So, $\angle CAB+\angle CQB={180}^0$ ...[Opposite angles of a cyclic quadrilateral]
$\therefore \angle CQB={180}^o-{30}^o={150}^0$.

Hence, option $B$ is correct.