Question

In the given figure, AB and CD are two equal chords of the circle with centre O. OP and OQ are perpendiculars of the chord AB and CD respectively. If $\angle POQ={80}^{\circ },$ then what is the value of $\angle APQ$ in degrees?

• A. 40

Answer 1

The correct option is A 40
We know that equal chords are equidistant from the centre.
Since $AB=CD,$ the perpendiculars from these chords to the centre of the circle are equal.
Hence, we get $OP=OQ.$

Since $OP\perp AB,$
$\angle OPQ+\angle QPA=\angle OPA={90}^{\circ }....\left(i\right)$
Now, using angle sum property in $△OPQ,$
$\angle OPQ+\angle PQO+\angle QOP={180}^{\circ }...\left(ii\right)$

We have $OP=OQ.$
Since angles opposite to equal sides are equal,
$\angle OPQ=\angle PQO=x\left(say\right).$
From (ii),
$\angle OPQ+\angle PQO+\angle QOP={180}^{\circ }.$
$⟹x+x+{80}^{\circ }={180}^{\circ }$
$⟹2x={180}^{\circ }-{80}^{\circ }={100}^{\circ }$
$⟹x={50}^{\circ }$
From (i),
$\angle APQ=\angle OPA-\angle OPQ$
i.e., $\angle APQ={90}^{\circ }-{50}^{\circ }={40}^{\circ }$