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 $\angle APQ$?

Answer 1

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 $\Delta OPQ$,
$\angle OPQ+\angle PQO+\angle QOP={180}^{\circ }...\left(ii\right)$

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