Question

In the adjoining figure, O is the centre of the circle and PQ, RS are its equal chords, $OD\perp PQ$ and $OE\perp RS$ . If $\angle DOE={130}^{\circ }$ then find $\angle PDE$

Answer 1

Since equal chords are equidistant from the centre, we have OD = OE.
$\Rightarrow \angle ODE=\angle OED=x$
$\therefore x+x+{130}^{\circ }={180}^{\circ }${Angle sum property}
$\Rightarrow 2x=50\Rightarrow x={25}^{\circ }$
$i.e.,\angle ODE={25}^{\circ }$
$OD\perp PQ\Rightarrow \angle PDO={90}^{\circ }$
$\therefore \angle PDE=\angle PDO-\angle ODE={90}^{\circ }-{25}^{\circ }={65}^{\circ }$
Thus, $\angle PDE={65}^{\circ }$