In a circle with centre P,AB and CD are equal | vedclass

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(B) $1$. In a circle,equal chords subtend equal angles at the centre. Since chord $AB = CD$,the angles subtended by them at the centre are equal,so $\

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) $1$. In a circle,equal chords subtend equal angles at the centre. Since chord $AB = CD$,the angles subtended by them at the centre are equal,so $\angle CPD = \angle APB = 100^{\circ}$.$2$. In $	riangle PCD$,$PC$ and $PD$ are radii of the circle,so $PC = PD$. This makes $	riangle PCD$ an isosceles triangle.$3$. In an isosceles triangle,the angles opposite to equal sides are equal,so $\angle PCD = \angle PDC$.$4$. The sum of angles in a triangle is $180^{\circ}$. Therefore,$\angle PCD + \angle PDC + \angle CPD = 180^{\circ}$.$5$. Substituting the known values: $2 	imes \angle PCD + 100^{\circ} = 180^{\circ}$.$6$. $2 	imes \angle PCD = 80^{\circ}$,which gives $\angle PCD = 40^{\circ}$.

In a circle with centre $P$,$AB$ and $CD$ are equal chords. If $\angle APB = 100^{\circ}$,then find $\angle PCD$. (in $^{\circ}$)

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