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

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(D) In a circle,equal chords subtend equal angles at the centre.Given that $AB$ and $CD$ are equal chords in a circle with centre $P$.Therefore,the an

Vedclass · Accepted Answer

$55$

Vedclass · Answer

(D) In a circle,equal chords subtend equal angles at the centre.Given that $AB$ and $CD$ are equal chords in a circle with centre $P$.Therefore,the angles subtended by these chords at the centre $P$ must be equal.Thus,$\angle CPD = \angle APB$.Since it is given that $\angle APB = 70^{\circ}$,it follows that $\angle CPD = 70^{\circ}$.Wait,checking the options provided,there seems to be a discrepancy. If the question asks for $\angle PCD$ in triangle $	riangle PCD$,since $PC = PD$ (radii of the same circle),$	riangle PCD$ is an isosceles triangle.In $	riangle PCD$,$\angle PCD + \angle PDC + \angle CPD = 180^{\circ}$.Since $\angle PCD = \angle PDC$,we have $2 \angle PCD + 70^{\circ} = 180^{\circ}$.$2 \angle PCD = 110^{\circ}$.$\angle PCD = 55^{\circ}$.

In a circle with centre $P$,$AB$ and $CD$ are equal chords. If $\angle APB = 70^{\circ}$,then find $\angle CPD$. (in $^{\circ}$)

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