In the figure,AB and CD are two equal chords | vedclass

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(B) In $	riangle OPQ$,since $OP$ and $OQ$ are perpendiculars from the center to equal chords,$OP = OQ$.Therefore,$	riangle OPQ$ is an isosceles tria

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) In $	riangle OPQ$,since $OP$ and $OQ$ are perpendiculars from the center to equal chords,$OP = OQ$.Therefore,$	riangle OPQ$ is an isosceles triangle,which implies $\angle OQP = \angle OPQ = k$.In $	riangle OPQ$,the sum of angles is $180^{\circ}$,so $\angle OQP + \angle OPQ + \angle POQ = 180^{\circ}$.$k + k + 150^{\circ} = 180^{\circ}$.$2k = 30^{\circ}$.$k = 15^{\circ}$.Thus,$\angle OPQ = 15^{\circ}$.Since $OP \perp AB$,$\angle OPA = 90^{\circ}$.From the figure,$\angle OPA = \angle OPQ + \angle APQ$.$90^{\circ} = 15^{\circ} + \angle APQ$.$\angle APQ = 90^{\circ} - 15^{\circ} = 75^{\circ}$.

In the figure,$AB$ and $CD$ are two equal chords of a circle with center $O$. $OP$ and $OQ$ are perpendiculars on chords $AB$ and $CD$ respectively. If $\angle POQ = 150^{\circ}$,then $\angle APQ$ is equal to: (in $^{\circ}$)

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