Two circles of radii 8 cm with centers P and | vedclass

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(A) In the circle with center $P$,$PA = PB = 8 \ cm$ (radii of the same circle). Since $PA = PB$,$	riangle PAB$ is an isosceles triangle. Therefore,$

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) In the circle with center $P$,$PA = PB = 8 \ cm$ (radii of the same circle). Since $PA = PB$,$	riangle PAB$ is an isosceles triangle. Therefore,$\angle PBA = \angle PAB = 40^{\circ}$. In $	riangle PAB$,the sum of angles is $180^{\circ}$,so $\angle APB = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 100^{\circ}$. Since the chords $AB$ and $CD$ are equal and the circles have the same radius $(8 \ cm)$,the chords subtend equal angles at their respective centers. Thus,$\angle CQD = \angle APB = 100^{\circ}$.

Two circles of radii $8 \ cm$ with centers $P$ and $Q$ are given. The chord $AB$ of the circle with center $P$ and the chord $CD$ of the circle with center $Q$ are equal. If $\angle PAB = 40^{\circ}$,then find $\angle CQD$. (in $^{\circ}$)

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