In a circle with centre P,AB is a chord and P | vedclass

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Question

(C) In a circle with centre $P$,$PA = PB = 4\, cm$ (radii). In $	riangle APB$,$PA = PB = 4\, cm$ and $AB = 5\, cm$. In a circle with centre $Q$,$QX =

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) In a circle with centre $P$,$PA = PB = 4\, cm$ (radii). In $	riangle APB$,$PA = PB = 4\, cm$ and $AB = 5\, cm$. In a circle with centre $Q$,$QX = QY = 4\, cm$ (radii). Given $\angle XQY = 50^{\circ}$. In $	riangle XQY$,$QX = QY = 4\, cm$. Using the Law of Cosines in $	riangle XQY$: $XY^2 = QX^2 + QY^2 - 2(QX)(QY) \cos(50^{\circ})$ $XY^2 = 4^2 + 4^2 - 2(4)(4) \cos(50^{\circ})$ $XY^2 = 16 + 16 - 32 \cos(50^{\circ}) = 32(1 - \cos(50^{\circ}))$ $XY^2 = 32(2 \sin^2(25^{\circ})) = 64 \sin^2(25^{\circ})$ $XY = 8 \sin(25^{\circ}) \approx 8 	imes 0.4226 \approx 3.38\, cm$. Since the question implies a comparison based on the chord length formula $L = 2r \sin(	heta/2)$,for the first circle $5 = 2(4) \sin(40^{\circ}) \implies 5 = 8 \sin(40^{\circ}) \approx 5.14$. Given the options provided,the intended answer is $3$ (closest integer).

In a circle with centre $P$,$AB$ is a chord and $PA = 4\, cm$. In a circle with centre $Q$,$XY$ is a chord and $QX = 4\, cm$. If $\angle APB = 80^{\circ}$,$\angle XQY = 50^{\circ}$ and $AB = 5\, cm$,then $XY = \dots\dots\dots\, cm$.

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