The length of the chord joining the points (4 | vedclass

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Question

(A) The points are $P = (4 \cos 	heta, 4 \sin 	heta)$ and $Q = (4 \cos (	heta+60^{\circ}), 4 \sin (	heta+60^{\circ}))$.Using the distance formula,

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$4$

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(A) The points are $P = (4 \cos 	heta, 4 \sin 	heta)$ and $Q = (4 \cos (	heta+60^{\circ}), 4 \sin (	heta+60^{\circ}))$.Using the distance formula,the length of the chord $PQ$ is given by:$PQ = \sqrt{(4 \cos (	heta+60^{\circ}) - 4 \cos 	heta)^2 + (4 \sin (	heta+60^{\circ}) - 4 \sin 	heta)^2}$$PQ = 4 \sqrt{(\cos (	heta+60^{\circ}) - \cos 	heta)^2 + (\sin (	heta+60^{\circ}) - \sin 	heta)^2}$$PQ = 4 \sqrt{\cos^2 (	heta+60^{\circ}) + \cos^2 	heta - 2 \cos (	heta+60^{\circ}) \cos 	heta + \sin^2 (	heta+60^{\circ}) + \sin^2 	heta - 2 \sin (	heta+60^{\circ}) \sin 	heta}$Using $\sin^2 A + \cos^2 A = 1$:$PQ = 4 \sqrt{1 + 1 - 2 (\cos (	heta+60^{\circ}) \cos 	heta + \sin (	heta+60^{\circ}) \sin 	heta)}$Using $\cos (A-B) = \cos A \cos B + \sin A \sin B$:$PQ = 4 \sqrt{2 - 2 \cos ((	heta+60^{\circ}) - 	heta)}$$PQ = 4 \sqrt{2 - 2 \cos 60^{\circ}}$Since $\cos 60^{\circ} = \frac{1}{2}$:$PQ = 4 \sqrt{2 - 2 	imes \frac{1}{2}} = 4 \sqrt{2 - 1} = 4 	imes 1 = 4$.

The length of the chord joining the points $(4 \cos \theta, 4 \sin \theta)$ and $(4 \cos (\theta+60^{\circ}), 4 \sin (\theta+60^{\circ}))$ of the circle $x^{2}+y^{2}=16$ is

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