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(C) Let the distance between the parallel lines be $5$. Point $P$ is at a distance of $1$ from one line and $4$ from the other.Let $	heta$ be the ang

Vedclass · Accepted Answer

$28$

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(C) Let the distance between the parallel lines be $5$. Point $P$ is at a distance of $1$ from one line and $4$ from the other.Let $	heta$ be the angle $PR$ makes with the line at distance $4$ from $P$.From the geometry of the triangle,$PR = \frac{4}{\sin 	heta} = 4 \operatorname{cosec} 	heta$.Similarly,$PQ = \frac{1}{\sin(90^{\circ} - (	heta + 30^{\circ}))} = \frac{1}{\cos(	heta + 30^{\circ})}$.Since $	riangle PQR$ is equilateral,$PR = PQ = d$.Thus,$4 \operatorname{cosec} 	heta = \frac{1}{\cos(	heta + 30^{\circ})}$.$4 \cos(	heta + 30^{\circ}) = \sin 	heta$.$4(\cos 	heta \cos 30^{\circ} - \sin 	heta \sin 30^{\circ}) = \sin 	heta$.$4(\frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta) = \sin 	heta$.$2\sqrt{3} \cos 	heta - 2 \sin 	heta = \sin 	heta$.$2\sqrt{3} \cos 	heta = 3 \sin 	heta \Rightarrow 	an 	heta = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.Then $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta} = \frac{4/3}{1 + 4/3} = \frac{4/3}{7/3} = \frac{4}{7}$.$d^2 = PR^2 = (4 \operatorname{cosec} 	heta)^2 = 16 \operatorname{cosec}^2 	heta = 16 	imes \frac{1}{\sin^2 	heta} = 16 	imes \frac{7}{4} = 28$.

Let the distance between two parallel lines be $5$ units and a point $P$ lie between the lines at a unit distance from one of them. An equilateral triangle $PQR$ is formed such that $Q$ lies on one of the parallel lines,while $R$ lies on the other. Then $(QR)^2$ is equal to . . . . . . .

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