JEE Main 2022 · 27 Jul · Shift 2 · Mathematics Q18

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MathematicsSingle correct+4 / -1Question 18 of 90

The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is 45°. Let R be a point on AQ and from a point B, vertically above R, the angle of elevation of P is 60°. If $\angle BAQ=30°$ , AB = d and the area of the trapezium PQRB is $\alpha$ , then the ordered pair $(d,\alpha)$ is :

\left(10(\sqrt3-1),25\right)

\left(10(\sqrt3-1),\dfrac{25}2\right)

\left(10(\sqrt3+1),25\right)

\left(10(\sqrt3+1),\dfrac{25}2\right)

Answer and worked solution

Official answer: A

Heights. $AQ=10$ . B is at height $\frac d2$ with $AR=\frac{\sqrt3}2d$ . Then $\frac{10-d/2}{10-\sqrt3d/2}=\sqrt3$ gives $d=10(\sqrt3-1)$ .

Trapezium. Parallel sides 10 and $5(\sqrt3-1)$ , width $5(\sqrt3-1)$ . Area $\frac12\cdot5(\sqrt3+1)\cdot5(\sqrt3-1)=25$ .

(A)

Topics:Heights and distances

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JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 18