JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q78

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MathematicsSingle correct+4 / -1Question 78 of 90

A horizontal park is in the shape of a triangle OAB with AB = 16. A vertical lamp post OP is erected at the point O such that $\angle PAO=\angle PBO=15^\circ$ and $\angle PCO=45^\circ$ , where C is the midpoint of AB. Then $(OP)^2$ is equal to

\dfrac{32}{\sqrt3}\left(\sqrt3-1\right)

\dfrac{32}{\sqrt3}\left(2-\sqrt3\right)

\dfrac{16}{\sqrt3}\left(\sqrt3-1\right)

\dfrac{16}{\sqrt3}\left(2-\sqrt3\right)

Answer and worked solution

Official answer: B

Distances. $OA=OB=h\cot15^\circ$ and $OC=h$ , with $OC^2=OA^2-64$ .

Solve. $h^2=\frac{64}{\cot^215^\circ-1}=\frac{64}{6+4\sqrt3}=\frac{32}{\sqrt3}\left(2-\sqrt3\right)$ .

(B)

Topics:Heights and distances

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JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 78