JEE Main 2022 · 27 Jul · Shift 1 · Mathematics Q15

← JEE Main 2022 · 27 Jul · Shift 1

MathematicsSingle correct+4 / -1Question 15 of 90

Let a vertical tower AB of height $2h$ stands on a horizontal ground. Let from a point P on the ground a man can see upto height $h$ of the tower with an angle of elevation $2\alpha$ . When from P, he moves a distance $d$ in the direction of $\overrightarrow{AP}$ , he can see the top B of the tower with an angle of elevation $\alpha$ . If $d=\sqrt7h$ , then $\tan\alpha$ is equal to

\sqrt5-2

\sqrt3-1

\sqrt7-2

\sqrt7-\sqrt3

Answer and worked solution

Official answer: C

Set up. With $t=\tan\alpha$ and P at distance $x$ : $x=\frac h{\tan2\alpha}=\frac{h(1-t^2)}{2t}$ and $\frac{2h}{x+\sqrt7h}=t$ .

Solve. $4=1-t^2+2\sqrt7t$ , so $t^2-2\sqrt7t+3=0$ and $t=\sqrt7\pm2$ . Since $2\alpha$ is acute, $t<1$ .

$\tan\alpha=\sqrt7-2$ . (C)

Topics:Heights and distances Trigonometric ratios and identities

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JEE Main 2022 · 27 Jul · Shift 1, Mathematics question 15