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 $d=\sqrt{7}h$, then $\tan \alpha$ is equal to

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Heights and Distances

Let a vertical tower AB of height

2 h

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 α

. When from

P

, he moves a distance d in the direction of

A P

, he can see the top

B

of the tower with an angle of elevation

α

. If

d = 7 h

, then

tan α

is equal to

Answer: (C)

From the geometry,

tan 2 α = \frac{h}{x}

where

x

is the initial distance from the tower.

After moving distance

d = 7 h

away:

tan α = \frac{2 h}{x + 7 h}

.

Substituting

x = h cot 2 α = \frac{h ( 1 - t a n ^{2} α )}{2 t a n α}

tan α = \frac{2}{\frac{1 - t a n ^{2} α}{2 t a n α} + 7}

Letting

t = tan α

and simplifying:

t = \frac{2}{\frac{1 - t ^{2}}{2 t} + 7}

Solving this equation yields

tan α = 7 - 2

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option