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A man observes a tower $AB$ of height $h$ from a point $P$ on the ground. He moves a distance $d$ towards the foot of the tower and finds that the angle of elevation is doubled. He further moves a distance $\frac{3 d}{4}$ in the same direction and the angle of elevation is three times that at $P$ . What is the value of $\frac{h^{2}}{d^{2}} ?$

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\frac{35}{9}

\frac{35}{36}

\frac{36}{5}

\frac{36}{35}

answer is B.

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Detailed Solution

It is given that a man observes a tower AB of height

h

from a point

P

on the ground.
He moves a distance

d

towards the foot of the tower and the angle of elevation is doubled.
Question Image

∠ BQR

is an exterior angle of

△ PBQ

.
⇒

∠ PBQ = 2 α - α = α

△ PBQ

∠ PBQ = ∠ QPB = α

\Rightarrow PQ = QB = d

Also,

∠ BRA

is an exterior angle

of △ BQR

∴ ∠ BQR = 3 α - 2 α = α

And

∠ QBR = π - 3 α

.
In

△ BQR

apply sine formula,

\frac{d}{\sin \sin (π - 3 α)} = \frac{\frac{3}{4} d}{sinα}

⇒

4 sinα = - 3 \sin 3 α

⇒

4 sinα = 3 (3 sinα - 4 \sin^{3} α)

… (

\sin 3 α = 3 sinα - 4 \sin^{3} α)

.
⇒

4 = 9 - 12 \sin^{2} α

⇒

\sin^{2} α = \frac{5}{12}

… (1).
From

△ BQA

h = BQ sin 2 α

⇒

h = 2 d sin α cos α

… (

\sin 2 α = 2 sin α cos α

).
⇒

h = 2 d sin α \sqrt{1 - \sin^{2} α}

⇒

h = 2 d \sqrt{\frac{5}{12}} \sqrt{1 - \frac{5}{12}}

… (from 1).
⇒

h = \frac{2 d}{12} \sqrt{35}

⇒

6 h = d \sqrt{35}

⇒

36 h^{2} = 35 d^{2}

⇒

\frac{h^{2}}{d^{2}} = \frac{35}{36}

Hence, the correct option is 2.

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