A man observes that the angle of elevation
of the top of a tower from a point $P$ on the ground is $θ$ .
He moves a certain distance towards the foot of the tower
and finds that the angle of elevation of the top has doubled.
He further moves a distance 3/4 of the previous and finds
that the angle of elevation is three times that at $P$ . The angle
$θ$ is given by.

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$\cos θ = 3 / 8$

$\sin θ = 3 / 4$

$\cos θ = \sqrt{5 / 12}$

$\sin θ = \sqrt{5 / 12}$

answer is A.

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Detailed Solution

Let $A B$ be the tower,

$∠ A P B = θ, ∠ A Q B = 2 θ and ∠ A R B = 3 θ$

Then $Q R = (3 / 4) P Q and ∠ P B Q = ∠ Q B R = θ$

$\Rightarrow B Q$ is the bisector of $∠ P B R$ (Fig. 27.15)

$\begin{aligned} \Rightarrow \frac{P B}{B R} = \frac{P Q}{Q ∣ R} \Rightarrow \frac{A B cosec θ}{A B cosec 3 θ} == \frac{4}{3} \Rightarrow \frac{\sin 3 θ}{\sin θ} = \frac{4}{3} \\ \Rightarrow 3 - 4 \sin^{2} θ = 4 / 3 \Rightarrow 12 \sin^{2} θ = 5 \Rightarrow \sin θ = \sqrt{5 / 12} \end{aligned}$