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From the bottom of a pole of height $h$ , the angle of elevation of the top of a tower is $α$ . The pole subtends an angle $β$ at the top of a tower. The height of the tower is:

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\frac{hsinαcos (α + β)}{cosβ}

\frac{hsinαcos (α - β)}{sinβ}

\frac{hsinαsin (α + β)}{cosβ}

\frac{hsinαsin (α - β)}{sinβ}

answer is B.

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Detailed Solution

It is given that the bottom of a pole of height

h

, the angle of elevation of the top of a tower is

α

. The pole subtends an angle

β

at the top of a tower.
Question Image

Let PQ be the tower and OA be the pole.
In △OPQ we have,

tanα = \frac{PQ}{OP} = \frac{PQ}{x}

⇒

PQ = xtanα . . . . (1)

⇒

h + QR = xtanα

⇒

QR = xtanα - h . . . . (2)

For triangle ARQ, we have,

\tan (α - β) = \frac{QR}{x} ​

⇒

\tan (α - β) = \frac{xtanα - h}{x}

… (from 2).
⇒

\tan (α - β) = tanα - \frac{h}{x}

⇒

\frac{h}{x} = tanα - \tan (α - β)

⇒

x = \frac{h}{tanα - \tan (α - β)}

Therefore, equation 1 becomes,

PQ = xtanα

⇒

PQ = \frac{htanα}{tanα - \tan (α - β)}

⇒

PQ = \frac{h \times \frac{sinα}{cosα}}{\frac{sinαcos (α - β) - \sin (α - β)}{sinαcos (α - β)}}

⇒ PQ =

\frac{hsinαcos (α - β)}{sinβ}

Hence, the correct option is 2.

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