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From the top of a tower $h m$ high, the angles of depression of two objects, which are in line with the foot of the tower are $α$ and $β (β > α)$ , the distance between the two objects is

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= (c o t α − c o t β)

= − h (c o t α − c o t β)

= h (c o t α + c o t β)

= h (c o t α − c o t β)

answer is B.

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Detailed Solution

Given, that a tower

h m

high and the angles of depression of two objects, which are in line with the foot of the tower are

α

and

β (β > α)

.
Let us consider that the distance between the two objects is x meter and let CD=y m.

∠ B A X = ∠ A B D = α … … … . a l t e r n a t e angle ∠ C A Y = ∠ A C D = β … … … … . a l t e r n a t e angles n o w, in triangle ACD t a n β = A D C D = h y ⇒ y = h tan β … … … . (1)

Now in triangle ABD,

tanα = \frac{AD}{BD}

\Rightarrow tanα = \frac{AD}{BC + CD}

\Rightarrow tanα = \frac{h}{x + y}

\Rightarrow x + y = \frac{h}{tanα}

\Rightarrow y = \frac{h}{tanα} - x . . . . . . . . . . . . . . . . . . . . . . . . (2)

Therefore after equating 1 and 2,

\frac{h}{tanβ} = \frac{h}{tanα} - x

\Rightarrow x = \frac{h}{tanα} - \frac{h}{tanβ}

\Rightarrow x = h (\frac{1}{tanα} - \frac{1}{tanβ})

\Rightarrow x = h (cotα - cotβ)

which is the required distance between the two objects.
The distance between two objects is x

= h (c o t α − c o t β)

Therefore, option 2 is correct.

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