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An observer finds that the angular elevation of a tower is $θ$ . On advancing $a$ meters towards the tower, the elevation is $45 °$ and on advancing $b$ meters nearer the elevation is ( $90 ° - θ)$ , then the height of the tower (in meters) is

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\frac{ab}{a + b}

\frac{ab}{a - b}

\frac{2 ab}{a + b}

\frac{2 ab}{a - b}

answer is B.

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Detailed Solution

It is given an observer finds that the angular elevation of a tower is

θ

. After moving

a

meters towards the tower, the elevation is

45 °

.
According to the given data, the figure has been drawn below.
Question Image

Let

b

be the initial point and and AE to be the height of the tower which is h m.
In,

△ ACE,

tan θ = P e r p e n d i c u l a r b a s e

\Rightarrow \tan 45 ° = \frac{AE}{CE}

\Rightarrow

= \frac{h}{b + x}

∵ tan 45 ° = 1

\Rightarrow b + x = h

……(1)
In,

△ ABE,

tan θ = P e r p e n d i c u l a r b a s e

\Rightarrow tanθ = \frac{AE}{BE}

\Rightarrow tanθ = \frac{h}{a + b + x}

Now replacing value of

b + x = h

we get,

\Rightarrow tanθ = \frac{h}{a + h} . . . (2)

△ ADE,

tan θ = P e r p e n d i c u l a r b a s e

\frac{AE}{DE} = \tan (90 ° - θ)

\Rightarrow \frac{h}{x} = cotθ

∵ tan 90 ° − θ = cot θ

\Rightarrow tanθ = \frac{x}{h}

∵ 1 tan θ = cot θ

Now putting x = h-b we get,

\Rightarrow tanθ = \frac{h - b}{h} . . .

…..(3)
From equation 2 and 3 we get,

\Rightarrow \frac{h - b}{h} = \frac{h}{a + h}

\Rightarrow (h - b) (a + h) = h^{2}

\Rightarrow ah - ab + h^{2} - hb = h^{2}

\Rightarrow h (a - b) = ab

\Rightarrow h = \frac{ab}{a - b}

Hence, the height of the tower is

\frac{ab}{a - b}

m.
So, the correct option is 2.

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