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The angle of elevation of the top of a tower as observed from a point on the horizontal ground is x. If we move a distance d towards the foot of the tower, the angle of elevation increases to y, then

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\frac{dtanxtany}{tany - tanx}

d (tany + tanx)

d (tany - tanx)

\frac{dtanxtany}{tany + tanx}

answer is A.

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Detailed Solution

It is given the figure is,
Question Image

Let AB be the height of the tower, CD=d,

∠ ACB = x

and

∠ ADB = y

...(given)
Consider triangle

Δ

ADB,

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

tany = \frac{AB}{BD}

BD = ABcoty

...(i)

∵ 1 tan θ = cot θ

Consider triangle

Δ

ACD,

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

tanx = \frac{AB}{BC}

\Rightarrow tanx = \frac{AB}{BD + CD}

\Rightarrow

BD + CD = ABcotx

∵ 1 tan θ = cot θ

\Rightarrow

ABcoty + CD = ABcotx

…[from(i)]

\Rightarrow

d = AB (cotx - coty)

….[CD=d]

\Rightarrow

AB = \frac{d}{cotx - coty}

Upon simplifying we get

\Rightarrow AB = \frac{d}{\frac{1}{tanx} - \frac{1}{tany}}

∵ 1 tan θ = cot θ

\Rightarrow

AB = \frac{d (tanxtany)}{tany - tanx}

So, the height of the tower is

\frac{d (tanxtany)}{tany - tanx}

units.
Hence answer is 1.

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