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A tower of height 30 metres casts a shadow of length 40 metres at a certain instant , when the sun’s elevation increases by $\tan^{- 1} (\frac{1}{7})$ , the length of the shadow cast by that tower , in metre is

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answer is C.

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Detailed Solution

In $Δ A B C, \tan α = \frac{30}{40} = \frac{3}{4}$

$θ = α + \tan^{- 1} (\frac{1}{7})$

In $Δ A B D, \tan θ = \frac{30}{x}$

$\tan (α + \tan^{- 1} (\frac{1}{7})) = \frac{30}{x}$

$\frac{\tan α + \frac{1}{7}}{1 - \tan α . \frac{1}{7}} = \frac{30}{x} \Rightarrow \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4} \times \frac{1}{7}} = \frac{30}{x}$

$\Rightarrow x = 30 m$

$∴$ The length of this shadow cast by the towel is $x = 30 m$

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