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The upper three-quarters of a vertical pole subtends an angle $\tan^{- 1} (3 / 5)$ at a point in the horizontal plane through its foot and distant $40 m$ from it. The height of the pole is

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$80 m$

$100 m$

$160 m$

$200 m$

answer is C.

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Detailed Solution

$\tan θ = \frac{h}{160}, \tan (α + θ) = \frac{h}{40}$ .

$\Rightarrow \frac{\tan α + \frac{h}{160}}{1 - \frac{h}{160} \cdot \tan α} = \frac{h}{40}$

$\begin{array}{l} \Rightarrow \frac{\frac{3}{5} + \frac{h}{160}}{1 - \frac{h}{160} \times \frac{3}{5}} = \frac{h}{40} \\ \Rightarrow 3 h^{2} - 600 h + 19200 = 0 \end{array}$

$\Rightarrow h = 40$ or $160 .$

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