Let a vertical tower AB have its end A on the level ground. Let C be the mid‑point of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β, then tan β is equal to

Engineering

Mathematics

problems on height and distance

Question

$\frac{4}{9}$

$\frac{1}{4}$

$\frac{6}{7}$

$\frac{2}{9}$

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$tan α = \frac{x}{4 x} = \frac{1}{4}$

$tan (α + β) = \frac{2 x}{4 x} = \frac{1}{2}$

$\frac{tanα + tanβ}{1 - tanα \cdot tanβ} = \frac{1}{2}$

⇒ 2 (tan α + tan β) = 1 – tan α · tan β

$\Rightarrow 2 (\frac{1}{4} + tanβ) = 1 - \frac{\tan β}{4} \Rightarrow \frac{1}{2} + 2tan β = 1 - \frac{\tan β}{4}$

$\Rightarrow 2tan β + \frac{tanβ}{4} = \frac{1}{2} \Rightarrow \frac{9}{4} \Rightarrow tan β = \frac{1}{2} \Rightarrow tan β = \frac{2}{9} .$