A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If $α$ and $β$ be the base angles and $θ$ the angle of projection, then

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$\tan^{2} θ = \tan^{2} α + \tan^{2} β$

$\tan θ = \tan α + \tan β$

$\tan θ = \sqrt{\tan αtan β}$

$\tan θ = \frac{\tan αtan β}{\tan α + \tan β}$

answer is B.

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Detailed Solution

$\begin{array}{l} y = xtan θ (1 - \frac{x}{R}) . . . . . . . (i) \\ \tan α = \frac{y}{x} . . . . . . . (ii) \\ \tan β = \frac{y}{R - x} . . . . . (iii) \end{array}$

Solving Eqs. (i), (ii) and (iii), we get

$\tan θ = \tan α + \tan β$

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