Question

A particle is thrown over a triangle from one end of a horizontal base and after grazing the vertex falls on the other end of the base. If $\alpha and\beta$ be the base angles and $\theta$ the angle of projection, prove that $\tan \theta =\tan \alpha +\tan \beta .$

Answer 1

$\tan \alpha +\tan \beta =\frac{y}{x}+\frac{y}{R-x}$

$\tan \alpha +\tan \beta =\frac{yR}{x(R-x)}$ ...(i)

Equation of trajectory is

$y=x\tan \theta [1-\frac{x}{R}]$

or, $\tan \theta =\frac{yR}{x(R-x)}$ ...(ii)

From Eqs. (i) and (ii), we have

$\tan \theta =\tan \alpha +\tan \beta$