Question

A particle is projected over a triangle from one end of a horizontal base and 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, then the correct relation between $\alpha ,\beta$ and $\theta$ is

• A. $\sin \theta =\cos \alpha +\tan \beta$
• B. $\tan \theta =\tan \alpha +\tan \beta$
• C. $\cos \theta =\cos \alpha +\cos \beta$
• D. $\sin \alpha =\sin \theta +\sin \beta$

Answer 1

Answer: (B)

For any point $P(x,y)$,
we have $y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

$=x\tan \theta \left[1-\frac{gx}{2u^2{\cos }^2\theta \tan \theta }\right]$
$=x\tan \theta \left[1-\frac{gx}{u^2(2\sin \theta \cos \theta )}\right]=x\tan \theta \left[1-\frac{x}{R}\right]$
$=x\tan \theta \left[\frac{R-x}{R}\right]\Rightarrow \tan \theta =\frac{yR}{x(R-x)}$
From figure;
$\tan \alpha +\tan \beta =\frac{y}{x}+\frac{y}{R-x}=\frac{y(R-x)+xy}{x(R-x)}=\frac{yR}{x(R-x)}$
Hence, $\tan \theta =\tan \alpha +\tan \beta$