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 $y=x \tan \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}$

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