Question

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 $\alpha$ and $\beta$ be the base angles and $\theta$ the angle of projection then correct. Relation between $\left(\theta \right)$, $\left(\alpha \right)$ and $\left(\beta \right)$ is:

• A. $\tan \theta =\tan \alpha -\tan \beta$
• B. $\tan \theta =\tan \alpha +\tan \beta$
• C. $\tan \beta =\tan \theta -2\tan \alpha$
• D. None of these

Answer 1

Answer: (B)

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

Let $BD=x,CD=R-x,AD=y$

$BAC$ is path followed by particle.

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

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

$=y\left[\frac{R}{x(R-x)}\right]$

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

$y=x(\tan \alpha +\tan \beta )(1-\frac{x}{R})$..(1)

Now ,if we find the equation of trajectory of projectile motion,

$y=u\sin \theta t-\frac{1}{2}g{t}^2$

$x=u\cos \theta t$

Put $t=\frac{x}{u\cos \theta }$ in y,

$y=\frac{u\sin \theta x}{u\cos \theta }-\frac{1}{2}g-\frac{x^2}{u^2{\cos }^2\theta }$

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

$=x\tan \theta (1-\frac{g{x}^2}{2u^2{\cos }^2\theta .x\tan \theta }$)

$y=x\tan \theta (1-\frac{gx}{u^2\sin 2\theta }$)

$y=x\tan \theta (1-\frac{x}{R}$) ..(2)

comparing (1) and (2)

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