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

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May 29, 2017

Given that a particle is thrown with angle of projection $\theta$ over a triangle $\Delta A C B$ from one of its end $A$ of the horizontal base $A B$ aligned along X-axis and it finally falls at the other end $B$of the base,grazing the vertex $C \left(x , y\right)$

Let $u$ be the velocity of projection, $T$ be the time of flight, $R = A B$ be the horizontal range and $t$ be the time taken by the particle to reach at C $\left(x , y\right)$

The horizontal component of velocity of projection $\to u \cos \theta$

The vertical component of velocity of projection $\to u \sin \theta$

Considering motion under gravity without any air resistance we can write

$y = u \sin \theta t - \frac{1}{2} g {t}^{2.} \ldots . \left[1\right]$

$x = u \cos \theta t \ldots \ldots \ldots \ldots \ldots \ldots . \left[2\right]$

combining [1] and [2] we get

$y = u \sin \theta \times \frac{x}{u \cos \theta} - \frac{1}{2} \times g \times {x}^{2} / \left({u}^{2} {\cos}^{2} \theta\right)$

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

$\implies \textcolor{b l u e}{\frac{y}{x} = \tan \theta - \left(\frac{g {\sec}^{2} \theta}{2 {u}^{2}}\right) x \ldots \ldots . . \left[3\right]}$

Now during time of flight $T$ the vertical displacement is zero
So

$0 = u \sin \theta T - \frac{1}{2} g {T}^{2}$

$\implies T = \frac{2 u \sin \theta}{g}$

Hence horizontal displacement during time of flight i.e. range is given by

$R = u \cos \theta \times T = u \cos \theta \times \frac{2 u \sin \theta}{g} = \frac{{u}^{2} \sin 2 \theta}{g}$

$\implies R = \frac{2 {u}^{2} \tan \theta}{g \left(1 + {\tan}^{2} \theta\right)}$

$\implies R = \frac{2 {u}^{2} \tan \theta}{g {\sec}^{2} \theta}$

$\implies \textcolor{b l u e}{\frac{g {\sec}^{2} \theta}{2 {u}^{2}} = \tan \frac{\theta}{R} \ldots \ldots \left[4\right]}$

Combining [3] and [4] we get

$\frac{y}{x} = \tan \theta - \frac{1}{2} \times \frac{g x}{u} ^ 2 \times {\sec}^{2} \theta$

$\implies \frac{y}{x} = \tan \theta - \frac{x \tan \theta}{R}$

$\implies \tan \alpha = \tan \theta - \frac{x \tan \theta}{R}$ [since $\textcolor{red}{\frac{y}{x} = \tan \alpha}$ from figure]

So $\tan \theta = \tan \alpha \times \left(\frac{R}{R - x}\right)$

$\implies \tan \theta = \tan \alpha \times \left(\frac{R - x + x}{R - x}\right)$

$\implies \tan \theta = \tan \alpha \times \left(1 + \frac{x}{R - x}\right)$

$\implies \tan \theta = \tan \alpha + \frac{x \tan \alpha}{R - x}$

$\implies \tan \theta = \tan \alpha + \frac{y}{R - x}$ [putting $\textcolor{red}{x \tan \alpha = y}$]

Finally we have from figure $\textcolor{m a \ge n t a}{\frac{y}{R - x} = \tan \beta}$

Hence we get our required relation

$\textcolor{g r e e n}{\tan \theta = \tan \alpha + \tan \beta}$

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