Question

A particle is projected over a triangle from one extremity of its horizontal base. Grazing over the vertex, it falls on the other extremity of the base. If $\alpha$ and $\beta$ be the base angles of the triangle and 𝜃 the angle of projection, then which of the following holds true ?

• A. $tan\theta =\tan \alpha +tan\beta$
• B. $tan\alpha =\tan \theta +tan\beta$
• C. $tan\beta =\tan \theta +tan\alpha$​​​​​​​
• D. $tan\theta =\tan \alpha -tan\beta$

Answer 1

The correct option is A $tan\theta =\tan \alpha +tan\beta$

From geometry,
$asin\alpha =bsin\beta$
Equation of trajectory ,
$y=xtan\theta -\frac{g{x}^2}{2u^{2}co{s}^2\theta }$
Since $y=asin\alpha$ and $x=acos\alpha$ from figure
$asin\alpha =acos\alpha tan\theta -\frac{g{a}^{2}co{s}^2\alpha }{2u^{2}co{s}^2\theta }$
Dividing the whole equation by $cos\alpha$,
$\begin{array}{}tan\alpha =tan\theta -\frac{gacos\alpha }{2u^{2}co{s}^2\theta }& \text{(1)}\end{array}$
Since range, $R=\frac{u^{2}sin2\theta }{g}$
Using the above equation in equation $\left(1\right)$ we get,
$\begin{array}{}tan\alpha =tan\theta -\frac{acos\alpha tan\theta }{R}& \text{(2)}\end{array}$
From geometry, range can be written as,
$R=acos\alpha +bcos\beta$
Substituting this range value in $\left(2\right)$ we get,
$\begin{array}{}tan\alpha =tan\theta [1-\frac{acos\alpha }{acos\alpha +bcos\beta }]& \text{(3)}\end{array}$
Since $asin\alpha =bsin\beta$
$\Rightarrow \frac{a}{b}=\frac{sin\beta }{sin\alpha }$
Using this in equation $\left(3\right)$,
$tan\alpha =tan\theta [1-\frac{1}{1+\frac{sin\alpha }{sin\beta }\frac{cos\beta }{cos\alpha }}]$
$\begin{array}{}tan\alpha =tan\theta [1-\frac{tan\beta }{tan\alpha +tan\beta }]& \text{(4)}\end{array}$
On rearranging equation $\left(4\right)$ becomes,
$tan\theta =\tan \alpha +tan\beta$