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 $a$ and $b$ are the base angles of the triangle and $\theta$ the angle of projection, prove that $tan\theta =tan\alpha +\tan \beta$

Answer 1

Let $ABC$ be the triangle with base $BC$. Let $h$ be the height of the vertex $A$ above $BC$. If $AM$ is the perpendicular drawn on base $BC$ from vertex $A$, then $tan\alpha =h/a$ and $tan\beta =h/b$, where $BM=a$ and $CM=b$.
Since $A(a,h)$ lies on the trajectory of the projectile,
$y=xtan\theta (1-\frac{x}{R})$ ..(i)
Therefore, it should satisfy (i) i.e.,
$h=atan\theta (1-\frac{a}{a+b})$
[$\therefore$ Range, $R=a+b$]
$\frac{h}{a}=tan\theta \left[\frac{b}{a+b}\right]$
$\Rightarrow tan\theta =\frac{h}{a}+\frac{h}{b}=tan\alpha +tan\beta$
Hence, proved.