projectile..

let one end of horizontal base from wer partical is projected be origin(0,0)
let K(x,y) be vertex of triangle
so tan α=y/x
tanβ=y/(R-x)....R is lenght of base of traingle

use eq of projectile
vertex of traingle will satisfy that eq of projectile
so y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}
\Rightarrow \;\;\;\;\; y = x\tan \theta \left[ {1 - \frac{{gx}}{{2{u^2}\cos \theta .\sin \theta }}} \right]}

\Rightarrow \;\;\;\;\; \frac{y}{x} = \left[ {1 - \frac{x}{R}} \right]\tan \theta
\Rightarrow \;\;\;\;\; \tan \theta = \frac{{yR}}{{x(R - x)}} = \frac{y}{x} + \frac{y}{{R - x}}

here theta is q