A particle is thrown over a triangle from one end of a horizontal base and after 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, prove that `tan theta = tan alpha + tan beta` .

Explanation

The situation is shown in the figure.
`tanalpha+tanbeta=(y)/(x)+(y)/((R-x))`
where R is the range. `tanalpha+tan beta=(y(R-x)+xy)/(x(R-x))`
or `tan alpha+tan beta=(y)/(x)xx(R)/((r-x))`.....(1)
we know `y=xtan theta(1-(x)/(R))`
or `tantheta=(y)/(x)xx(R)/((R-x))`......(2)
From equation (1) and (2), we have `tantheta=tanalpha+tanbeta`