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 equation of trajectory is
`y=x tan alpha-(gx^(2))/(2u^(2)cos ^(2)alpha)`
`y=x tan alpha(1-x//R)` where R=range
`[because R=(2u^(2) sin alpha cos alpha)/(g)]`
The co-ordinates of A are `(h cot theta,h)` and range=OB `=h cot theta+ cot phi`. Substituting the co-ordinatteso fA in th equation of trajectory.
`h=h cot theta alpha[1-(h cot theta)/(h cot theta+h cot phi)]`
`tan theta=(tan alpa cot phi)/(cot theta+cot phi)`
`rArr tan theta cot theta+tan theta cot phi=tan alpha cot phi`
`rArr 1+(tan theta)/(tan phi)=(tan alpha)/(tan phi)`
`rArr tan phi+tan theta=tan alpha`

For a particle projected with an initial velocity u at an angle `theta` trajectory is
`y=x tan alpha-(gx^(2))/(2u^(2)cos^(2)alpha)`