A particle is projected over a traingle from one end of a horizontal base and 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`.

Options

A. `sin theta = cos alpha + tan beta`
B. `tan theta = tan alpha + tan beta`
C. `cos theta = cos alpha + cos beta`
D. `sin alpha + sin theta + sin beta`

Answer

B

Explanation

For any point P (x,y),
we have
`y=x tan theta - (gx^2)/(2u^2 cos^2 theta)`
`=x tan theta [1-(gx)/(2u^2 cos^2 theta tan theta)]`
`=x tan theta [1-(gx)/(u^2(2 sin theta cos theta))]= x tan theta [1-x/R]`
`=x tan theta [(R-x)/R] rArr tan theta =(yR)/(x(R-x))`
From figure ,
`tan alpha + tan beta=y/x + y/(R-x) = (y(R-x)+xy)/(x(R-x))=(yR)/(x(R-x))`
Hence , `tan theta =tan alpha + tanbeta`