A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If alpha and betabe the base angles and thetabe the angle of projection, prove that tan theta = tan alpha+ tan beta

A particle is thrown over a triangle from one end of a horizontal base

1.	A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If alpha and betabe the base angles and thetabe the angle of projection, prove that tan theta = tan alpha+ tan beta
Answer» Solution :The SITUATION is SHOWN in figure. From figure, we have `tan alpha + tan beta = y/x + y/(R-x), tan alpha + tan beta = (yR)/(x(R-x))…(i)` But equation of trajectory is `Y = x tan THETA [1-x/R]` `tan theta =(yR)/(x(R-x))...(ii)` From Eqs. (i) and (ii), `tan theta = tan alpha + tan beta`

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A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If alpha and betabe the base angles and thetabe the angle of projection, prove that tan theta = tan alpha+ tan beta

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