Given that a particle is thrown with angle of projection #theta# over a triangle #DeltaACB# from one of its end #A# of the horizontal base #AB# aligned along X-axis and it finally falls at the other end #B#of the base,grazing the vertex #C(x,y)#
Let #u# be the velocity of projection, #T# be the time of flight, #R=AB# be the horizontal range and #t# be the time taken by the particle to reach at C #(x,y)#
The horizontal component of velocity of projection #->ucostheta#
The vertical component of velocity of projection #->usintheta#
Considering motion under gravity without any air resistance we can write
#y=usinthetat-1/2 g t^2.....[1]#
#x=ucosthetat...................[2]#
combining [1] and [2] we get
#y=usinthetaxxx/(ucostheta)-1/2 xxgxxx^2/(u^2cos^2theta)#
#=>y=usinthetaxxx/(ucostheta)-1/2 xxgxxx^2/u^2xxsec^2theta#
#=>color(blue)(y/x=tantheta-((gsec^2theta)/(2u^2))x........[3])#
Now during time of flight #T# the vertical displacement is zero
So
#0=usinthetaT-1/2 g T^2#
#=>T=(2usintheta)/g#
Hence horizontal displacement during time of flight i.e. range is given by
#R=ucosthetaxxT=ucosthetaxx(2usintheta)/g=(u^2sin2theta)/g#
#=>R=(2u^2tantheta)/(g(1+tan^2theta))#
#=>R=(2u^2tantheta)/(gsec^2theta)#
#=>color(blue)((gsec^2theta)/(2u^2)=tantheta/R......[4])#
Combining [3] and [4] we get
#y/x=tantheta-1/2 xx(gx)/u^2xxsec^2theta#
#=>y/x=tantheta-(xtantheta)/R#
#=>tanalpha=tantheta-(xtantheta)/R# [since #color(red)(y/x=tanalpha)# from figure]
So #tantheta=tanalphaxx(R/(R-x))#
#=>tantheta=tanalphaxx((R-x+x)/(R-x))#
#=>tantheta=tanalphaxx(1+x/(R-x))#
#=>tantheta=tanalpha+(xtanalpha)/(R-x)#
#=>tantheta=tanalpha+y/(R-x)# [putting #color(red)(xtanalpha=y)#]
Finally we have from figure #color(magenta)(y/(R-x)=tanbeta)#
Hence we get our required relation
#color(green)(tantheta=tanalpha+tanbeta)#
