Let CB be the cliff . From point A the angle of elevation of the peak C of the cliff is #/_CAB=theta#'. After going up a distance , #AD=k# #m# towards the top of the cliff at an angle,#/_DAE =phi#,it ls found the angle of elevation #/_CDF=alpha#.
DF and DE are perpendiculars drawn from D on CB and AB.
Now #DE =ksinphi and AE =kcosphi#
Let
#h="CB the height of the cliff"
and b=BA#
For #DeltaCAB," "(CB)/(BA)=tantheta#
#=>b/h=cottheta=>b=hcottheta#
Now #DF=BE=BA-AE=b-kcosphi#
#CF=CB-FB=CB-DE=h-ksinphi#
For #DeltaCDF," "(CF)/(DF)=tanalpha#
#=>(h-ksinphi)/(b-kcosphi)=tanalpha#
#=>(h-ksinphi)/(hcottheta-kcosphi)=tanalpha#
#=>h-htanalphacottheta=ksinphi-kcosphitanalpha#
#=>h=(k(sinphi-cosphitanalpha))/(1-tanalphacottheta)#
#=>h=(ksintheta(sinphicosalpha-cosphisinalpha))/(sinthetacosalpha-costhetasinalpha)#
#=>h=(ksinthetasin(phi-alpha))/(sin(theta-alpha))#
#=>h=(ksinthetasin(alpha-phi))/(sin(alpha-theta))#
