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))#