Example: An Observer standing on the ground at a certain distance from a tower AB of height 9√3 meters, makes an angle of 60° with the horizontal and another tower CD at a distance of 9 meters from the tower AB makes an angle of 45° with the horizontal . Find the height of tower CD.

Concept for solving the problems

Let the height of tower CD be h2 meter.
In ∆ AOB, tan 60° = ${AB}/{OB}$ = ${9√3}/{x1}$
          $√3$ = ${9√3}/{x1}$
          $x1$ = ${9√3}/{√3}$
		  $x1$ = $9$ meter
		  
In ∆ODC , base OD = OB+BD = 9+9 = 18 meters.
    Hence , tan 45° = ${CD}/{OD}$ = ${H2}/{18}$
	        H2= 18 meters