Given ,distance between two observer P and Q be $a$ meters.
Let the height of tower be $h$ meter
Let the distance between observer Q and base of tower be $x$ meter

In ∆QAB , tan 65° = $h/x$ 
           $√3$ = $h/x$
		   $x$ = $h/√3$ => ${√3h}/3$  ......(1)
		   
In ∆PAB ,  tan 45° = $h/{a+x}$
               1 = $h/{a+x}$
			   $h$ = $a+x$
substitute the value of $x$ from eqn (1) in above eqn.
 Therefore    $h$ = $a+ {√3h}/3$
          =>  $h - {√3h}/3$  = $a$
		  =>  ${3h-√3h}/3$  = $a$
		  => $h$ = ${3a}/{3-√3}$ => ${(3+√3)a}/2$