From triangle BCD, tan $\alpha$ = ${{80} \over x}$ and from triangle BFE, tan $\alpha$ = ${{h} \over y}$ $\therefore$ ${{80} \over x}$ = ${{h} \over y}$ $\Rightarrow$ $y = {{hx} \over {80}}$ ........ (1) From triangle ABC, tan $\beta$ = ${{20} \over x}$ and from triangle EFC, tan $\beta$ = ${{h} \over {x-y}}$ $\therefore$ ${{20} \over x}$ = ${{h} \over {x-y}}$ $\Rightarrow$ $x-y = {{hx} \over {20}}$ ........ (2) By adding equation (1) and (2) we get, x = ${{hx} \over {80}}$ + ${{hx} \over {20}}$ $\Rightarrow$ 1 = ${{h} \over {80}}$ + ${{h} \over {20}}$ $\Rightarrow$ h = 16 m $\therefore$ Height of intersection point is 16 m