Two poles of equal heights are standing opposite to each other on either side of a 100 m wide road. From a point between them on the road, the angles of elevation of the top of the poles are 60^o and 30^o, respectively. Find the height of the poles and the distance of the point from the nearer of the two poles.

Let AC be the road and poles AB and CD are standing opposite to each other.

AB = CD  = x m Let    PC  =   y m   AP  = (100 - y) m,                  Angle  CPD  = 60° and angle APB = 30° In ^_$\triangle$DCP, DC/PC = tan 60° x/y = tan 60° = ^{_{$\sqrt 3$}} x = ^{$\sqrt 3$_y}.........(1) In ^_$\triangle$ABP, AB/AP = tan 30° x/(100-y) = 1/^{_{$\sqrt 3$} $\sqrt 3$}x = 100-y ...... (2) Substituting the value of x from (1) into (2), we get^{_{$\sqrt 3$}} x ^{$\sqrt 3$_y} = 100 - y 3y = 100 - y y =25. Putting the value of y in (1), we get x = ^{$\sqrt 3$} x 25 = 25 (1.73) = 43.3 m The height of pole is 43.3 m and distance of point P from the nearer of the 2 poles is 25 m.