Two pillars of equal heights stand on either side of a roadway, which is 120 m wide. At a point on the road lying between the pillars, the elevations of the pillars are 60° and 30° respectively. Find the height of each pillar and the position of the point.

Given,

AB and CD are the two towers of height h meters. E is a point in the roadway BD such that BD = 120 m, ∠AEB = 60° and ∠CED = 30°.

In ∆ABE,

$\Rightarrow \tan 60° = \dfrac{AB}{BE} \\[1em] \Rightarrow \sqrt3 = \dfrac{h}{BE} \\[1em] \Rightarrow BE = \dfrac{h}{\sqrt3} \\[1em] \Rightarrow BE = \dfrac{h}{\sqrt3} \text{ …..(1)}$

In △CDE,

$\Rightarrow \tan 30° = \dfrac{CD}{ED} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{CD}{ED} \\[1em] \Rightarrow ED = \sqrt3CD \\[1em] \Rightarrow ED = \sqrt3h \text{ ….(2)}$

We know that,

⇒ BD = 120 m

⇒ BE + ED = 120 m

From (1) and (2), we get :

$\Rightarrow \dfrac{h}{\sqrt3} + \sqrt{3}h = 120 \\[1em] \Rightarrow \dfrac{h + 3h}{\sqrt3} = 120 \\[1em] \Rightarrow \dfrac{4h}{\sqrt3} = 120 \\[1em] \Rightarrow 4h = 120\sqrt3 \\[1em] \Rightarrow h = \dfrac{120 \times 1.732}{4} \\[1em] \Rightarrow h = 51.96 \text{ meters}$

From equation (1),

$BE = \dfrac{h}{\sqrt3} = \dfrac{51.96}{1.732}$ = 30 meters.

Hence, height of each pillar is 51.96 m and the point E is 30 m from the pillar AB.