Question

From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be ${30}^{\circ }$ and ${60}^{\circ }$ respectively. Find
(i) the horizontal distance between AB and CD.
(ii) the height of the lamp post.
(iii) the difference between the heights of the building and the lamp post.

Answer 1

We have,

AB=60 m,

∠ACE=30° and ∠ADB=60°

Let BD=CE=x and CD=BE=y

⇒AE=AB-BE=60-y

In ∆ACE,

$tan30=\frac{AE}{EC}$

$\frac{1}{\sqrt{3}}=\frac{60-y}{x}$

$x=60\sqrt{3}-\sqrt{3}y$---(1)

Also, in $△ABD$,

$tan60=\frac{AB}{BD}$

$\sqrt{3}=\frac{60}{x}$

$x=\frac{60}{\sqrt{3}}$

$x=20\sqrt{3}$

Subtitute this value in (1)

we get,

$20\sqrt{3}=60\sqrt{3}-\sqrt{3}y$

$y=40$

(i) the horizontal distance between AB and

CD=BD=x=203=20×1.732=34.64 m. (ii) the height of the lamp post=CD=y=40 m
(iii) the difference between the heights of the building and the lamp post=AB-CD=60-40=20 m