**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° and 60° 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.

**Solution:**

Let $A B$ be the building of height 60 and $C D$ be the lamp post of height $h$, an angle of depression of the top and bottom of vertical lamp post are $30^{\circ}$ and $60^{\circ}$ respectively. Let $A E=h, A C=x$ and $A C=E D$. It is also given $A B=60 \mathrm{~m}$. Then $B E=60-h$ And $\angle A C B=60^{\circ}, \angle B D E=30^{\circ}$

We have to find the following

(i) The horizontal distance between *AB* and *CD*

(ii) The height of lamp post

(iii) The difference between the heights of building and the lamp post

We have the corresponding figure as follows

(*i*) So we use trigonometric ratios.

$\ln \Delta A B C$

$\Rightarrow \quad \tan 60^{\circ}=\frac{A B}{A C}$

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

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

$\Rightarrow \quad x=34.64$

Hence the distance between $A B$ and $C D$ is $34.64$

(*ii*) Again in

$\Rightarrow \quad \tan 30^{\circ}=\frac{B E}{D E}$

$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{60-h}{x}$

$\Rightarrow \quad \frac{60}{\sqrt{3}}=(60-h) \sqrt{3}$

$\Rightarrow \quad 60=180-3 h$

$\Rightarrow \quad 60=180-3 h$

$\Rightarrow \quad 3 h=120$

$\Rightarrow \quad 3 h=120$

$\Rightarrow \quad h=40$

Hence the height of lamp post is $40 \mathrm{~m}$.

(iii) Since BE $=60-h$

$\Rightarrow \quad B E=60-40$

$\Rightarrow \quad B E=20$

Hence the difference between height of building and lamp post is $20 \mathrm{~m}$.

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