From the top of a building 60 m high, the angles of depression of the top and bottom of a vertical lamp post are observed to be 30° and 60° respectively. Find the height of the lamp post.

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Solution

Let $A B$ be the building and $C D$ be the vertical lamppost.
Here,
$A B = 60$ m, ∠ $A C B = 60^{o}$ and ∠ $A D E = 30^{o}$
Let $C D = h$ m such that $B E = h$ m. Thus, $A E = (A B - B E) = (60 - h)$ m and $B C = D E = x$ m.

From ∆ $A B C$ , we have:
$\frac{A B}{B C} = \tan 60^{o} = \sqrt{3}$

⇒ $\frac{60}{x}$ = $\sqrt{3}$
⇒ $x = \frac{60}{\sqrt{3}}$ ...(i)

From ∆ $A E D$ , we have:
$\frac{A E}{D E} = \tan 30^{o} = \frac{1}{\sqrt{3}}$

⇒ $\frac{(60 - h)}{x} = \frac{1}{\sqrt{3}}$

⇒ $x = (60 - h) \sqrt{3}$ ...(ii)

From (i) and (ii), we have:
$\frac{60}{\sqrt{3}} = (60 - h) \sqrt{3}$
⇒ $60 = 3 (60 - h)$
⇒ $60 = 180 - 3 h$
⇒ $3 h = 180 - 60 = 120$
⇒ $h = \frac{120}{3} = 40$ m.
∴ Height of the lamppost = $C D = h = 40$ m

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Similar questions

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.

In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top & bottom of a vertical lamp post CD are observed to be 30^o& 60^o respectively. Find:

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