A tower subtends an angle of 30° at a point on the same level as its foot.

Question:

A tower subtends an angle of 30° at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is 60°. The height of the tower is

(a) $\frac{h}{2} m$

(b) $\sqrt{3 h} m$

(c) $\frac{h}{3} m$

 

(d) $\frac{h}{\sqrt{3}} m$

Solution:

Let AB be the tower and C is a point on the same level as its foot such that ACB = 30°

The given situation can be represented as,

Here D is a point h m above the point C.

In ΔBCD,

$\Rightarrow \tan B=\frac{C D}{C B}$

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

$\Rightarrow \sqrt{3}=\frac{h}{C B}$

$\Rightarrow C B=\frac{h}{\sqrt{3}}$

Again in triangle ABC,

$\tan C=\frac{A B}{C B}$

$\Rightarrow \tan 30^{\circ}=\frac{A B}{\left(\frac{h}{\sqrt{3}}\right)} \quad[$ Using (1) $]$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{\left(\frac{h}{\sqrt{3}}\right)}$

$\Rightarrow A B=\frac{h}{3}$

Hence the correct option is $c$.

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