Question

From the top of a tower, the angle of depression of a point on the ground is $60°$. If the distance of this point from the tower is $\frac{1}{(\sqrt{3}+1)}m$, then the height of the tower is

• A. $\frac{\left(4\sqrt{3}\right)}{2}$m
• B. $\frac{(\sqrt{3}+3)}{2}$m
• C. $\frac{(3–\sqrt{3})}{2}$m
• D. $\frac{\sqrt{3}}{2}$m
• E. $\sqrt{3}+1$m

Answer 1

The correct option is C

$\frac{(3–\sqrt{3})}{2}$m

Explanation for the correct option:

Step 1. Draw the figure according the question:

From the figure, we get

$QR=\frac{1}{(\sqrt{3}+1)}m$

Step 2. Let the height of the tower be $h$ m.

Given, $\angle APR=60°$

$\angle PRQ=\angle APR$

$=60°$ [$\because$$\angle APR$ and $\angle PRQ$ are alternate interior angles]

Step 3. In the right-angled $△PQR$, we get

$\frac{PQ}{QR}=\tan 60°$

$\Rightarrow$$\frac{h}{\left[{\displaystyle \frac{1}{(\sqrt{3}+1)}}\right]}=\sqrt{3}$

$\Rightarrow$ $h=\frac{\sqrt{3}}{(\sqrt{3}+1)}$

$\Rightarrow$ $h=\frac{\sqrt{3}}{(\sqrt{3}+1)}\times \frac{(\sqrt{3}–1)}{(\sqrt{3}–1)}$ [Rationalizing denominator]

$=\frac{[3–\sqrt{3}]}{2}$

Thus, the height of the tower is $\frac{\mathbf{[}\mathbf{3}\mathbf{}\mathbf{–}\mathbf{}\sqrt{\mathbf{3}}\mathbf{]}}{\mathbf{2}}$ m

Hence, Option ‘C’ is Correct.