A lower 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. $\frac{h}{\sqrt{3}} m$ Answer: Option CShow Answer
Solution(By Apex Team)
Let CD is the tower and A is a point such that the angle of elevation of C is 30°
B is and their point h m high of A and angle of depression of D is 60°
$\begin{array}{l}\text{AB}=h\text{ metre}\\
\text{Let CD}=\text{h}\ \text{metre}\\
\text{and AD}=x\\
\text{Now in right }\triangle\ \text{ABD,}\\
\tan\theta=\frac{AB}{AD}\\
\Rightarrow\tan60^{\circ}=\frac{h}{x}\\
\Rightarrow\sqrt{3}=\frac{h}{x}\\
\Rightarrow x=\frac{h}{\sqrt{3}}\ldots\ldots\text{ (i) }\\
\text{Similarily in right }\triangle\ \text{ACB,}\\
\tan30^{\circ}=\frac{CD}{AD}\\
\Rightarrow\frac{1}{\sqrt{3}}=\frac{H}{x}\\
\Rightarrow x=\sqrt{3}H\ldots\ldots.\\
\text{ From (i) and (ii) }\\
\sqrt{3}H=\frac{h}{\sqrt{3}}\\
H=\frac{h}{\sqrt{3}\times\sqrt{3}}=\frac{h}{3}\\
\therefore\text{Height of tower}\\
=\frac{h}{3}\end{array}$
