Question

A vertical pole consists of two parts, the lower part being one-third of the whole. At a point in the horizontal plane through the base of the pole and distance $20$ meters from it, the upper part of the pole subtends an angle whose tangent is $\frac{1}{2}$. The possible heights of the pole are

• A. $20m$ and $20\sqrt{3}m$
• B. $20m$ and $60m$
• C. $16m$ and $48m$
• D. none of these

Answer 1

The correct option is B

$20m$ and $60m$

Explanation for correct option:

Step-1: Draw the figure with the help of given data.

Let, the height of pole, $AB=H$. Let point $D$ be at the height of $\frac{H}{3}$.

Then, $AD=\frac{2H}{3}$ and $BD=\frac{H}{3}$

The distance $BC=20m$.

Step-2: Finding height of the pole.

In $∆DBC,$

$$\begin{gathered} \tan \left(\alpha \right)=\frac{{\displaystyle \frac{H}{3}}}{20} \\ \tan \left(\alpha \right)=\frac{H}{60} \end{gathered}$$

Also,

$\tan \left(\theta \right)=\frac{1}{2}$ (given)

In $∆ABC,$

$\Rightarrow$ $\tan \left(\alpha +\theta \right)=\frac{H}{20}$

$\Rightarrow$ $\frac{\tan \left(\alpha \right)+\tan \left(\theta \right)}{1-\tan \left(\alpha \right)\tan \left(\theta \right)}=\frac{H}{20}$ $\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{(}\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{)}\mathbf{=}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\mathbf{B}\mathbf{\right)}}$

$\Rightarrow$ $\frac{\left({\displaystyle \frac{H}{60}}+{\displaystyle \frac{1}{2}}\right)}{\left(1-{\displaystyle \frac{H}{60}}\times {\displaystyle \frac{1}{2}}\right)}=\frac{H}{20}$

$\Rightarrow$ $\frac{\left({\displaystyle \frac{H+30}{60}}\right)}{\left({\displaystyle \frac{120-H}{120}}\right)}=\frac{H}{20}$

$\Rightarrow$ $\frac{2H+60}{120-H}=\frac{H}{20}$

$\Rightarrow$ $H^2-80H+1200=0$

$\Rightarrow$ $H^2-60H-20H+1200=0$

$\Rightarrow$ $\left(H-60\right)\left(H-20\right)=0$

$\Rightarrow$ $H=60m\mathrm{and}\mathit{20}m$

Hence, correct answer is option $\left(B\right)$.