Question

The distance between two vertical poles is $60$ m. The height of one of the poles is double the height of the other. The angles of elevation of the top of the top of the poles from the middle point of the line segment joining their feet are complementary to each other. The heights of the poles are :

• A. $10$ m and $20$ m
• B. $20$ m and $40$ m
• C. $20.9$ m and $41.8$ m
• D. $15\sqrt{2}$ m and $30\sqrt{2}$ m

Answer 1

The correct option is D $15\sqrt{2}$ m and $30\sqrt{2}$ m

Given that, the distance between the two poles is $60m$ and the height of one pole is double that of the other.

Let the height of pole $CD=hm$

Then, the height of the pole$AB=2hm$

The above figure can be drawn using all the information.

Let $X$ be the midpoint of $BD$

The angles of elevation of the top of the poles from the midpoint of $BD$ are complementary.

Hence, if $\angle DXC=\theta ,\text{then,}\angle BXA={90}^{\circ }-\theta$

We know that, $\tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}$

So, in $△CDX$

$\tan \theta =\frac{CD}{DX}=\frac{h}{30}$ ......$\left(1\right)$

In $△ABX$

$\tan ({90}^{\circ }-\theta )=\frac{AB}{BX}=\frac{2h}{30}$

We know that, $\tan ({90}^{\circ }-\theta )=\cot \theta$

Hence, $\cot \theta =\frac{AB}{BX}=\frac{2h}{30}$ ......$\left(2\right)$

Multiplying $\left(1\right)$ and $\left(2\right)$, we get:

$\tan \theta \times \cot \theta =\frac{h}{30}\times \frac{2h}{30}$

$\Rightarrow 1=\frac{h^2}{450}$

$\Rightarrow h=15\sqrt{2}m$

$\therefore 2h=2\times 15\sqrt{2}$

$=30\sqrt{2}m$

Hence, the height of smaller poleis $15\sqrt{2}m$ and the height of the bigger pole is $30\sqrt{2}m$.