Question

A tower $T_1$ of height $60m$ is located exactly opposite to a tower $T_2$ of height $80m$ on a straight road. From the top of $T_2$, if the angle of depression of the foot of $T_1$ is twice the angle of elevation of the top of $T_2$ from top of $T_1,$ then find the width of the road between the feet of the towers $T_1$ and $T_2$.$(\text{Use}\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta })$

• A. $20\sqrt{2}$
• B. $10\sqrt{2}$
• C. $10\sqrt{3}$
• D. $20\sqrt{3}$

Answer 1

Answer: (A)

$20\sqrt{2}$

Let the width of the road between the feet of the towers $T_1$ and $T_2$ be $w.$

In $△ABC,$

$\tan \theta =\frac{BC}{AC}$

$\Rightarrow \tan \theta =\frac{20}{w}\dots \left(i\right)$

Now, in $△BOD,$

$\tan 2\theta =\frac{BO}{OD}$

$\Rightarrow \tan 2\theta =\frac{80}{w}\dots \left(ii\right)$

Now, by using the given formula of $\tan 2\theta$ in the question,

$\begin{array}{cc}\tan 2\theta & =\frac{2\tan \theta }{1-{\tan }^2\theta }\\ \frac{80}{w}& =\frac{2\times \frac{20}{w}}{1-{\left(\frac{20}{w}\right)}^2}\\ \left[1-{\left(\frac{20}{w}\right)}^2\right]\times \frac{80}{w}& =\frac{40}{w}\\ 1-{\left(\frac{20}{w}\right)}^2& =\frac{1}{2}\\ {\left(\frac{20}{w}\right)}^2& =\frac{1}{2}\\ \frac{20}{w}& =\frac{1}{\sqrt{2}}\\ w& =20\sqrt{2}m\end{array}$

Hence, the distance between the two towers is equal to $20\sqrt{2}m.$