Question

A flag is standing vertically on a tower of height $b.$ On a point at a distance$a$ from the foot of the tower, the flag and the tower subtend equal angles. The height of the flag is.

• A. $\frac{b.[a^2+b^2]}{[a^2–b^2]}$
• B. $\frac{a.[a^2-b^2]}{[a^2+b^2]}$
• C. $\frac{b.[a^2-b^2]}{[a^2+b^2]}$
• D. $\frac{a.[a^2+b^2]}{[a^2-b^2]}$

Answer 1

The correct option is A

$\frac{b.[a^2+b^2]}{[a^2–b^2]}$

Explanation for the correct answer.

Step 1: Find the value of $\tan \left(\theta \right)$:

As per the diagram

$$\begin{gathered} In∆QRS, \\ \tan \left(\theta \right)=\frac{QR}{RS}\mathbf{[}\mathbf{\because }\mathbf{tan}\left(\theta \right)\mathbf{=}\frac{\mathbf{Opposite}\mathbf{}\mathbf{side}}{\mathbf{Adjacent}\mathbf{}\mathbf{side}}\mathbf{]} \\ \Rightarrow \tan \left(\theta \right)=\frac{b}{a}\_\_\_\_\_\left(1\right) \end{gathered}$$

Step 2: Find the height of the flag:

$\begin{array}{rcl}& & \mathrm{In}∆\mathrm{PRS},\end{array}$

$$\begin{gathered} \tan \left(2\mathrm{\theta }\right)=\frac{\mathrm{PR}}{\mathrm{RS}}\mathbf{[}\mathbf{\because }\mathbf{\angle }\mathbf{PSR}\mathbf{}\mathbf{=}\mathbf{\theta }\mathbf{}\mathbf{+}\mathbf{\theta }\mathbf{\Rightarrow }\mathbf{\angle }\mathbf{PSR}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{\theta }\mathbf{]} \\ \Rightarrow \tan \left(2\mathrm{\theta }\right)=\frac{\mathrm{h}+\mathrm{b}}{\mathrm{a}}\left[\mathbf{\because }\mathbf{PR}\mathbf{=}\mathbf{h}\mathbf{+}\mathbf{b}\right] \\ \Rightarrow \frac{2\tan \left(\mathrm{\theta }\right)}{1-{\tan }^2\left(\mathrm{\theta }\right)}=\frac{\mathrm{h}+\mathrm{b}}{\mathrm{a}}\left[\mathbf{\because }\mathbf{tan}\mathbf{\left(}\mathbf{2}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{tan}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}}{\mathbf{1}\mathbf{-}{\mathbf{tan}}^{\mathbf{2}}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}}\right] \\ \Rightarrow \frac{2\left({\displaystyle \frac{\mathrm{b}}{\mathrm{a}}}\right)}{1-{\left({\displaystyle \frac{\mathrm{b}}{\mathrm{a}}}\right)}^2}=\frac{\mathrm{h}+\mathrm{b}}{\mathrm{a}}\left[\mathbf{Substitute}\mathbf{}\mathbf{the}\mathbf{}\mathbf{value}\mathbf{}\mathbf{of}\mathbf{}\mathbf{tan}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}\mathbf{from}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\right] \\ \Rightarrow \frac{2\mathrm{ab}}{{\mathrm{a}}^2-{\mathrm{b}}^2}=\frac{\mathrm{h}+\mathrm{b}}{\mathrm{a}} \\ \Rightarrow \mathrm{h}+\mathrm{b}=\frac{2{\mathrm{a}}^2\mathrm{b}}{{\mathrm{a}}^2-{\mathrm{b}}^2} \\ \Rightarrow \mathrm{h}=\frac{2{\mathrm{a}}^2\mathrm{b}}{{\mathrm{a}}^2-{\mathrm{b}}^2}-\mathrm{b} \\ \Rightarrow \mathrm{h}=\frac{2{\mathrm{a}}^2\mathrm{b}-{\mathrm{ba}}^2+{\mathrm{b}}^3}{{\mathrm{a}}^2-{\mathrm{b}}^2} \\ \Rightarrow \mathrm{h}=\frac{\mathrm{b}({\mathrm{a}}^2+{\mathrm{b}}^2)}{({\mathrm{a}}^2-{\mathrm{b}}^2)} \end{gathered}$$

Hence, Option(A) is the correct answer.