A tower, of $x$ metres high, has a flagstaff at its top. The tower and the flagstaff subtend equal angles at a point distant $y$ metres from the foot of the tower. Then, the length of the flagstaff (in metres), is

• A

  $\frac{y\left(x^2-y^2\right)}{\left(x^2+y^2\right)}$
• Correct answer
  B

  $\frac{x\left(y^2+x^2\right)}{\left(y^2-x^2\right)}$
• C

  $\frac{x\left(x^2+y^2\right)}{\left(x^2-y^2\right)}$
• D

  $\frac{x\left(x^2-y^2\right)}{\left(x^2+y^2\right)}$

Solution

Let $B C$ be the height of tower and $C D$ be height of the flagstaff,
Let $C D=h$ Since, the tower and flagstaff makes equal angle, i.e. $\theta$ In $\triangle B A C$,
In $\triangle D A B$, $\begin{aligned} \tan 2 \theta & =\frac{x+h}{y} \\ \Rightarrow \quad \frac{2 \tan \theta}{1-\tan ^2 \theta} & =\frac{x+h}{y} \\ \Rightarrow \quad \frac{2\left(\frac{x}{y}\right)}{1-\frac{x^2}{y^2}} & =\frac{x+h}{y} \\ \Rightarrow \quad 2 x y^2 & =\left(y^2-x^2\right)(x+h) \\ \Rightarrow \quad & \text { [from Eq. (i)] } \\ \Rightarrow \quad x^2+x^3 & =\left(y^2-x^2\right) h \\ h & =\frac{x\left(x^2+y^2\right)}{\left(y^2-x^2\right)} \end{aligned}$

Hence, the correct answer is option: 'B'.