Question

$A$ tower of $x$ meters high has a flagstaff at its top. The tower and the flagstaff subtend equal angles at a point distant $y$ meters from the foot of the tower then the length of the flagstaff in meters is

• A. $\frac{y(x^2-y^2)}{x^2+y^2}$
• B. $\frac{x(y^2+x^2)}{y^2-x^2}$
• C. $\frac{x(x^2+y^2)}{x^2-y^2}$
• D. $\frac{x(x^2-y^2)}{x^2+y^2}$

Answer 1

Answer: (B)

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

Let $BD$ be the height of tower and $DA$ be the height of flagstaff.

Let $DA=h$

Since, the tower and flagstaff makes equal angle i.e,

In $△BCD,\tan \theta =\frac{x}{y}⟶\left(1\right)$

In $△BCA,\tan 2\theta =\frac{x+h}{y}$

$\Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{x+h}{y}$

From equation $\left(1\right),$ put $\tan \theta =\frac{x}{y},$ we get

$\Rightarrow \frac{2\left(\frac{x}{y}\right)}{1-{\left(\frac{x}{y}\right)}^2}=\frac{x+h}{y}$

$\Rightarrow \frac{\frac{2x}{y}}{\frac{y^2-x^2}{y^2}}\frac{x+h}{y}$

$\Rightarrow 2x{y}^2=(y^2-x^2)(x+h)$

$\Rightarrow 2x{y}^2=x{y}^2-x^3+h(y^2-x^2)$

$\Rightarrow x{y}^2+x^3=h(y^2-x^2)$

$\Rightarrow x(y^2+x^2)=h(y^2-x^2)$

$\Rightarrow h=\frac{x(y^2+x^2)}{(y^2-x^2)}$

Hence, the answer is $\frac{x(y^2+x^2)}{(y^2-x^2)}.$