Question

An observer finds that the angular elevation of a tower is $\theta$. On advancing $a$ metres towards the tower, the elevation is ${45}^0$ and on advancing $b$ metres nearer the elevation is ${90}^0-\theta$, then the height of the tower (in metres) is

• A. $\frac{ab}{a+b}$
• B. $\frac{ab}{a-b}$
• C. $\frac{2ab}{a+b}$
• D. $\frac{2ab}{a-b}$

Answer 1

The correct option is B $\frac{ab}{a-b}$

Let $b$ be the initial point and $AE$ be the height of the tower.
In, $△ADE,$

$\Rightarrow \tan (90°-\theta )=\frac{h}{x}$

$\Rightarrow \cot \theta =\frac{h}{x}$

$\Rightarrow x=h\tan \theta ⟶\left(1\right)$

$\Rightarrow \tan \theta =\frac{x}{h}$

In, $△ACE,$

$\Rightarrow \tan 45°=\frac{h}{b+x}$

$\Rightarrow b+x=h$

$\Rightarrow b+h\tan \theta =h$ [From $\left(1\right)$ ]

$\Rightarrow b=h(1-\tan \theta )⟶\left(2\right)$

In, $△ABE,$

$\Rightarrow \tan \theta =\frac{h}{a+b+x}$

$\Rightarrow \tan \theta =\frac{h}{a+h}$

$\Rightarrow \frac{x}{h}=\frac{h}{a+h}$

$\Rightarrow \frac{h-b}{h}=\frac{h}{a+h}$

$\Rightarrow (h-b)(a+h)=h^2$

$\Rightarrow ah-ab+h^2-hb=h^2$

$\Rightarrow h(a-b)=ab$

$\Rightarrow h=\frac{ab}{a-b}$

Hence, the answer is $\frac{ab}{a-b}.$