Question

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ meters on the ground towards the object, the angle of elevation is found to be $\beta$, then the height (in meters) of the object is

• A. $d\tan \alpha$
• B. $d\cot \beta$
• C. $\frac{d}{\cot \alpha +\cot \beta }$
• D. $\frac{d}{\cot \alpha -\cot \beta }$

Answer 1

Answer: (D)

$\frac{d}{\cot \alpha -\cot \beta }$

Let the height of the object $AB$ be $h$ metres.

Given, $\angle ACB=\alpha ,\angle ADB=\beta ,CD=d$ metres

Let $DB=x$ metres.

Then, in right-angled $△ABD,$
$\cot \beta =\frac{x}{h}$

$\Rightarrow x=h\cot \beta$ $...\left(i\right)$

In right-angled $△ACB,$

$\cot \alpha =\frac{x+d}{h}$

$\Rightarrow x+d=h\cot \alpha$ $...\left(ii\right)$

Let us subtract the equations: $\left(ii\right)–\left(i\right)$

$\Rightarrow d=h\cot \alpha –h\cot \beta$

$\Rightarrow d=h(\cot \alpha -\cot \beta )$

$\Rightarrow h=\frac{d}{\cot \alpha -\cot \beta }$