Question

A man observes two objects in a line in the west. On walking a distance $x$ towards the north, the objects subtends an angle $\alpha$ in front of him and on walking a further distance $x$ to north they subtend an angle, $\beta$, then the distance between the objects is

• A. $\frac{2x}{2\cot \beta -\cot \alpha }$
• B. $\frac{3x}{2\cot \beta -\cot \alpha }$
• C. $\frac{3x}{2\cot \beta +\cot \alpha }$
• D. none of these

Answer 1

Answer: (B)

$\frac{3x}{2\cot \beta -\cot \alpha }$

$\tan p=\frac{2x}{c}$ $\tan q=\frac{x}{c}$

$\tan a=\frac{2x}{c+d}$ $\tan b=\frac{x}{c+d}$

$\beta +a=p$

$\alpha +b=q$

hence $\tan \beta$ = $\tan \left(p-a\right)$

$\tan \alpha$ = $\tan \left(q-b\right)$

so

$\tan \beta =\frac{\tan p-\tan a}{1-\tan p\tan a}$

$=\frac{\frac{2x}{c}-\frac{2x}{c+d}}{1+\frac{2x}{c}\frac{2x}{c+d}}$

$=\frac{2xd}{c^2+dc+4x^2}$

$c^2+dc+4x^2=2xd\cot \beta$ ...........(1)

$\tan \alpha =\frac{\tan q-\tan b}{1-\tan q\tan b}$

$=\frac{\frac{x}{c}-\frac{x}{c+d}}{1+\frac{x}{c}\frac{x}{c+d}}$

$=\frac{xd}{c^2+dc+x^2}$

$c^2+dc+x^2=xd\cot \alpha$ ...........(2)

From 1 and 2

$3x^2=xd(2\cot \beta -\cot \alpha )$

$\Rightarrow d=\frac{3x}{2\cot \beta -\cot \alpha }$