Question : Two posts are $x$ metres apart and the height of one is double that of the other. If, from the midpoint of the line joining their feet, an observer finds the angular elevation of their tops to be complementary, then the height (in metres) of the shorter posts is:
Option 1: $\frac{x}{2\sqrt{2}}$
Option 2: $\frac{x}{4}$
Option 3: $x\sqrt{2}$
Option 4: $\frac{x}{\sqrt{2}}$
Correct Answer: $\frac{x}{2\sqrt{2}}$
Solution :
Let BD be the distance between two posts is $x$ metres.
$\therefore$ OB = OD = $\frac{x}{2}$
From $\triangle$OCD
$\tan\theta$ = $\frac{CD}{OD}$
$\tan\theta$ = $\frac{h}{\frac{x}{2}}$= ${\frac{2h}{x}}$............(equation 1)
From $\triangle$OAB
$\tan(90–\theta)=\frac{AB}{OB}$
$\cot\theta$ = $\frac{2h}{\frac{x}{2}}$= ${\frac{4h}{x}}$............(equation 2)
Multiplying both equations, we get:
$\tan\theta×\cot\theta$ = ${\frac{2h}{x}}×{\frac{4h}{x}}$
⇒ 1 = $\frac{8h^2}{x^2}$
$\therefore h=\frac{x}{2\sqrt2}$
Hence, the correct answer is $\frac{x}{2\sqrt2}$.
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