Question

Two persons are ${}^{\prime }{a}^{\prime }$ metres apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer who looks at the top of their head , finds the angular elevations to be complementary, then the height of the shorter person in metres is :

• A. $\frac{a}{4}$
• B. $\frac{a}{\sqrt{2}}$
• C. $a\sqrt{2}$
• D. $\frac{a}{2\sqrt{2}}$

Answer 1

The correct option is D $\frac{a}{2\sqrt{2}}$

Let AB and CD denote the height of the two persons and CA the distance between them. Then, according to the question, we get the following figure.

In $\Delta CDE$,
$tan({90}^{\circ }-\theta )=\frac{h}{\frac{a}{2}}$
$\Rightarrow cot\theta =\frac{2h}{a}$ - - - (1)

$In\Delta ABE,tan\theta =\frac{AB}{EA}=\frac{2h}{\frac{a}{2}}$
$\Rightarrow tan\theta =\frac{4h}{a}$ - - - (2)

Multiplying (1) by (2), we get
$cot\theta \times tan\theta =\frac{2h}{a}\times \frac{4h}{a}\Rightarrow 1=\frac{8h^2}{a^2}\Rightarrow h=\frac{a}{2\sqrt{2}}$