Two posts are k 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 finds the angular elevations of their tops to be complementary, then the height (in metres) of the shorter post is:

1. $\Big(\dfrac{k}{2\sqrt{2}}\Big)$

2. $\Big(\dfrac{k}{4}\Big)$

3. $k\sqrt{2}$

4. $\Big(\dfrac{k}{\sqrt{2}}\Big)$

Let the height of the shorter post (AB) be h meters and the height of the the taller post (CD) be 2h meters..

Let the distance between the post be k meters.

Let M be the point of observation,

BM = MD = $\dfrac{k}{2}$

In triangle ABM,

$\Rightarrow \tan \theta = \dfrac{h}{\dfrac{k}{2}} \\[1em] \Rightarrow \tan \theta = \dfrac{2h}{k} \text{ ….(1)}$

In triangle CDM,

$\Rightarrow \tan (90^{\circ} - \theta) = \dfrac{2h}{\dfrac{k}{2}} \\[1em] \Rightarrow \tan (90^{\circ} - \theta) = \dfrac{4h}{k} \\[1em] \Rightarrow \cot \theta = \dfrac{4h}{k} \\[1em] \Rightarrow \tan \theta = \dfrac{k}{4h} \text{ ….(2)}$

From (1) and (2), we get :

$\dfrac{2h}{k} = \dfrac{k}{4h}$

$\Rightarrow 8h^2 = k^2 \\[1em] \Rightarrow h^2 = \dfrac{k^2}{8} \\[1em] \Rightarrow h = \sqrt{\frac{k^2}{8}} \\[1em] \Rightarrow h = \frac{k}{\sqrt{4 \times 2}} \\[1em] \Rightarrow h = \frac{k}{2\sqrt{2}} \text{ m}$

Hence, option 1 is the correct option.