Two persons are standing '$x$' meter apart from each other. The height of the first person 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, find the height of the shorter person.
$P_1, \; P_2 =$ Two persons standing at $O$ and $Q$ respectively
$OP_1 =$ Height of $P_1 = 2h$ meter
$OP_2 =$ Height of $P_2 = h$ meter
$M =$ Midpoint between $O$ and $Q$
Let $\;$ $\angle P_1MO = \theta$
Then, $\;$ $\angle P_2MQ = 90 - \theta$
$OQ =$ Distance between the two people $= x$ meter
In $\triangle MP_2Q$, $\;$ $\dfrac{P_2Q}{MQ} = \tan \left(90 - \theta\right) = \cot \theta$
i.e. $\;$ $\cot \theta = \dfrac{h}{x / 2}$ $\;\;\; \cdots \; (1)$
In $\triangle MPO$, $\;$ $\dfrac{PO}{OM} = \tan \theta$
i.e. $\;$ $\tan \theta = \dfrac{2h}{x/2}$ $\;\;\; \cdots \; (2)$
Now, $\;$ $\tan \theta \times \cot \theta = 1$ $\;\;\; \cdots \; (3)$
In view of equations $(1)$ and $(2)$ equation $(3)$ becomes
$\dfrac{2h}{x / 2} \times \dfrac{h}{x / 2} = 1$
i.e. $\;$ $8h^2 = x^2$ $\implies$ $h = \dfrac{x}{2 \sqrt{2}}$
$\therefore \;$ The height of the shorter person $= \dfrac{x}{2 \sqrt{2}}$ meter