Two vertical poles are 40 metres apart and the height of one is double that of the other. From the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary. Find their heights.

Let AB and CD be the vertical poles such that the height of one is double to that of the other

It is given that the angles of elevation are complimentary. Thus let us assume the angle made by smaller tower be $$\theta\ $$. Then angle made by DE is $$90-\theta\ $$

In $$\bigtriangleup\ $$ ABC,  $$\tan\left(\theta\ \right) = \frac{\ h}{20}$$

In $$\bigtriangleup\ $$ CDE, $$\tan\left(\theta\ \right) = \frac{\ 20}{2h}$$

Equating the values of$$\tan\left(\theta\right)$$ we get, $$\ \frac{\ h}{20}=\ \frac{\ 20}{2h}$$ 

$$h^2\ =\ 200$$ $$\Rightarrow$$ $$h=14.14$$

Hence height of poles are $$14.14$$ and $$28.28$$