Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is:

Let the height of the shorter pole be $$h$$ metres. Then the height of the taller pole is $$3h$$ metres. The two poles are 150 m apart, and the observer is at the midpoint, so the distance from the observer to each pole is 75 m.

Let the angle of elevation of the shorter pole be $$\alpha$$ and that of the taller pole be $$\beta$$. We are given that $$\alpha + \beta = 90°$$.

We have $$\tan \alpha = \frac{h}{75}$$ and $$\tan \beta = \frac{3h}{75}$$.

Since $$\beta = 90° - \alpha$$, we get $$\tan \beta = \cot \alpha = \frac{75}{h}$$.

So $$\frac{3h}{75} = \frac{75}{h}$$, which gives $$3h^2 = 75 \times 75 = 5625$$.

Therefore $$h^2 = 1875 = 625 \times 3$$, and $$h = 25\sqrt{3}$$ metres.

Hence, the correct answer is Option D.