Discussion
A vertical pole fixed to the ground is divided in the ratio 1 : 9 by a mark on it, the lower part being the shorter. If the two parts subtend equal angles at a point on the ground 15 m away from the base of the pole, what is the height of the pole?
Answer: A
Let the lower part be \( h \), so the upper part is \( 9h \) and the pole is \( 10h \). From a point 15 m away, let \( \alpha \) be the angle subtended by the lower part. The upper part subtends the same angle, so the whole pole subtends \( 2\alpha \).
\[ \tan\alpha = \frac{h}{15}, \qquad \tan 2\alpha = \frac{10h}{15} \]
Applying the double-angle identity \( \tan 2\alpha = \dfrac{2\tan\alpha}{1-\tan^{2}\alpha} \) with \( t = \dfrac{h}{15} \):
\[ 10t = \frac{2t}{1 - t^{2}} \quad\Longrightarrow\quad 10(1 - t^{2}) = 2 \]
\[ t^{2} = \frac{8}{10} = \frac{4}{5} \quad\Longrightarrow\quad \frac{h^{2}}{225} = \frac{4}{5} \quad\Longrightarrow\quad h^{2} = 180 \]
\[ h = 6\sqrt{5} \quad\Longrightarrow\quad \text{pole} = 10h = 60\sqrt{5} \text{ m} \]
The key step is recognising that "equal angles" makes the total angle exactly double the lower one, which is what lets the double-angle formula in.
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