Discussion
A vertical pole fixed to the ground is divided in the ratio 1:4 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 16 m away from the base of the pole, what is the height of the pole?
Answer: C
Let the lower part be \( h \), so the upper is \( 4h \) and the pole is \( 5h \). If the lower part subtends \( \alpha \) at the observation point, the whole pole subtends \( 2\alpha \).
\[ \tan\alpha = \frac{h}{16}, \qquad \tan 2\alpha = \frac{5h}{16} \]
Writing \( t = \dfrac{h}{16} \) and using \( \tan 2\alpha = \dfrac{2t}{1-t^{2}} \):
\[ 5t = \frac{2t}{1-t^{2}} \quad\Longrightarrow\quad 5(1-t^{2}) = 2 \quad\Longrightarrow\quad t^{2} = \frac{3}{5} \]
\[ \frac{h^{2}}{256} = \frac{3}{5} \quad\Longrightarrow\quad h = 16\sqrt{\frac{3}{5}} \]
\[ \text{pole} = 5h = 80\sqrt{\frac{3}{5}} = 16\sqrt{15} \approx 61.97 \text{ m} \]
"Equal angles" is what makes the total exactly \( 2\alpha \), and that is the hook the double-angle identity hangs on.
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