Discussion
From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:
Answer: C
The tower is opposite the \( 30^{\circ} \) angle and the distance is adjacent, so:
\[ \tan 30^{\circ} = \frac{100}{d} \quad\Longrightarrow\quad d = \frac{100}{\tan 30^{\circ}} = 100\sqrt{3} \]
\[ d = 100 \times 1.732 = 173.2 \approx 173 \text{ m} \]
A \( 30^{\circ} \) elevation always puts the observer \( \sqrt{3} \) times the height away. Multiplying by \( \tan 30^{\circ} \) instead of dividing gives about 58 m - the classic inversion.
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