Discussion
An observer 1.6 m tall is 20√3 away from a tower. If angle of elevation from his eye to the top of tower is 30°. The heights of the tower is:
Answer: A
The tower is 21.6 m high.
Measure from the observer's eye, 1.6 m above the ground, at a horizontal distance of \(20\sqrt{3}\) m (the paper omits the unit):
\[ \begin{aligned} h' &= 20\sqrt{3} \tan 30^\circ \\ &= 20\sqrt{3} \times \frac{1}{\sqrt{3}} = 20\ \text{m} \\ h &= 20 + 1.6 = 21.6\ \text{m} \end{aligned} \]
Leaving out the eye height gives 20 m, which is not offered. 23.2 m adds the 1.6 m twice. 24.72 m and 20.3 m do not follow from \(\tan 30^\circ\).
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