Discussion
The angle of elevation of the top of a tower from a point on the ground is \( \sin^{-1}\!\left(\dfrac{3}{5}\right) \). If the point of observation is 20 m away from the foot of the tower, what is the height of the tower?
Answer: B
The elevation is given as an inverse sine, so convert it to a tangent before using it with the horizontal distance.
\[ \sin\theta = \frac{3}{5} \quad\Longrightarrow\quad \cos\theta = \sqrt{1 - \left(\tfrac{3}{5}\right)^{2}} = \frac{4}{5} \]
\[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3}{4} \]
\[ h = 20\tan\theta = 20 \times \frac{3}{4} = 15 \text{ m} \]
This is the 3–4–5 triangle. Using \( \tfrac{3}{5} \) directly with the 20 m base gives 12 m, the error the question is testing for.
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