Discussion
Two persons are on either sides of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is
Answer: D
The two observers are on opposite sides, so the car's path is the sum of the two horizontal distances from the 50 m tower.
\[ d = \frac{50}{\tan 30^{\circ}} + \frac{50}{\tan 60^{\circ}} = 50\sqrt{3} + \frac{50}{\sqrt{3}} \]
\[ d = \frac{150 + 50}{\sqrt{3}} = \frac{200}{\sqrt{3}} = \frac{200\sqrt{3}}{3} \text{ m} \]
That is covered in 10 seconds, so
\[ v = \frac{200\sqrt{3}}{3 \times 10} = \frac{20\sqrt{3}}{3} \text{ m/s} \]
\[ v = \frac{20\sqrt{3}}{3} \times \frac{18}{5} = \frac{360\sqrt{3}}{15} = 24\sqrt{3} \text{ km/h} \]
Numerically that is about 41.6 km/h. Converting m/s to km/h with the factor \( \tfrac{18}{5} \) is the step the distractors are built around.
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