Discussion
A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
Answer: C
Let the tower be \( h \) and let the car's distance be measured from its foot. Because the angle of depression from the top equals the angle of elevation from the car:
\[ d_{30} = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}, \qquad d_{45} = \frac{h}{\tan 45^{\circ}} = h \]
So in 8 minutes the car covers \( h\sqrt{3} - h = h(\sqrt{3}-1) \), and the distance still to go is \( h \). At constant speed, time is proportional to distance:
\[ t = 8 \times \frac{h}{h(\sqrt{3}-1)} = \frac{8}{\sqrt{3}-1} \]
\[ t = \frac{8(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{8(\sqrt{3}+1)}{2} = 4(\sqrt{3}+1) \]
\[ t = 4 \times 2.732 = 10.93 \text{ min} = 10 \text{ min } 56 \text{ s} \approx 10 \text{ min } 57 \text{ s} \]
The tower height cancels, so it never needed to be known.
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