Question

A person is standing at a distance of $1800	ext{ meters}$ facing a giant clock at the top of a tower. At 5.00 p.m., he can see the tip of the minute hand of the clock at $30^\circ$ degree elevation from his eye-level. Immediately, the person starts walking towards the tower. At 5.10 p.m., the person noticed that the tip of the minute hand made an angle of $60^\circ$ degrees with respect to his eye-level. Using three-dimensional vision, find the speed at which the person is walking. The length o

Accepted Answer

D. At 5:00 p.m., minute hand points vertically UP ($P$). Distance to tower base $AM = 1800	ext{ m}$. $PM = AM 	an 30^\circ = \frac{1800}{\sqrt{3}} = 600\sqrt{3}	ext{ m}$. Center of clock $O$: $OM = PM - OP = 600\sqrt{3} - 200\sqrt{3} = 400\sqrt{3}	ext{ m}$. At 5:10 p.m., minute hand is at $60^\circ$ angle from vertical, but its horizontal distance projection $BM = \frac{OM}{	an 60^\circ} = \frac{400\sqrt{3}}{\sqrt{3}} = 400	ext{ m}$. Distance walked $AB = 1800 - 400 = 1400	ext{ m}$ in 10 minutes ($600	ext{ s}$). Speed $= \frac{1400}{600} = \frac{7}{3}	ext{ m/s} = \frac{7}{3} 	imes \frac{18}{5} = 8.4	ext{ km/hr}$.

Walking Speed and Elevation Angles to Clock Tower

Explanation

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