Question

Peter was standing on top of a rock cliff facing the sea. He saw a boat coming towards the shore. As he kept watching, time passed quickly. Ten minutes less than half an hour (that is, 20 minutes) after his first sighting, the angle of depression to the boat changed from 30 degrees to 60 degrees. How much more time, in minutes, will the boat take to reach the shore?

• A. 5
• B. 10
• C. 15
• D. 20

Hint:

Find BD and BC using tan 30 and tan 60, then use the known 20 minute travel time for CD to scale down to BC.

Answer 1

The Correct Option is B

Step 1: Use a plain ratio idea instead of computing speed.
At constant speed, time taken is directly proportional to distance covered. So if we find the ratio of the leftover distance BC to the distance CD already covered in 20 minutes, we can scale 20 minutes by that ratio to get the remaining time.

Step 2: Find BD and BC using the tangent ratio.
With cliff height h: $BD = h \cot 30^{\circ} = h\sqrt{3}$ and $BC = h \cot 60^{\circ} = \frac{h}{\sqrt{3}}$.

Step 3: Express BC as a fraction of CD.
$CD = BD - BC = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$. The ratio $\frac{BC}{CD} = \frac{h/\sqrt{3}}{2h/\sqrt{3}} = \frac{1}{2}$. So the remaining stretch BC is exactly half as long as the stretch CD already traveled.

Step 4: Scale the known time by this ratio.
Since CD took 20 minutes at constant speed, and BC is half of CD in length, BC takes half of 20 minutes, which is 10 minutes.

Final Answer:
The remaining time for the boat to reach the shore is 10 minutes. \[ \boxed{10 \text{ minutes}} \]