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

To solve this problem, we need to determine how much more time the boat will take to reach the shore after the angle of depression changes from 60 degrees. Here's a step-by-step explanation: 

1. Let's define the variables:
   • Height of the cliff (H): It remains constant throughout.
   • Initial distance from the base of the cliff (D₁): When the angle of depression is 30 degrees.
   • Final distance from the base of the cliff (D₂): When the angle of depression is 60 degrees.
2. We know that:
   • Tan θ is defined as the opposite side over the adjacent side in a right triangle.
3. At 30 degrees:
   • \(\tan 30^\circ = \frac{H}{D_1}\)
   • Using \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), we get \(\frac{1}{\sqrt{3}} = \frac{H}{D_1}\).
   • Simplifying gives: \(D_1 = H\sqrt{3}\).
4. At 60 degrees:
   • \(\tan 60^\circ = \frac{H}{D_2}\)
   • Using \(\tan 60^\circ = \sqrt{3}\), we get \(\sqrt{3} = \frac{H}{D_2}\).
   • Simplifying gives: \(D_2 = \frac{H}{\sqrt{3}}\).
5. Time taken for the angle to change from 30 degrees to 60 degrees is 20 minutes.
6. The boat travels the distance from \(D_1\) to \(D_2\) in 20 minutes. The distance is:
   • \(D_1 - D_2 = H\sqrt{3} - \frac{H}{\sqrt{3}}\).
   • Simplifying: \(D_1 - D_2 = H \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = H \frac{2}{\sqrt{3}}\).
7. Now, we calculate the time for the boat to travel from \(D_2\) to the shore (distance is \(D_2\)):
   • Velocity of the boat: \(v = \frac{H \frac{2}{\sqrt{3}}}{20}\).
   • Time to travel \(D_2\): \(t = \frac{D_2}{v} = \frac{\frac{H}{\sqrt{3}}}{\frac{H \frac{2}{\sqrt{3}}}{20}}\).
   • Simplifying: \(t = \frac{20}{2} = 10\) minutes.
8. Thus, the boat will take 10 more minutes to reach the shore.

The correct answer is therefore 10 minutes.

Answer 2

Step 1: Set up the diagram.
Let A be the top of the cliff, B be the foot of the cliff directly below A, and h be the cliff height AB. Let D be where the boat first was, when the angle of depression was \(30^{\circ}\), and C be where the boat was 20 minutes later, when the angle of depression became \(60^{\circ}\). Points B, C, D lie in a line along the sea, with C between B and D since the boat moved closer to shore.

Step 2: Write the two right triangle relations.
In right triangle ABD, \(\tan 30^{\circ} = \frac{h}{BD}\), so \(BD = \frac{h}{\tan 30^{\circ}} = h\sqrt{3}\). In right triangle ABC, \(\tan 60^{\circ} = \frac{h}{BC}\), so \(BC = \frac{h}{\tan 60^{\circ}} = \frac{h}{\sqrt{3}}\).

Step 3: Find the distance covered in the 20 minutes.
The distance the boat covered between the two sightings is \(CD = BD - BC = h\sqrt{3} - \frac{h}{\sqrt{3}} = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}\), and this took 20 minutes.

Step 4: Find the speed and apply it to the remaining distance.
Speed = \(\frac{2h/\sqrt{3}}{20}\). The remaining distance to shore from C is \(BC = \frac{h}{\sqrt{3}}\). Time to cover BC = \(\frac{BC}{\text{speed}} = \frac{h/\sqrt{3}}{(2h/\sqrt{3})/20} = \frac{h}{\sqrt{3}} \times \frac{20\sqrt{3}}{2h} = \frac{20}{2} = 10\) minutes.

Final Answer:
The boat takes 10 more minutes to reach the shore. \[ \boxed{10 \text{ minutes}} \]