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Let's go through the questions one by one.

### (a) Find the horizontal distance of the boat, \( S \), from a point directly below the capsule at sea level.

We know the following:
- The angle of depression from point \( P \) to the boat \( S \) is \( 8.9^\circ \).
- The vertical height from \( P \) to the sea level is \( 165 \, \text{m} \).

Using the trigonometric relationship:

\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]

Here, \( \theta = 8.9^\circ \), the opposite side (vertical distance from \( P \) to sea level) is \( 165 \, \text{m} \), and we need to find the adjacent side, which is the horizontal distance \( d \) to the boat \( S \).

\[
\tan(8.9^\circ) = \frac{165}{d}
\]

Solving for \( d \):

\[
d = \frac{165}{\tan(8.9^\circ)} \approx \frac{165}{0.1566} \approx 1054.2 \, \text{m}
\]

So, the horizontal distance of the boat from the point directly below the capsule is approximately **1054 m**.

### (b) After 5 minutes, the capsule moved from point \( P \) to a point \( A \).

The complete rotation of the wheel takes **30 minutes**, and the wheel has a diameter of **150 m**, which gives a radius of \( r = 75 \, \text{m} \).

#### (i) Find the arc length \( PA \) in terms of \( \pi \).

In 5 minutes, the capsule covers \( \frac{5}{30} = \frac{1}{6} \) of the full circle. The total circumference of the Ferris wheel is:

\[
C = 2 \pi r = 2 \pi \times 75 = 150 \pi \, \text{m}
\]

The arc length is then:

\[
\text{Arc length } PA = \frac{1}{6} \times 150 \pi = 25 \pi \, \text{m}
\]

#### (ii) Find the angle \( \angle POA \) in radians.

Since the capsule has traveled \( \frac{1}{6} \) of the full circle, the angle is also \( \frac{1}{6} \) of the full angle of a circle, which is \( 2\pi \) radians:

\[
\angle POA = \frac{1}{6} \times 2\pi = \frac{\pi}{3} \, \text{radians}
\]

#### Find the angle of depression of the boat from point \( A \).

At point \( A \), the capsule has moved down from the highest point. The vertical height from \( A \) to the sea level is now \( 75 \cos\left(\frac{\pi}{3}\right) = 75 \times \frac{1}{2} = 37.5 \, \text{m} \).

Thus, the new vertical height from \( A \) to the sea level is:

\[
165 - 37.5 = 127.5 \, \text{m}
\]

We again use the tangent function to find the new angle of depression, \( \theta_A \), using the same horizontal distance \( d \approx 1054 \, \text{m} \):

\[
\tan(\theta_A) = \frac{127.5}{1054}
\]

\[
\theta_A = \tan^{-1}\left(\frac{127.5}{1054}\right) \approx \tan^{-1}(0.121) \approx 6.91^\circ
\]

So, the angle of depression from point \( A \) is approximately **6.9°**.

### (c) State and justify the macro-concept applied in this question.

The macro-concept applied here is **trigonometry** and **circular motion**. We use trigonometric relationships to solve for distances and angles of depression, and the concept of circular motion is used to compute the arc length and angular displacement as the capsule moves along the Ferris wheel.

---

Let me know if you need any further clarification on any part!

### Related Questions:
1. What is the horizontal distance if the angle of depression was increased to 12° instead of 8.9°?
2. How would the arc length change if the rotation time was reduced to 20 minutes?
3. Can you derive the time it would take for the capsule to travel an arc length of \( 50 \pi \) m?
4. How would the angle of depression change if the capsule stopped 10 meters higher than point \( A \)?
5. What would be the new distance if the diameter of the Ferris wheel was increased to 200 m?

### Tip:
Always ensure you understand the relationship between angles of depression and elevation in trigonometric problems—it’s key for solving height and distance problems!

Looking out of a capsule at the highest point, P, the angle of depression of a stationary boat, S, in the sea is 8.9°. Find the horizontal distance of the boat from a point directly below the capsule at sea level. Additionally, after 5 minutes, the capsule moved from the highest point, P, to a point A. Find in terms of π the arc length PA and ∠POA, and determine the new angle of depression.

Learn how to calculate the horizontal distance of a boat and the angle of depression using trigonometry, as well as the arc length traveled by a capsule on the Singapore Flyer. This problem covers core concepts of circular motion and trigonometric functions, ideal for students in Grades 9-12.

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