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Let's break down the questions:

### d) 
The task involves calculating the vertical distance between a person at point B on a Ferris wheel, which has rotated 30°, and the center of the wheel. A diagram is requested as part of the solution.

- Assuming the Ferris wheel is a circle, when a point is rotated 30°, the vertical distance between the person and the center of the wheel can be calculated using trigonometry. If the radius of the Ferris wheel is \( r \), the vertical distance \( h \) can be determined using the sine of the angle:
\[
h = r \times \sin(30^\circ)
\]
\[
h = r \times 0.5
\]
Thus, the vertical distance is half the radius.

### e)
In this part, we use the result from part d) to calculate how high the person is above the ground. If the radius of the wheel is \( r \), and the wheel's center is at height \( r \) above the ground, the height above the ground at 30° is the sum of the center height and the vertical distance calculated earlier:
\[
\text{Height above ground} = r + (r \times \sin(30^\circ))
\]
Substitute \( \sin(30^\circ) = 0.5 \), the formula becomes:
\[
\text{Height above ground} = r + \frac{r}{2} = \frac{3r}{2}
\]

### f)
Here, we are tasked with calculating the height above the ground when the wheel has rotated 255°. This involves using the sine function again:
\[
\text{Vertical distance from center} = r \times \sin(255^\circ)
\]
Since \( 255^\circ = 255^\circ - 180^\circ = 75^\circ \) in the third quadrant, where sine is negative:
\[
\sin(255^\circ) = -\sin(75^\circ) = -0.9659
\]
Then, the height above the ground would be:
\[
\text{Height above ground} = r + (r \times -0.9659) = r \times (1 - 0.9659) = r \times 0.0341
\]
Thus, the height off the ground is a small fraction of the wheel’s radius.

Would you like a more detailed explanation of the steps involved? Here are 5 related questions to expand your understanding:

1. How do you find the vertical height of a point on a circle using different angles?
2. What is the relationship between sine and cosine when solving circular motion problems?
3. How would the calculation change if the Ferris wheel started rotating in the opposite direction?
4. How can trigonometric identities help simplify circular motion problems?
5. What other real-world examples rely on trigonometry for solving circular motion?

**Tip**: Always visualize trigonometric problems on a unit circle to better understand how angles affect the calculation of vertical and horizontal distances.

Point B on the diagram shows the position of a person when the Ferris wheel has rotated 30°. Calculate the vertical distance between this person and the center of the wheel. Then, calculate how high above the ground the person is after rotation, for both 30° and 255°. Diagrams and explanations are required.

Learn how to calculate the vertical distance and height above the ground for a person on a Ferris wheel rotated by 30° and 255°. This problem covers key concepts in trigonometry and circular motion, using the sine function. Suitable for students in Grades 9-11.

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