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Let's break down and solve the problem step by step.

### Given Information:
- The radius of the circle is \( r = 13 \) cm.
- The length of the chord \( AB = 24 \) cm.
- The pole is 150 cm long.

### Part (a) — Show that the angle \( A\hat{O}B = 134.8^\circ \) (correct to 1 decimal place):

We can use the **cosine rule** for a circle to find the angle \( A\hat{O}B \). In any circle, if you know the radius and the chord length, the angle subtended by the chord at the center can be calculated using the following formula derived from the cosine rule:

\[
\cos(\theta) = \frac{2r^2 - AB^2}{2r^2}
\]

Substitute the values:
\[
\cos(\theta) = \frac{2(13)^2 - 24^2}{2(13)^2} = \frac{2(169) - 576}{2(169)} = \frac{338 - 576}{338} = \frac{-238}{338}
\]

\[
\cos(\theta) = -0.7041
\]

Now, taking the inverse cosine:

\[
\theta = \cos^{-1}(-0.7041) \approx 134.8^\circ
\]

So, \( A\hat{O}B \approx 134.8^\circ \), as required.

### Part (b) — Calculate:
#### (i) The length of the arc \( AB \):

The formula for the length of an arc is:
\[
\text{Arc Length} = r \theta
\]
where \( \theta \) is in radians.

To convert \( 134.8^\circ \) to radians:
\[
\theta = 134.8^\circ \times \frac{\pi}{180} \approx 2.3527 \text{ radians}
\]

Now, the arc length:
\[
\text{Arc Length} = 13 \times 2.3527 \approx 30.59 \text{ cm}
\]

#### (ii) The area of the cross-section of the pole:

The area of the cross-section of the pole consists of the area of the sector minus the area of the triangle \( OAB \).

- **Area of the sector**:
\[
\text{Area of sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 13^2 \times 2.3527 \approx 198.93 \text{ cm}^2
\]

- **Area of the triangle**:
We can use the formula for the area of a triangle:
\[
\text{Area of triangle} = \frac{1}{2} r^2 \sin(\theta)
\]
Substituting the values:
\[
\text{Area of triangle} = \frac{1}{2} \times 13^2 \times \sin(2.3527) \approx 85.79 \text{ cm}^2
\]

Now, subtract the area of the triangle from the sector:
\[
\text{Area of cross-section} = 198.93 - 85.79 = 113.14 \text{ cm}^2
\]

### Part (c) — The whole pole is to be painted:

The pole is 150 cm long, and we need to calculate the total surface area, which includes the **lateral surface area** and the **two circular cross-sections** at the ends.

- **Lateral surface area**:
The lateral surface area of a cylinder is given by:
\[
\text{Lateral surface area} = 2 \pi r h
\]
where \( h = 150 \) cm and \( r = 13 \) cm.

\[
\text{Lateral surface area} = 2 \pi \times 13 \times 150 \approx 12245.85 \text{ cm}^2
\]

- **Area of the two cross-sections**:
We already found the area of one cross-section as \( 113.14 \text{ cm}^2 \), so for both cross-sections:
\[
\text{Total area of cross-sections} = 2 \times 113.14 = 226.28 \text{ cm}^2
\]

- **Total area to be painted**:
\[
\text{Total area} = 12245.85 + 226.28 = 12472.13 \text{ cm}^2
\]

### Summary of answers:
- (a) \( A\hat{O}B = 134.8^\circ \)
- (b)(i) Arc length = 30.59 cm
- (b)(ii) Area of cross-section = 113.14 cm²
- (c) Total area to be painted = 12472.13 cm²

Would you like more details or clarifications? 

Here are 5 related questions you can explore:
1. How do you convert degrees to radians?
2. Can you derive the formula for the area of a segment of a circle?
3. What is the formula for finding the surface area of a cylinder?
4. How does the cosine rule apply to non-right triangles?
5. How do you calculate the area of a sector using both degrees and radians?

**Tip:** Always double-check your angle conversions, especially when switching between degrees and radians!

The diagram shows a cross-section of a wooden pole. The cross-section is a segment of a circle with center O, radius 13 cm, and chord AB = 24 cm. The pole is 150 cm long. (a) Show that the size of ∠AÔB = 134.8°, correct to 1 decimal place. (b) Calculate (i) the length of the arc AB, (ii) the area of the cross-section of the pole. (c) The whole pole is to be painted. Calculate the total area to be painted.

Learn how to calculate the angle, arc length, and area of a circular cross-section of a wooden pole with a radius of 13 cm and a chord length of 24 cm. Explore cosine rule applications, sector areas, and surface area calculations. Suitable for Grades 10-12.

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