15. Find the angle at the centre of these sectors to the nearest degree.: A- area of 104.72cm square and a radius of 10cm. B- area of 139.63m square and a radius of 40m. 16. An area to be landscaped is a sector with a radius of 19 m and an angle at the centre of 145 degrees. The area is to be covered with turf at $12.50 per square m and then top-dressed with soil at $2.30 per square m. Find the total cost of the landscaping to the nearest dollar.

### Problem 15: Finding the angle at the center of two sectors

The formula for the area of a sector of a circle is given by:

\[
\text{Area of sector} = \frac{\theta}{360} \times \pi r^2
\]

where:
- $\theta$ is the angle of the sector in degrees.
- $r$ is the radius of the circle.

#### A. Sector with area of $104.72 \, \text{cm}^2$ and radius of $10 \, \text{cm}$

1. We know:
   \[
   \text{Area} = 104.72 \, \text{cm}^2, \, r = 10 \, \text{cm}
   \]

   Plugging into the formula for the area of the sector:

   \[
   104.72 = \frac{\theta}{360} \times \pi \times 10^2
   \]

   Solving for $\theta$:

   \[
   \theta = \frac{104.72 \times 360}{\pi \times 100}
   \]

   Let's calculate $\theta$.

#### B. Sector with area of $139.63 \, \text{m}^2$ and radius of $40 \, \text{m}$

1. We know:
   \[
   \text{Area} = 139.63 \, \text{m}^2, \, r = 40 \, \text{m}
   \]

   Plugging into the formula for the area of the sector:

   \[
   139.63 = \frac{\theta}{360} \times \pi \times 40^2
   \]

   Solving for $\theta$:

   \[
   \theta = \frac{139.63 \times 360}{\pi \times 1600}
   \]

   Let's calculate this $\theta$ as well.

---

### Problem 16: Landscaping cost

The formula for the area of a sector is the same as before:

\[
\text{Area of sector} = \frac{\theta}{360} \times \pi r^2
\]

Given:
- Radius $r = 19 \, \text{m}$
- Angle $\theta = 145^\circ$
- Cost of turfing is $12.50 \, \text{per square meter}$
- Cost of top-dressing is $2.30 \, \text{per square meter}$

#### Step 1: Find the area of the sector

Using the formula for the area of the sector:

\[
\text{Area} = \frac{145}{360} \times \pi \times 19^2
\]

Let's calculate this area.

#### Step 2: Calculate total cost

1. **Cost for turfing**: Multiply the area of the sector by the cost of turfing per square meter:
   \[
   \text{Turfing cost} = \text{Area} \times 12.50
   \]

2. **Cost for top-dressing**: Multiply the area of the sector by the cost of top-dressing per square meter:
   \[
   \text{Top-dressing cost} = \text{Area} \times 2.30
   \]

3. **Total cost**: Add both costs to get the total landscaping cost:
   \[
   \text{Total cost} = \text{Turfing cost} + \text{Top-dressing cost}
   \]

Let's calculate the total cost.

I'll compute all of these values now.### Problem 15: Angles at the center of sectors

#### A. Sector with area $104.72 \, \text{cm}^2$ and radius $10 \, \text{cm}$
- The angle at the center is approximately $120^\circ$.

#### B. Sector with area $139.63 \, \text{m}^2$ and radius $40 \, \text{m}$
- The angle at the center is approximately $10^\circ$.

---

### Problem 16: Landscaping cost

- The area of the sector to be landscaped is approximately $456.80 \, \text{m}^2$.
- The total cost for turfing and top-dressing the area is **\$6761**.

---

Would you like more details or clarifications on any part of the solution? Here are some related questions for deeper understanding:

1. How does changing the radius affect the angle for a given sector area?
2. What are the units for the area in different sectors, and how do they change with radius?
3. How is the formula for the area of a sector derived from the full circle's area?
4. How would the cost change if the angle at the center were increased?
5. If the cost of turfing or top-dressing were to increase, how would you calculate the new total?

**Tip:** Always keep track of the units (cm, m) when solving area-related problems, as mixing them up can lead to incorrect results.

15. Find the angle at the centre of these sectors to the nearest degree: A- area of 104.72cm² and a radius of 10cm. B- area of 139.63m² and a radius of 40m. 16. An area to be landscaped is a sector with a radius of 19 m and an angle at the centre of 145 degrees. The area is to be covered with turf at $12.50 per square m and then top-dressed with soil at $2.30 per square m. Find the total cost of the landscaping to the nearest dollar.

Solve problems related to finding the angle at the center of a sector with given area and radius, and calculate the total landscaping cost based on sector area and turfing/top-dressing rates. The solution covers geometry concepts and real-life applications, suitable for students in Grades 9-11.

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