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Let's solve the given problem step-by-step.

### a) Finding the angle \(\angle AOB\)

To find the angle \(\angle AOB\), we can use trigonometric relationships within the triangle \(\triangle AOB\). 

Given:
- \( OA = OB = 5 \, \text{m} \) (radii of the circle)
- \( AB = 4 \, \text{m} \)

Since \( \triangle AOB \) is an isosceles triangle (with \( OA = OB \)), we can use the Law of Cosines to find the angle \(\angle AOB\).

The Law of Cosines states:

\[
AB^2 = OA^2 + OB^2 - 2 \times OA \times OB \times \cos(\angle AOB)
\]

Substituting the known values:

\[
4^2 = 5^2 + 5^2 - 2 \times 5 \times 5 \times \cos(\angle AOB)
\]

Simplifying:

\[
16 = 25 + 25 - 50 \cos(\angle AOB)
\]

\[
16 = 50 - 50 \cos(\angle AOB)
\]

\[
50 \cos(\angle AOB) = 34
\]

\[
\cos(\angle AOB) = \frac{34}{50} = 0.68
\]

\[
\angle AOB = \cos^{-1}(0.68)
\]

Calculating \(\angle AOB\):

\[
\angle AOB \approx 47.19^\circ
\]

### b) Finding the area of the shaded region

The shaded region is a sector minus a triangle.

1. **Area of Sector \(AOB\)**:

   The area of the sector is given by:

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

   Here, \(\theta \approx 47.19^\circ\) and \( r = 5 \, \text{m} \).

   \[
   \text{Area of Sector} = \frac{47.19}{360} \times \pi \times 5^2
   \]

   \[
   \text{Area of Sector} \approx \frac{47.19}{360} \times \pi \times 25
   \]

2. **Area of Triangle \(AOB\)**:

   The area of the triangle can be found using the formula:

   \[
   \text{Area of Triangle} = \frac{1}{2} \times AB \times h
   \]

   The height \( h \) can be found using trigonometry:

   Since \( \sin(\angle AOB/2) = \frac{h}{5} \), find \( h \).

Let's calculate these values.It seems that the calculation took longer than expected. I'll simplify the computations and break them down to avoid the timeout issue. 

### Step-by-step Calculation:

1. **Angle \(\angle AOB\)**:
   \[
   \cos(\angle AOB) = \frac{34}{50} = 0.68 \implies \angle AOB \approx 47.19^\circ
   \]

2. **Area of the Sector \(AOB\)**:
   \[
   \text{Area of Sector} = \frac{47.19}{360} \times \pi \times 5^2
   \approx \frac{47.19}{360} \times 3.1416 \times 25 \approx 10.31 \, \text{m}^2
   \]

3. **Area of Triangle \(AOB\)**:
   - Half of \(\angle AOB\) in radians: \( \approx 47.19^\circ / 2 = 23.595^\circ\).
   - Convert this to radians: \( \approx 0.4117 \, \text{radians}\).
   - Height \( h \approx 5 \times \sin(0.4117) \approx 2 \, \text{m}\).
   \[
   \text{Area of Triangle} = 0.5 \times 4 \times 2 = 4 \, \text{m}^2
   \]

4. **Area of the Shaded Region**:
   \[
   \text{Shaded Area} = \text{Area of Sector} - \text{Area of Triangle} \approx 10.31 - 4 = 6.31 \, \text{m}^2
   \]

Would you like further clarifications or have any other questions? Here are some related questions:

1. How would the area change if the radius was doubled?
2. What if the angle \(\angle AOB\) was a right angle?
3. How does the Law of Cosines apply to non-isosceles triangles?
4. Can we find the area using integration methods?
5. How does the shaded area relate to the total circle area?

**Tip:** Always double-check your angle calculations when using trigonometric functions to ensure accuracy.

1. A sector of a circle with center at O and radius 5 m is shown below. a) Find the angle ∠AOB. b) Find the area of the shaded region.

Learn how to find the angle ∠AOB of a sector in a circle with radius 5 m and calculate the area of the shaded region using the Law of Cosines, sector area, and triangle area formulas. Suitable for Grades 9-12 students.

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