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### Problem Breakdown:
- **Given:**
  - \( OAB \) is a sector of a circle centered at \( O \) with radius \( 5 \) cm.
  - \( BN \) is perpendicular to \( OA \) and \( BN = 3 \) cm.
  - \( OB \) is a radius of the circle, so \( OB = 5 \) cm as well.
  
- **Tasks:**
  1. Find the angle \( \theta = \angle BON \) in radians.
  2. Find the area of the shaded region.

### Step 1: Find the angle \( \theta = \angle BON \) in radians.

Since \( BN \) is perpendicular to \( OA \), we can consider \( \triangle OBN \) as a right-angled triangle.

Using trigonometry:
\[
\cos(\theta) = \frac{ON}{OB}
\]

But \( ON \) can be found using the Pythagorean theorem:
\[
ON = \sqrt{OB^2 - BN^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \text{ cm}
\]

Thus:
\[
\cos(\theta) = \frac{ON}{OB} = \frac{4}{5}
\]

Now, solve for \( \theta \):
\[
\theta = \cos^{-1}\left(\frac{4}{5}\right)
\]

Using a calculator, this gives:
\[
\theta \approx 0.6435 \text{ radians}
\]

### Step 2: Find the Area of the Shaded Region

The shaded region is the area of the sector \( OAB \) minus the area of triangle \( OBN \).

#### 2a. Area of Sector \( OAB \):

The area of a sector is given by:
\[
\text{Area of sector } OAB = \frac{1}{2} r^2 \theta
\]
where \( r = 5 \) cm and \( \theta \approx 0.6435 \) radians.

So:
\[
\text{Area of sector } OAB = \frac{1}{2} \times 5^2 \times 0.6435 = \frac{25 \times 0.6435}{2} \approx 8.04375 \text{ cm}^2
\]

#### 2b. Area of Triangle \( OBN \):

The area of a right triangle is given by:
\[
\text{Area of triangle } OBN = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base \( ON = 4 \) cm and height \( BN = 3 \) cm.

So:
\[
\text{Area of triangle } OBN = \frac{1}{2} \times 4 \times 3 = 6 \text{ cm}^2
\]

#### 2c. Area of Shaded Region:

Finally, subtract the area of triangle \( OBN \) from the area of sector \( OAB \):
\[
\text{Area of shaded region} = \text{Area of sector } OAB - \text{Area of triangle } OBN
\]
\[
\text{Area of shaded region} \approx 8.04375 - 6 = 2.04375 \text{ cm}^2
\]

### Final Answer:
The area of the shaded region is approximately \( 2.044 \) cm².

---

Would you like more details or have any questions? Here are some related questions to consider:

1. How would the area of the shaded region change if the radius was doubled?
2. How can the angle \( \theta \) be found in degrees?
3. What is the perimeter of the shaded region?
4. Can the area of the shaded region be found using integration?
5. How would the problem change if \( BN \) were not perpendicular to \( OA \)?

**Tip:** Always double-check your trigonometric calculations to ensure accuracy in geometric problems.

Learn how to find the area of the shaded region step-by-step in a circle sector problem. This detailed solution covers geometric concepts such as the area of a sector and a triangle, involving trigonometric calculations. Suitable for high school students.

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