https://file.aiquickdraw.com/mathbot/file/173001236917427erohoi.jpg

To solve this problem, we’ll use the **formula for the area of a triangle** given two sides and the included angle between them.

The formula is:

\[
\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle ACB)
\]

### Given Information:
- \( AB = 4.6 \, \text{cm} \)
- \( AC = 7.1 \, \text{cm} \)
- \( \angle ACB = 33^\circ \)

### Solution:
1. First, convert \( \angle ACB \) to radians or use it directly in degrees (most calculators handle degrees).
2. Substitute the values into the formula:

   \[
   \text{Area} = \frac{1}{2} \times 4.6 \times 7.1 \times \sin(33^\circ)
   \]

3. Calculate \( \sin(33^\circ) \).
4. Then, perform the multiplications to find the area.

Let’s calculate:

\[
\sin(33^\circ) \approx 0.5446
\]

Now plug in the values:

\[
\text{Area} = \frac{1}{2} \times 4.6 \times 7.1 \times 0.5446
\]

\[
\text{Area} \approx \frac{1}{2} \times 4.6 \times 7.1 \times 0.5446
\]

\[
\text{Area} \approx 8.8876 \, \text{cm}^2
\]

### Final Answer:
The area of the triangle \( \triangle ABC \) is approximately \( 8.89 \, \text{cm}^2 \).

Would you like a deeper breakdown of any step or have further questions?

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### Related Questions to Explore:
1. How would you find the area if only the three sides of the triangle were given?
2. What is the cosine rule, and how is it applied in triangle problems?
3. How do you convert angles from degrees to radians?
4. Why does using sine with two sides work for finding the area in non-right triangles?
5. Can the same formula be used if the given angle were obtuse?

### Tip:
When using trigonometric functions, double-check your calculator mode to ensure it’s set to degrees if your angle is in degrees.

In a triangle ABC with AC = 7.1 cm, AB = 4.6 cm, and ∠ACB = 33°. Find the area of the triangle.

Find the area of triangle ABC using trigonometric methods with given sides AB = 4.6 cm, AC = 7.1 cm, and included angle ∠ACB = 33°. This problem demonstrates the trigonometric area formula for triangles, suitable for Grades 9-10.

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