This is a triangle problem where you are asked to find the area of the triangle $\triangle ABC$. To solve this, we can use the formula for the area of a triangle when two sides and the included angle are known:

$\text{Area} = \frac{1}{2}ab\sin(C)$

Here:

• $a = 8 \, \text{cm}$ (side $AC$)
• $b = 12 \, \text{cm}$ (side $AB$)
• $\angle C = 41^\circ$ (the angle between sides $a$ and $b$)

Now let's calculate:

$\text{Area} = \frac{1}{2} \times 8 \, \text{cm} \times 12 \, \text{cm} \times \sin(41^\circ)$

First, compute $\sin(41^\circ)$, which is approximately 0.6561. So:

$\text{Area} = \frac{1}{2} \times 8 \times 12 \times 0.6561 = 31.4928 \, \text{cm}^2$

Rounding to 1 decimal place:

$\text{Area} \approx 31.5 \, \text{cm}^2$

So, the area of the triangle is approximately 31.5 cm².

Would you like a more detailed explanation of the steps or further assistance?

Here are some related questions:

1. How do you calculate the sine of an angle using a calculator?
2. Can the area of a triangle be found using other methods?
3. What happens if the angle is not included between the two sides?
4. How would you find the missing side using the Law of Cosines?
5. What are the properties of an oblique triangle like this one?

Tip: Always double-check angle measurements in radians or degrees to ensure correct trigonometric function values!