To calculate the area of the triangle, we can use the formula for the area of a triangle when two sides and the included angle are given:

$\text{Area} = \frac{1}{2} \times a \times b \times \sin(C)$

where:

• $a$ and $b$ are the lengths of the two sides,
• $C$ is the included angle between those sides.

In this problem:

• $a = 6 \, \text{cm}$,
• $b = 6 \, \text{cm}$ (since it's an isosceles triangle),
• $C = 41^\circ$.

Step-by-Step Solution

1. Substitute the values into the formula:

   $\text{Area} = \frac{1}{2} \times 6 \times 6 \times \sin(41^\circ)$

2. Calculate $\sin(41^\circ)$: $\sin(41^\circ) \approx 0.6561$

3. Plugging this back in:

   $\text{Area} \approx \frac{1}{2} \times 6 \times 6 \times 0.6561$

4. Calculate the area:

   $\text{Area} \approx 11.809 \, \text{cm}^2$

5. Rounding to the nearest $0.1 \, \text{cm}^2$, we get:

   $\text{Area} \approx 11.8 \, \text{cm}^2$

Final Answer

The area of the triangle is approximately 11.8 cm².

Would you like further details, or do you have any questions?

Related Questions

1. How do you find the area of a triangle if you know all three sides?
2. How would you solve for the area if the angle given was in radians?
3. What are the methods for finding the area of non-right triangles?
4. Can you find the perimeter of this triangle?
5. How would the solution change if one side was different?

Tip:

In trigonometry, knowing one angle and the lengths of two adjacent sides in a triangle can always help calculate the area directly.