To calculate the area of triangle $ABC$, we can use the formula:

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

where:

• $a = 6 \, \text{cm}$ (side $AB$),
• $b = 5 \, \text{cm}$ (side $BC$),
• $C = 40^\circ$ (angle $\angle ACB$).

Now, let's compute the area:

$\text{Area} = \frac{1}{2} \times 6 \times 5 \times \sin(40^\circ)$

I'll calculate this for you.The area of triangle $ABC$ is approximately $9.64 \, \text{cm}^2$ when rounded to three significant figures.

Would you like more details or have any questions?

Here are some related questions:

1. How would the area change if angle $\angle ACB$ were increased to $50^\circ$?
2. Can you calculate the area if you were given the length of $AC$ instead of $BC$?
3. How can you use the Law of Sines to find the remaining angles in triangle $ABC$?
4. What is the perimeter of triangle $ABC$?
5. How would you calculate the area if $\angle BAC$ were given instead of $\angle ACB$?

Tip: Remember that the sine function reaches its maximum value at $90^\circ$. This implies that for a fixed base and height, the area of a triangle is maximized when the angle between them is $90^\circ$.