We are given a triangle $\triangle ABC$ with the following information:

• $m\angle BAC = 60^\circ$,
• $AB = 8 \, \text{cm}$,
• $BC = 10 \, \text{cm}$.

To find the area of the triangle, we can use the formula for the area of a triangle using two sides and the included angle:

$\text{Area} = \frac{1}{2} \times AB \times BC \times \sin(\angle BAC)$

Let's plug in the given values:

$\text{Area} = \frac{1}{2} \times 8 \times 10 \times \sin(60^\circ)$

We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so:

$\text{Area} = \frac{1}{2} \times 8 \times 10 \times \frac{\sqrt{3}}{2}$

Simplifying:

$\text{Area} = \frac{1}{2} \times 80 \times \frac{\sqrt{3}}{2} = \frac{80\sqrt{3}}{4} = 20\sqrt{3} \, \text{cm}^2$

Thus, the area of triangle $ABC$ is $20\sqrt{3} \, \text{cm}^2$, which is approximately $34.64 \, \text{cm}^2$.

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. How would the area change if $\angle BAC$ were 90° instead of 60°?
2. Can the triangle be classified as scalene, isosceles, or equilateral based on the given information?
3. What would be the height of the triangle dropped from vertex $B$ onto side $AC$?
4. How can you find the length of side $AC$ using the Law of Cosines?
5. What is the perimeter of triangle $ABC$ based on this information?

Tip: The sine rule or cosine rule can be used to solve many problems involving non-right triangles.