In triangle ABC, m∠BAC = 60° and length of the two sides AB and BC are 8 cm and 10 cm respectively. The area of the triangle ABC, in sq. cm, is

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.

In triangle ABC, m∠BAC = 60° and the lengths of the two sides AB and BC are 8 cm and 10 cm respectively. The area of the triangle ABC, in sq. cm, is

Learn how to calculate the area of triangle ABC given a 60° angle and sides AB = 8 cm and BC = 10 cm. This solution uses the formula for the area of a triangle with two sides and an included angle. Suitable for students in Grades 9-11.

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