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The problem is in Russian and translates to:

"Two sides of a triangle are 6 cm and 10 cm, and the angle between them is 120°.  
a) Find the perimeter of the triangle.  
b) Find the area of the triangle."

### a) Finding the Perimeter:
To find the perimeter, we first need to find the third side of the triangle using the **cosine rule**:
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
Where:
- \(a = 6 \, \text{cm}\),
- \(b = 10 \, \text{cm}\),
- \(\cos(C) = \cos(120^\circ) = -0.5\).

Substituting these values:
\[
c^2 = 6^2 + 10^2 - 2 \cdot 6 \cdot 10 \cdot (-0.5)
\]
\[
c^2 = 36 + 100 + 60 = 196
\]
\[
c = \sqrt{196} = 14 \, \text{cm}
\]

Now the perimeter \(P\) is:
\[
P = a + b + c = 6 + 10 + 14 = 30 \, \text{cm}
\]

### b) Finding the Area:
We can find the area using the formula for the area of a triangle with two sides and the included angle:
\[
\text{Area} = \frac{1}{2}ab \cdot \sin(C)
\]
Using \(\sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[
\text{Area} = \frac{1}{2} \cdot 6 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{cm}^2 \approx 25.98 \, \text{cm}^2
\]

### Final answers:
- The perimeter is **30 cm**.
- The area is approximately **25.98 cm²**.

Do you have any questions or need further clarification? 

Here are 5 related questions to explore:
1. How would the perimeter change if the angle were \(90^\circ\)?
2. Can the same formula be used for any type of triangle?
3. What happens if the angle is less than \(90^\circ\)?
4. How would you find the third side if the angle were not provided?
5. How does the value of the cosine affect the result for the third side?

**Tip:** When solving triangle problems, knowing both the cosine and sine rules can be helpful depending on the information provided.

Two sides of a triangle are 6 cm and 10 cm, and the angle between them is 120°. Find the perimeter and the area of the triangle.

Learn how to solve a triangle with sides of 6 cm and 10 cm, and an included angle of 120°. This solution explains how to calculate the perimeter using the cosine rule and the area using the sine rule. Suitable for Grades 9-11 students studying trigonometry and geometry.

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