To calculate the largest angle in a triangle with sides of 50m, 60m, and 70m, we can use the cosine rule. The cosine rule states:

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Where:

• $a$, $b$, and $c$ are the sides of the triangle.
• $C$ is the angle opposite the side $c$.

Since the largest angle is opposite the largest side, we are interested in the angle opposite the side measuring 70m (i.e., $C$).

Here:

• $a = 50 \, \text{m}$
• $b = 60 \, \text{m}$
• $c = 70 \, \text{m}$

Now, applying the cosine rule:

$\cos C = \frac{50^2 + 60^2 - 70^2}{2 \times 50 \times 60}$

Let's compute this step by step.The largest angle of the triangle, opposite the side measuring 70m, is approximately $78.46^\circ$.

Do you need further details or have any questions?

Here are five related questions to explore:

1. How do you calculate the other angles in the triangle using the cosine rule?
2. How would you calculate the area of the triangle using the sides?
3. How does the cosine rule differ from the sine rule?
4. Can the cosine rule be applied to non-triangular shapes?
5. What is the smallest angle in this triangle?

Tip: Always use the largest side when calculating the largest angle in a triangle using the cosine rule!