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To solve for \( m\angle C \) in the triangle, we can use **the law of cosines** since we are given all three sides of the triangle.

The law of cosines states:
\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]
where \( c \) is the side opposite the angle \( C \), and \( a \) and \( b \) are the other two sides.

In this case:
- \( c = 13 \) miles (opposite angle \( C \)),
- \( a = 12 \) miles,
- \( b = 9 \) miles.

Now we can plug the values into the formula:
\[
13^2 = 12^2 + 9^2 - 2 \cdot 12 \cdot 9 \cdot \cos(C)
\]

First, compute the squares:
\[
169 = 144 + 81 - 2 \cdot 12 \cdot 9 \cdot \cos(C)
\]
\[
169 = 225 - 216 \cdot \cos(C)
\]

Next, subtract 225 from both sides:
\[
169 - 225 = -216 \cdot \cos(C)
\]
\[
-56 = -216 \cdot \cos(C)
\]

Now, divide both sides by -216:
\[
\cos(C) = \frac{-56}{-216} = \frac{56}{216}
\]
\[
\cos(C) = \frac{7}{27}
\]

Finally, to find \( C \), take the inverse cosine (arccos):
\[
C = \cos^{-1}\left(\frac{7}{27}\right)
\]

Using a calculator:
\[
C \approx 74.07^\circ
\]

So, \( m\angle C \approx 74.07^\circ \).

Would you like more details on any part of the solution, or do you have any questions?

Here are 5 related questions to explore:
1. How would you solve for \( m\angle B \) using the same information?
2. What is the sum of all interior angles in this triangle?
3. Could the law of sines be used to verify the result?
4. How does the Pythagorean theorem differ from the law of cosines, and when is each applied?
5. How would you solve this if the angle \( A \) was given instead of the side lengths?

**Tip**: Always ensure that your calculator is in degree mode when working with trigonometric functions that require angle measurements in degrees!

Find the measure of angle C in a triangle with side lengths 12 mi, 9 mi, and 13 mi.

Learn how to find the measure of angle C in a triangle with side lengths 12 miles, 9 miles, and 13 miles using the Law of Cosines. This problem is suitable for students in Grades 10-12 and covers key trigonometric concepts.

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