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To find \( m \angle P \), we can use the **Law of Cosines**. This formula relates the sides of a triangle to the cosine of one of its angles. 

The Law of Cosines is given by:

\[
\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}
\]

Where:
- \(a\), \(b\) are the sides adjacent to the angle, and
- \(c\) is the side opposite the angle.

For \( \angle P \), we have:
- \( a = 9 \) (side \( PR \)),
- \( b = 7 \) (side \( PQ \)), and
- \( c = 11 \) (side \( RQ \)).

Let's plug these values into the Law of Cosines to solve for \( \theta = \angle P \).

\[
\cos(\theta) = \frac{9^2 + 7^2 - 11^2}{2 \cdot 9 \cdot 7}
\]

Now, I will calculate this for you.The measure of \( m \angle P \) is approximately \( 85.9^\circ \), rounded to the nearest tenth.

Do you have any further questions or would you like additional details?

Here are some related questions for practice:
1. How do you apply the Law of Sines to find unknown angles in a triangle?
2. What is the sum of the interior angles of any triangle?
3. How can you determine whether a triangle is obtuse, acute, or right-angled?
4. Can you use the Law of Cosines to solve for missing sides?
5. How do you find the area of a triangle using two sides and an included angle?

**Tip:** The Law of Cosines is especially useful in non-right triangles where the Pythagorean theorem does not apply.

Find m∠P in a triangle with sides PR = 9, PQ = 7, and RQ = 11 using the Law of Cosines.

Learn how to find the angle P in triangle PRQ, where PR = 9, PQ = 7, and RQ = 11, using the Law of Cosines. This problem involves applying the trigonometric formula to solve for an angle in a non-right triangle. Suitable for high school students.

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