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We need to find \(\cos P\) in simplest fractional form for the given right triangle \( \triangle PQR \).

### Step 1: Identify Sides Relative to \( \angle P \)
From the diagram:
- \( PQ = 18 \) (opposite side to \( \angle P \))
- \( RP = 30 \) (adjacent side to \( \angle P \))
- \( QR \) (hypotenuse) is unknown.

### Step 2: Use the Pythagorean Theorem
Since \( \triangle PQR \) is a right triangle:
\[
QR^2 = PQ^2 + RP^2
\]
\[
QR^2 = 18^2 + 30^2
\]
\[
QR^2 = 324 + 900 = 1224
\]
\[
QR = \sqrt{1224} = 2\sqrt{306}
\]

### Step 3: Compute \( \cos P \)
Cosine is defined as:
\[
\cos P = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{RP}{QR}
\]
\[
\cos P = \frac{30}{2\sqrt{306}} = \frac{15}{\sqrt{306}}
\]

Rationalizing the denominator:
\[
\cos P = \frac{15\sqrt{306}}{306}
\]

This is the simplified fractional form. Let me know if you need further simplification or explanation! 

### Related Questions:
1. How do you simplify square roots further in trigonometry?
2. What is the sine and tangent of angle \( P \)?
3. How does the Pythagorean theorem apply to non-right triangles?
4. Can cosine values be negative in right triangles?
5. What are the common trigonometric identities that help in solving problems like this?

### Tip:
Always check if the square root can be simplified further to make the fraction simpler!

Express cos P as a fraction in simplest terms.

Solve for cos(P) in the right triangle PQR using trigonometry and the Pythagorean Theorem. This problem demonstrates the process of finding the adjacent side ratio to the hypotenuse. Learn step-by-step how to express cos(P) in simplest fractional terms. Suitable for Grades 9-11 students.

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