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!