**Find the cosine of ∠P.**The image depicts a right triangle PQR with the right angle at vertex Q. The length of side PQ is 9, and the length of side QR is $\sqrt{17}$.To find the cosine of ∠P, use the definition of cosine for a right triangle:\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \]Here, the adjacent side to ∠P is PQ, the hypotenuse is PR, and PR can be found using the Pythagorean theorem:\[ PR = \sqrt{PQ^2 + QR^2} \]\[ PR = \sqrt{9^2 + (\sqrt{17})^2} \]\[ PR = \sqrt{81 + 17} \]\[ PR = \sqrt{98} \]\[ PR = 7\sqrt{2} \]Now, we can find $\cos(P)$:\[ \cos(P) = \frac{PQ}{PR} \]\[ \cos(P) = \frac{9}{7\sqrt{2}} \]To rationalize the denominator:\[ \cos(P) = \frac{9}{7\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \]\[ \cos(P) = \frac{9\sqrt{2}}{14} \]**Write your answer in simplified, rationalized form. Do not round.**\[ \cos(P) = \frac{9\sqrt{2}}{14} \][Input Box for Answer: Fraction Template with a Square Root Option]

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Description: **Find the cosine of ∠P.**The image depicts a right triangle PQR with the right angle at vertex Q. The length of side PQ is 9, and the length of side QR is $\sqrt{17}$.To find the cosine of ∠P, use the definition of cosine for a right triangle:\[

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Find the cosine of ZP.

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ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter2: Right Triangle Trigonometry

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Fundamentals of Trigonometric Identities

Trigonometric Identities are the equalities that involve trigonometry functions and hold for all the values of variables given in the equation.

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**Find the cosine of ∠P.**

The image depicts a right triangle PQR with the right angle at vertex Q. The length of side PQ is 9, and the length of side QR is $\sqrt{17}$.

To find the cosine of ∠P, use the definition of cosine for a right triangle:
\[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \]

Here, the adjacent side to ∠P is PQ, the hypotenuse is PR, and PR can be found using the Pythagorean theorem:
\[ PR = \sqrt{PQ^2 + QR^2} \]
\[ PR = \sqrt{9^2 + (\sqrt{17})^2} \]
\[ PR = \sqrt{81 + 17} \]
\[ PR = \sqrt{98} \]
\[ PR = 7\sqrt{2} \]

Now, we can find $\cos(P)$:
\[ \cos(P) = \frac{PQ}{PR} \]
\[ \cos(P) = \frac{9}{7\sqrt{2}} \]

To rationalize the denominator:
\[ \cos(P) = \frac{9}{7\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \]
\[ \cos(P) = \frac{9\sqrt{2}}{14} \]

**Write your answer in simplified, rationalized form. Do not round.**

\[ \cos(P) = \frac{9\sqrt{2}}{14} \]

[Input Box for Answer: Fraction Template with a Square Root Option]

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