Question

find the sine, cosine, and tangent of ∠p.
q 68 p
32
r
simplify your answers and write them as proper fractions, improper fractions, or whole numbers.
sin (p) =
cos (p) =
tan (p) =

Explanation:

Step1: Calculate side QR

First, use the Pythagorean theorem to find the length of side $QR$. For right triangle $QRP$, $QP^2 = QR^2 + RP^2$. Rearrange to solve for $QR$:
$$QR = \sqrt{QP^2 - RP^2} = \sqrt{68^2 - 32^2}$$
$$QR = \sqrt{4624 - 1024} = \sqrt{3600} = 60$$

Step2: Find $\sin(P)$

Sine of $\angle P$ is opposite over hypotenuse:
$$\sin(P) = \frac{\text{Opposite to } \angle P}{\text{Hypotenuse}} = \frac{QR}{QP} = \frac{60}{68} = \frac{15}{17}$$

Step3: Find $\cos(P)$

Cosine of $\angle P$ is adjacent over hypotenuse:
$$\cos(P) = \frac{\text{Adjacent to } \angle P}{\text{Hypotenuse}} = \frac{RP}{QP} = \frac{32}{68} = \frac{8}{17}$$

Step4: Find $\tan(P)$

Tangent of $\angle P$ is opposite over adjacent:
$$\tan(P) = \frac{\text{Opposite to } \angle P}{\text{Adjacent to } \angle P} = \frac{QR}{RP} = \frac{60}{32} = \frac{15}{8}$$

Answer:

$\sin(P) = \frac{15}{17}$
$\cos(P) = \frac{8}{17}$
$\tan(P) = \frac{15}{8}$