Hello, Chez!

Ha! . . . I mislabelled my diagram!

I have the triangle PQR: .$\displaystyle PQ=17\text{ cm, }\:QR=15\text{ cm, and }\:PR=8\text{ cm}$

Find: .$\displaystyle \sin Q,\;\cos Q,\;\tan Q$ Code:

P *
| *
| * 17
8 | *
| *
| *
R * - - - - - - - - * Q
15

Note that: .$\displaystyle 8^2 + 15^2 \:=\:17^2$

. . The sides satisfy the Pythgorean Theorem.

Hence, $\displaystyle \Delta PQR$ is a right triangle.

Therefore:

. . $\displaystyle \begin{array}{ccccc}\sin Q & = &\dfrac{8}{17} \\ \\[-2mm] \cos Q & = & \dfrac{15}{17} \\ \\[-2mm] \tan Q & = & \dfrac{8}{15} \end{array}\quad \hdots$ . corrected