We are given a right triangle △CDE \triangle CDE △CDE with the following side lengths: CE=15CE = 15CE=15 CD=8CD = 8CD=8 DE=17DE = 17DE=17 Angle CCC is a right angle.
sin(D)=oppositehypotenuse=CEDE=1517 \sin(D) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{CE}{DE} = \frac{15}{17} sin(D)=hypotenuseopposite=DECE=1715
cos(D)=adjacenthypotenuse=CDDE=817 \cos(D) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{CD}{DE} = \frac{8}{17} cos(D)=hypotenuseadjacent=DECD=178
tan(D)=oppositeadjacent=CECD=158 \tan(D) = \frac{\text{opposite}}{\text{adjacent}} = \frac{CE}{CD} = \frac{15}{8} tan(D)=adjacentopposite=CDCE=815
sin(E)=oppositehypotenuse=CDDE=817 \sin(E) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{CD}{DE} = \frac{8}{17} sin(E)=hypotenuseopposite=DECD=178
cos(E)=adjacenthypotenuse=CEDE=1517 \cos(E) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{CE}{DE} = \frac{15}{17} cos(E)=hypotenuseadjacent=DECE=1715
tan(E)=oppositeadjacent=CDCE=815 \tan(E) = \frac{\text{opposite}}{\text{adjacent}} = \frac{CD}{CE} = \frac{8}{15} tan(E)=adjacentopposite=CECD=158
sin(D)=1517 \sin(D) = \frac{15}{17} sin(D)=1715 cos(D)=817 \cos(D) = \frac{8}{17} cos(D)=178 tan(D)=158 \tan(D) = \frac{15}{8} tan(D)=815 sin(E)=817 \sin(E) = \frac{8}{17} sin(E)=178 cos(E)=1517 \cos(E) = \frac{15}{17} cos(E)=1715 tan(E)=815 \tan(E) = \frac{8}{15} tan(E)=158
Oops, Image-based questions are not yet availableUse Solvely.ai for full features.
Failed. You've reached the daily limit for free usage.Please come back tomorrow or visit Solvely.ai for additional homework help.