Question

what are sin d and cos d? sin d = 9/41 and cos d = 40/41 sin d = 9/41 and cos d = 41/40 sin d = 41/9 and cos d = 40/41 sin d = 41/9 and cos d = 41/40

Explanation:

Step1: Recall trigonometric - ratio definitions

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. For angle $D$, the opposite side to angle $D$ is $CE = 9$, the adjacent side to angle $D$ is $DE=40$, and the hypotenuse $CD = 41$.

Step2: Calculate $\sin D$

$\sin D=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{9}{41}$

Step3: Calculate $\cos D$

$\cos D=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{40}{41}$

Answer:

$\sin D=\frac{9}{41}$ and $\cos D=\frac{40}{41}$