Given the relationships \( \tan x^\circ = \frac{3}{y} \) and \( \cos x^\circ = \frac{y}{z} \), we want to express \( \sin x^\circ \) in terms of \( y \) and \( z \).
First, recall the definitions of the trigonometric ratios in terms of the opposite side, adjacent side, and hypotenuse in a right triangle:
- \( \tan x^\circ = \frac{\text{opposite}}{\text{adjacent}} \)
- \( \cos x^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- \( \sin x^\circ = \frac{\text{opposite}}{\text{hypotenuse}} \)
From the tangent relationship: \[ \tan x^\circ = \frac{\text{opposite}}{\text{adjacent}} \implies \frac{3}{y} = \frac{3}{\text{adjacent}} \] This suggests that the opposite side is \( 3 \) and the adjacent side is \( y \).
Next, using the cosine relationship: \[ \cos x^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} \implies \frac{y}{z} = \frac{y}{\text{hypotenuse}} \] Here, since \( y \) is the adjacent side and \( z \) is the hypotenuse, we have the hypotenuse as \( z \).
To find \( \sin x^\circ \), we use the opposite side (3) and the hypotenuse (z): \[ \sin x^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{z}. \]
Thus, \( \sin x^\circ = \frac{3}{z} \).
The correct answer is: \[ \sin x^\circ = \frac{3}{z}. \]