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practice using trigonometric ratios to solve for missing lengths. which…

Question

practice using trigonometric ratios to solve for missing lengths. which equation could be used to solve for the length of xy? \\(\circ\\) \\(xy = (22)\sin(41^{\circ})\\) \\(\circ\\) \\(xy = (22)\cos(41^{\circ})\\) \\(\circ\\) \\(xy = \frac{22}{\cos(41^{\circ})}\\) \\(\circ\\) \\(xy = \frac{22}{\sin(41^{\circ})}\\)

Explanation:

Step1: Identify sides relative to $\angle Y$

In right $\triangle XYZ$, $\angle Y=41^\circ$, side $XZ=22$ (opposite $\angle Y$), $XY$ is the hypotenuse.

Step2: Choose correct trigonometric ratio

Use sine: $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$
$\sin(41^\circ)=\frac{22}{XY}$

Step3: Rearrange to solve for $XY$

Rearrange the formula: $XY=\frac{22}{\sin(41^\circ)}$

Answer:

D. $XY = \frac{22}{\sin(41^\circ)}$