Question

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

Explanation:

Step1: Recall cosine ratio

In right - triangle \(XYZ\) with right - angle at \(Z\), \(\cos(Y)=\frac{YZ}{XY}\). Here, \(YZ = 22\) and \(Y = 41^{\circ}\), so \(\cos(41^{\circ})=\frac{22}{XY}\).

Step2: Solve for \(XY\)

Cross - multiply the equation \(\cos(41^{\circ})=\frac{22}{XY}\) to get \(XY=\frac{22}{\cos(41^{\circ})}\).

Answer:

\(XY=\frac{22}{\cos(41^{\circ})}\)