Question

solve for $x$. round to the nearest tenth, if necessary.

Explanation:

Step1: Identify the trigonometric ratio

In right - triangle \(PQO\) with \(\angle Q = 24^{\circ}\), \(\angle P=90^{\circ}\), and we know the side \(PO = 5.3\) (opposite to \(\angle Q\)) and we need to find \(x = PQ\) (hypotenuse). We use the sine ratio: \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\).
So, \(\sin(24^{\circ})=\frac{5.3}{x}\).

Step2: Solve for \(x\)

Rearrange the formula \(\sin(24^{\circ})=\frac{5.3}{x}\) to get \(x=\frac{5.3}{\sin(24^{\circ})}\).
We know that \(\sin(24^{\circ})\approx0.4067\).
Then \(x=\frac{5.3}{0.4067}\approx13.0\).

Answer:

\(x\approx13.0\)