Question

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

Explanation:

Step1: Identify the trigonometric ratio

In right - triangle \(OPQ\) with \(\angle O = 53^{\circ}\), \(\angle P=90^{\circ}\), and we know the adjacent side \(x\) (to \(\angle O\)) and the hypotenuse \(OQ = 1.8\). We use the cosine ratio. The cosine of an angle in a right - triangle is defined as \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). So, \(\cos\angle O=\frac{OP}{OQ}\).

Step2: Substitute the values

Since \(\angle O = 53^{\circ}\) and \(OQ = 1.8\), we have \(\cos(53^{\circ})=\frac{x}{1.8}\).

Step3: Solve for \(x\)

We know that \(\cos(53^{\circ})\approx0.6\). Then \(x = 1.8\times\cos(53^{\circ})\). Substituting the value of \(\cos(53^{\circ})\), we get \(x=1.8\times0.6 = 1.08\approx1.1\).

Answer:

\(x\approx1.1\)