Question

length of side x in simplest radical form with a rational denominator.

Explanation:

Step1: Use cosine function

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here \(\theta = 30^{\circ}\), adjacent side \(=\sqrt{6}\), and hypotenuse \(=x\). So \(\cos30^{\circ}=\frac{\sqrt{6}}{x}\).
Since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{x}\).

Step2: Solve for \(x\)

Cross - multiply: \(\sqrt{3}x = 2\sqrt{6}\).
Then \(x=\frac{2\sqrt{6}}{\sqrt{3}}\).
Rationalize the denominator: \(x=\frac{2\sqrt{6}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{2\sqrt{18}}{3}\).
Simplify \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\). So \(x = 2\sqrt{2}\).

Answer:

\(2\sqrt{2}\)