Question

find the length of side ( x ) in simplest radical form with a rational denominator.

Explanation:

Step1: Identify triangle type and trigonometric ratio

This is a right - triangle with angles \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). We know that for a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, for the \(30^{\circ}\) angle, the opposite side to \(30^{\circ}\) is \(6\) and the adjacent side is \(x\). So \(\tan(30^{\circ})=\frac{6}{x}\).
We know that \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\). So we have the equation \(\frac{1}{\sqrt{3}}=\frac{6}{x}\).

Step2: Solve for \(x\)

Cross - multiply the equation \(\frac{1}{\sqrt{3}}=\frac{6}{x}\) to get \(x = 6\sqrt{3}\). We can also use the properties of \(30 - 60-90\) triangles. In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (where the side opposite \(30^{\circ}\) is the shortest side, opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shortest side, and the hypotenuse is twice the shortest side). The side of length \(6\) is opposite the \(30^{\circ}\) angle? Wait, no. Wait, the right angle is between the side of length \(6\) and side \(x\). So the angle of \(30^{\circ}\) has adjacent side \(x\) and opposite side \(6\). Also, in a \(30 - 60 - 90\) triangle, if the side opposite \(30^{\circ}\) is \(a\), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, the side of length \(6\) is opposite the \(60^{\circ}\) angle? Wait, no. Let's re - examine the triangle. The angles are \(90^{\circ}\), \(30^{\circ}\), and \(60^{\circ}\). The side with length \(6\) is adjacent to the \(60^{\circ}\) angle and opposite to the \(30^{\circ}\) angle? Wait, no. Let's use trigonometry correctly. Let's consider the angle of \(30^{\circ}\). The tangent of \(30^{\circ}\) is \(\frac{\text{opposite}}{\text{adjacent}}=\frac{6}{x}\). Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), then \(\frac{6}{x}=\frac{1}{\sqrt{3}}\), so \(x = 6\sqrt{3}\). Alternatively, using cotangent: \(\cot(30^{\circ})=\frac{\text{adjacent}}{\text{opposite}}=\frac{x}{6}\), and \(\cot(30^{\circ})=\sqrt{3}\), so \(x = 6\sqrt{3}\).

Answer:

\(6\sqrt{3}\)