Question

find the length of side x in simplest radical form with a rational denominator. triangle image with right angle, 30°, 60°, side x, and base 9 answer attempt 1 out of 2 x = input box submit answer

Explanation:

Step1: Identify the triangle type

This is a 30 - 60 - 90 right triangle. 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, opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shortest, and the hypotenuse is twice the shortest. Here, the side opposite \(60^{\circ}\) is 9, and \(x\) is adjacent to \(60^{\circ}\) (opposite \(30^{\circ}\)? Wait, no. Wait, the right angle, \(30^{\circ}\), \(60^{\circ}\). Let's label: the right angle, angle \(A = 90^{\circ}\), angle \(B=30^{\circ}\), angle \(C = 60^{\circ}\). Then side opposite \(B\) (30°) is \(AC\)? Wait, no, the side labeled 9 is adjacent to \(30^{\circ}\) and opposite \(60^{\circ}\), and \(x\) is adjacent to \(60^{\circ}\) and opposite \(30^{\circ}\)? Wait, no, let's use trigonometry. \(\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{9}{x}\). Since \(\tan(60^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{9}{x}\).

Step2: Solve for \(x\)

From \(\sqrt{3}=\frac{9}{x}\), we can cross - multiply to get \(x\sqrt{3}=9\). Then \(x = \frac{9}{\sqrt{3}}\). To rationalize the denominator, multiply numerator and denominator by \(\sqrt{3}\): \(x=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)? Wait, no, wait. Wait, maybe I mixed up the angles. Wait, in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is \(s\), opposite \(60^{\circ}\) is \(s\sqrt{3}\), hypotenuse \(2s\). Let's see: the side with length 9 is opposite \(60^{\circ}\), so \(s\sqrt{3}=9\), so \(s=\frac{9}{\sqrt{3}} = 3\sqrt{3}\)? Wait, no, \(x\) is the side opposite \(30^{\circ}\), so \(x = s\), and the side opposite \(60^{\circ}\) is \(s\sqrt{3}=9\). So \(s=\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)? Wait, no, that can't be. Wait, no, if the side opposite \(60^{\circ}\) is \(9\), then \(s\sqrt{3}=9\), so \(s=\frac{9}{\sqrt{3}} = 3\sqrt{3}\), and \(x\) is the side opposite \(30^{\circ}\), so \(x = s = 3\sqrt{3}\)? Wait, no, let's use cotangent. \(\cot(60^{\circ})=\frac{x}{9}\), and \(\cot(60^{\circ})=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\). So \(\frac{\sqrt{3}}{3}=\frac{x}{9}\), then \(x=\frac{9\sqrt{3}}{3}=3\sqrt{3}\). Wait, yes. So \(x = 3\sqrt{3}\).

Wait, let's re - check. In a right triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). For \(\theta = 60^{\circ}\), opposite side is 9, adjacent side is \(x\). So \(\tan(60^{\circ})=\frac{9}{x}\), \(\tan(60^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{9}{x}\), so \(x=\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\).

Answer:

\(3\sqrt{3}\)