Question

find s.
write your answer in simplest radical form.
millimeters

Explanation:

Step1: Identify 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 (let's call it \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\).
Here, the side opposite \(60^{\circ}\) is \(6\sqrt{3}\) mm. Let the side opposite \(30^{\circ}\) be \(x\), then the side opposite \(60^{\circ}\) is \(x\sqrt{3}\). So we have \(x\sqrt{3}=6\sqrt{3}\). Solving for \(x\), we divide both sides by \(\sqrt{3}\), getting \(x = 6\) mm.

Step2: Find the hypotenuse \(s\)

The hypotenuse \(s\) (opposite the right angle) in a 30 - 60 - 90 triangle is \(2x\). Since \(x = 6\) mm, then \(s=2\times6 = 12\) mm? Wait, no, wait. Wait, let's re - check. Wait, the side given is \(6\sqrt{3}\) mm. Let's see the angles: the right angle, \(60^{\circ}\), and \(30^{\circ}\). So the side adjacent to \(30^{\circ}\) and opposite \(60^{\circ}\) is \(6\sqrt{3}\), the side opposite \(30^{\circ}\) is \(x\), and hypotenuse is \(s\).
In a 30 - 60 - 90 triangle, \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{s}\), \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{6\sqrt{3}}{s}\). We know that \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\). So \(\frac{6\sqrt{3}}{s}=\frac{\sqrt{3}}{2}\). Cross - multiply: \(\sqrt{3}\times s=6\sqrt{3}\times2\). Then \(s=\frac{12\sqrt{3}}{\sqrt{3}} = 12\)? Wait, no, that's not right. Wait, another approach: in a 30 - 60 - 90 triangle, the sides are \(x\) (opposite \(30^{\circ}\)), \(x\sqrt{3}\) (opposite \(60^{\circ}\)), and \(2x\) (hypotenuse). Here, the side opposite \(60^{\circ}\) is \(6\sqrt{3}\), so \(x\sqrt{3}=6\sqrt{3}\), so \(x = 6\). Then the hypotenuse \(s = 2x\)? Wait, no, wait, the hypotenuse is opposite the right angle. Wait, the right angle is between the two legs. So the leg opposite \(30^{\circ}\) is the shorter leg, leg opposite \(60^{\circ}\) is the longer leg. The hypotenuse is the longest side. So if the longer leg (opposite \(60^{\circ}\)) is \(6\sqrt{3}\), then the shorter leg (opposite \(30^{\circ}\)) is \(x\), and longer leg is \(x\sqrt{3}\). So \(x\sqrt{3}=6\sqrt{3}\implies x = 6\). Then hypotenuse \(s = 2x=12\)? Wait, but let's use trigonometry. Let's take the angle of \(30^{\circ}\). The side adjacent to \(30^{\circ}\) is \(6\sqrt{3}\), and the hypotenuse is \(s\). \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{6\sqrt{3}}{s}\). Since \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), we have \(\frac{6\sqrt{3}}{s}=\frac{\sqrt{3}}{2}\). Cross - multiply: \(\sqrt{3}\times s=6\sqrt{3}\times2\). Then \(s=\frac{12\sqrt{3}}{\sqrt{3}} = 12\). Wait, but let's check with the other angle. \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{6\sqrt{3}}{s}\), \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{6\sqrt{3}}{s}=\frac{\sqrt{3}}{2}\), same result. So \(s = 12\)? Wait, no, wait, maybe I mixed up the sides. Wait, the right angle, \(60^{\circ}\), \(30^{\circ}\). So the sides: the side opposite \(30^{\circ}\) is the shortest, then opposite \(60^{\circ}\), then hypotenuse. So if the side opposite \(60^{\circ}\) is \(6\sqrt{3}\), then the side opposite \(30^{\circ}\) is \(6\), and hypotenuse is \(12\). Yes, that's correct. Because in a 30 - 60 - 90 triangle, hypotenuse is twice the shorter leg. The shorter leg is opposite \(30^{\circ}\), longer leg is shorter leg times \(\sqrt{3}\). So longer leg \(=x\sqrt{3}=6\sqrt{3}\implies x = 6\), hypotenuse \(=2x = 12\).

Answer:

\(12\)