Question

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

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 side (let's call it \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, the side opposite \(60^{\circ}\) is 2? Wait, no, wait. Wait, the right angle, the angle of \(30^{\circ}\), \(60^{\circ}\). Let's label the triangle: the right angle is between the side of length 2 and the other leg, and the hypotenuse is \(x\). Wait, in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is the shortest leg, the side opposite \(60^{\circ}\) is the longer leg, and the hypotenuse is twice the shortest leg. Wait, let's see: the angle of \(30^{\circ}\) is opposite the side of length 2? No, wait, the side with length 2 is adjacent to the \(30^{\circ}\) angle? Wait, no, let's look at the triangle. The right angle is at the bottom right, the \(30^{\circ}\) angle is at the bottom left, the \(60^{\circ}\) angle is at the top. So the side opposite \(30^{\circ}\) is the side with length 2? Wait, no, the side opposite \(30^{\circ}\) would be the side opposite the \(30^{\circ}\) angle, which is the vertical side (length 2). Wait, no, in a right triangle, the hypotenuse is opposite the right angle. So the hypotenuse is \(x\). The angle of \(30^{\circ}\) is at the bottom left, so the side opposite \(30^{\circ}\) is the side opposite that angle, which is the vertical leg (length 2). Then, in a 30 - 60 - 90 triangle, the hypotenuse is twice the length of the side opposite \(30^{\circ}\)? Wait, no, that's if the side opposite \(30^{\circ}\) is the shorter leg. Wait, no, the side opposite \(30^{\circ}\) is the shorter leg, and the hypotenuse is twice that. Wait, let's recall: in a 30 - 60 - 90 triangle, the ratios are: shorter leg (opposite \(30^{\circ}\)): \(a\), longer leg (opposite \(60^{\circ}\)): \(a\sqrt{3}\), hypotenuse: \(2a\). So if the side opposite \(30^{\circ}\) is \(a\), then hypotenuse is \(2a\). Wait, in our triangle, the side with length 2: is that the shorter leg (opposite \(30^{\circ}\))? Let's check the angles. The \(30^{\circ}\) angle is at the bottom left, so the side opposite \(30^{\circ}\) is the side going up from the right angle (length 2). So that's the shorter leg (opposite \(30^{\circ}\)), so \(a = 2\). Then the hypotenuse \(x\) would be \(2a = 4\)? Wait, no, that can't be. Wait, no, maybe I got the sides mixed up. Wait, no, wait, the side opposite \(60^{\circ}\) is the longer leg. Wait, let's use trigonometry. Let's use cosine or sine. Let's take the \(30^{\circ}\) angle. The cosine of \(30^{\circ}\) is adjacent over hypotenuse. The adjacent side to \(30^{\circ}\) is the horizontal leg, and the hypotenuse is \(x\). The opposite side to \(30^{\circ}\) is the vertical leg (length 2). So \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{2}{x}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{2}{x}\). Solving for \(x\): \(x = 2\times2 = 4\)? Wait, that seems off. Wait, no, maybe the side with length 2 is the longer leg (opposite \(60^{\circ}\)). Let's try that. So if the side opposite \(60^{\circ}\) is 2, then in the 30 - 60 - 90 triangle, the longer leg (opposite \(60^{\circ}\)) is \(a\sqrt{3}\), and the shorter leg (opposite \(30^{\circ}\)) is \(a\), hypotenuse is \(2a\). So if \(a\sqrt{3}=2\), then \(a=\frac{2}{\sqrt{3}}\), and hypotenuse \(x = 2a=\frac{4}{\sqrt{3}}\), but we need to rationalize the denominator: \(\frac{4\sqrt{3}}{3}\). Wait, but that contradicts the earlier thought. Wait, let's check the angles again. The right angle, \(30^{\circ}\), \(60^{\circ}\). So the sides: hypotenuse is \(x\), one leg is 2, the other leg is, say, \(b\). Let's use trigonometry. Let's take the angle of \(60^{\circ}\). The side adjacent to \(60^{\circ}\) is 2 (the vertical leg), and the hypotenuse is \(x\). So \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{2}{x}\). Since \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{2}{x}\), so \(x = 4\). Wait, that's different. Wait, no, \(\cos(60^{\circ})\) is adjacent over hypotenuse. The angle of \(60^{\circ}\) is at the top, so the adjacent side to \(60^{\circ}\) is the vertical leg (length 2), and the hypotenuse is \(x\). So \(\cos(60^{\circ})=\frac{2}{x}\), and \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{2}{x}=\frac{1}{2}\), so \(x = 4\). Alternatively, using the 30 - 60 - 90 ratios: the side opposite \(30^{\circ}\) is the shorter leg. The angle of \(30^{\circ}\) is at the bottom left, so the side opposite \(30^{\circ}\) is the vertical leg (length 2)? Wait, no, the side opposite \(30^{\circ}\) would be the side opposite that angle, which is the vertical leg (length 2). Then the hypotenuse is twice that, so \(x = 2\times2 = 4\). Wait, that makes sense. Because in a 30 - 60 - 90 triangle, hypotenuse is twice the shorter leg (opposite \(30^{\circ}\)). So if the shorter leg (opposite \(30^{\circ}\)) is 2, then hypotenuse is 4. Alternatively, using sine: \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), where the opposite side to \(60^{\circ}\) is the horizontal leg, but we don't know that. Wait, maybe I confused the sides. Let's draw the triangle: right angle at (0,0), \(30^{\circ}\) at (c,0), \(60^{\circ}\) at (0,b), hypotenuse from (c,0) to (0,b) is \(x\). The side from (0,0) to (0,b) is length 2 (vertical leg), the side from (0,0) to (c,0) is horizontal leg, hypotenuse is \(x\) from (c,0) to (0,b). Then angle at (c,0) is \(30^{\circ}\), so the angle between the hypotenuse and the horizontal leg is \(30^{\circ}\). So the vertical leg (length 2) is opposite the \(30^{\circ}\) angle, so it's the shorter leg. Then hypotenuse is twice that, so \(x = 2\times2 = 4\). Yes, that makes sense. So the length of \(x\) is 4. Wait, but let's confirm with trigonometry. \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{2}{x}\). \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{2}{x}\) → \(x = 4\). Yep, that works.

Step2: Confirm the ratio

In a 30 - 60 - 90 right triangle, the hypotenuse is twice the length of the side opposite the \(30^{\circ}\) angle. Here, the side opposite \(30^{\circ}\) is 2, so hypotenuse \(x = 2\times2 = 4\).

Answer:

\(4\)