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° is the shortest, opposite 60° is \(\sqrt{3}\) times that, and the hypotenuse is twice the shortest side. Here, the side opposite 60° is 4, and \(x\) is opposite 30°? Wait, no—wait, the right angle, 60°, 30°. Let's label: the right angle, angle with 60°: the side adjacent to 60° (wait, no, let's see: the side of length 4 is adjacent to 60°? Wait, no, in the triangle, the right angle, one angle 60°, one 30°. So the side opposite 30° is the shortest, opposite 60° is \(\sqrt{3}\) times that, hypotenuse is twice the shortest. Wait, the side of length 4: let's see, the angle of 60°: the side opposite 30° is \(x\), the side opposite 60° is 4? Wait, no, wait: in a right triangle, \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}\). So \(\tan(60^\circ)=\frac{4}{x}\)? Wait, no, angle 30°: \(\tan(30^\circ)=\frac{4}{x}\)? Wait, no, let's clarify the angles. The triangle has angles 30°, 60°, 90°. So the sides: let's denote the side opposite 30° as \(a\), opposite 60° as \(a\sqrt{3}\), hypotenuse \(2a\). Now, looking at the triangle: the side with length 4 is opposite 60°? Wait, the angle of 60°: the side opposite to 60° would be \(a\sqrt{3}\), and the side opposite 30° is \(a\) (which is \(x\)), and hypotenuse \(2a\). Wait, but in the diagram, the side labeled 4 is adjacent to 30°? Wait, no, let's use trigonometry. \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{x}\)? Wait, no, angle 60°: the opposite side is 4? Wait, no, the right angle is between \(x\) and 4? Wait, the right angle is at the corner where the two legs meet: one leg is \(x\), one leg is 4, hypotenuse is the other side. So angle 30° is at the bottom, angle 60° at the top. So the leg opposite 30° is 4? No, wait, no: in a 30-60-90 triangle, the side opposite 30° is the shorter leg. So if the angle at the bottom is 30°, then the side opposite to it (the leg) would be the shorter one. Wait, maybe I mixed up. Let's use \(\tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{x}\)? Wait, no, \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\), so \(\frac{4}{x}=\frac{1}{\sqrt{3}}\)? No, that would make \(x = 4\sqrt{3}\), but that seems off. Wait, no, maybe \(\tan(60^\circ)=\frac{4}{x}\), since \(\tan(60^\circ)=\sqrt{3}\). So \(\sqrt{3}=\frac{4}{x}\), so \(x=\frac{4}{\sqrt{3}}\). But we need to rationalize the denominator: multiply numerator and denominator by \(\sqrt{3}\), so \(x=\frac{4\sqrt{3}}{3}\)? Wait, no, that can't be. Wait, no, maybe I got the opposite and adjacent wrong. Let's re-express: the angle of 30°: the side adjacent to 30° is 4, and the side opposite is \(x\)? No, wait, the right angle is between \(x\) and 4, so \(x\) and 4 are the legs, hypotenuse is the other side. So angle 30°: \(\tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{x}\), because opposite 30° is 4, adjacent is \(x\)? Wait, no, angle 30°: the side opposite is 4, adjacent is \(x\). Then \(\tan(30^\circ)=\frac{4}{x}\), so \(\frac{1}{\sqrt{3}}=\frac{4}{x}\), so \(x = 4\sqrt{3}\). Wait, but that contradicts the earlier thought. Wait, no, let's use sine or cosine. \(\sin(60^\circ)=\frac{4}{\text{hypotenuse}}\), but we need \(x\). Alternatively, in a 30-60-90 triangle, the sides are in ratio \(1 : \sqrt{3} : 2\). So if the side opposite 60° is \(4\), then the side opposite 30° (which is \(x\)) would be \(\frac{4}{\sqrt{3}}\), but rationalized, that's \(\frac{4\sqrt{3}}{3}\)? Wait, no, wait: the side opposite 30° is \(a\), opposite 60° is \(a\sqrt{3}\), hypotenuse \(2a\). So if \(a\sqrt{3}=4\), then \(a=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\)? But that seems small. Wait, no, maybe the side of length 4 is the hypotenuse? No, the hypotenuse is the longest side. Wait, the side labeled 4: is it a leg or hypotenuse? The triangle has a right angle, so the hypotenuse is opposite the right angle. So the two legs are \(x\) and 4, hypotenuse is the other side. So angle 30°: the side opposite 30° is the shorter leg. So if angle 30° is at the bottom, then the side opposite to it (the leg) is 4? No, that would mean the shorter leg is 4, so the longer leg (opposite 60°) is \(4\sqrt{3}\), but that's not \(x\). Wait, I think I messed up the labels. Let's look at the diagram again: the triangle has a right angle, a 60° angle at the top, 30° at the bottom. So the leg adjacent to 30° is \(x\), and the leg opposite to 30° is 4? No, wait, no: in a right triangle, the side opposite 30° is the shortest leg. So if the angle at the bottom is 30°, then the side opposite to it (the leg) is the shortest, so that would be 4? But then the side opposite 60° (which is \(x\)) would be \(4\sqrt{3}\). Wait, that makes more sense. Wait, no, the side labeled 4 is adjacent to 60°? Wait, maybe using trigonometric ratios: \(\tan(60^\circ)=\frac{x}{4}\), because \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). So \(\theta = 60^\circ\), opposite side is \(x\), adjacent is 4. So \(\tan(60^\circ)=\sqrt{3}=\frac{x}{4}\), so \(x = 4\sqrt{3}\). Wait, but that would be if \(x\) is opposite 60°. But the angle at the bottom is 30°, so the side opposite 30° is 4, side opposite 60° is \(x\). Then \(\tan(30^\circ)=\frac{4}{x}\), so \(\frac{1}{\sqrt{3}}=\frac{4}{x}\), so \(x = 4\sqrt{3}\). Wait, that's the same result. Wait, no, \(\tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{x}\), so \(x=\frac{4}{\tan(30^\circ)}=\frac{4}{\frac{1}{\sqrt{3}}}=4\sqrt{3}\). Yes, that's correct. So \(x = 4\sqrt{3}\). Wait, but let's confirm with 30-60-90 ratios. In a 30-60-90 triangle, the sides are \(a\) (opposite 30°), \(a\sqrt{3}\) (opposite 60°), \(2a\) (hypotenuse). So if the side opposite 30° is \(a = 4\), then the side opposite 60° (which is \(x\)) would be \(4\sqrt{3}\), and hypotenuse \(8\). That matches. So yes, \(x = 4\sqrt{3}\). Wait, but earlier I thought maybe I mixed up, but this makes sense. So the key is recognizing the 30-60-90 triangle ratios, or using trigonometric functions. So using \(\tan(60^\circ)=\sqrt{3}=\frac{x}{4}\), so \(x = 4\sqrt{3}\).

Step2: Rationalize (if needed)

Wait, the problem says "simplest radical form with a rational denominator". But \(4\sqrt{3}\) already has a rational denominator (the denominator is 1). Wait, did I make a mistake? Wait, maybe the side of length 4 is the hypotenuse? Wait, no, the hypotenuse is opposite the right angle, so it's the longest side. If 4 were the hypotenuse, then the side opposite 30° would be \(2\) (since hypotenuse is \(2a\), so \(a = 2\)), and side opposite 60° would be \(2\sqrt{3}\). But in the diagram, the side of length 4 is a leg, not the hypotenuse, because the hypotenuse would be longer than both legs. So 4 is a leg, \(x\) is the other leg. So using \(\tan(60^\circ)=\frac{x}{4}\), so \(x = 4\sqrt{3}\), which is in simplest radical form with a rational denominator (denominator 1).

Answer:

\(4\sqrt{3}\)