Question

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

Explanation:

Step1: Identify triangle type

This is a 30-60-90 right triangle. In such a triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\) (opposite 30°, 60°, 90° respectively). The side of length 7 is opposite the 60° angle? Wait, no: wait, the right angle is between \(x\) and 7? Wait, no, let's check angles. The 30° angle: the side opposite 30° is \(x\)? Wait, no, let's label the triangle. Let's denote the right angle as \(C\), 30° as \(A\), 60° as \(B\). Then side opposite \(A\) (30°) is \(BC = x\), side opposite \(B\) (60°) is \(AC = 7\), and hypotenuse \(AB\). Wait, in 30-60-90 triangle, the side opposite 30° is the shortest leg, opposite 60° is the longer leg (times \(\sqrt{3}\) of the shorter leg), hypotenuse is twice the shorter leg. Wait, so if the side opposite 60° is 7, then the shorter leg (opposite 30°) is \( \frac{7}{\sqrt{3}} \)? No, wait, maybe I mixed up. Wait, let's use trigonometry. Let's take angle 30°: \(\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{7}\). Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so \(x = 7 \times \tan(30^\circ)\)? Wait, no, wait: angle 30°: adjacent side is 7, opposite side is \(x\). So \(\tan(30^\circ) = \frac{x}{7}\), so \(x = 7 \times \frac{1}{\sqrt{3}}\). But we need rational denominator: multiply numerator and denominator by \(\sqrt{3}\), so \(x = \frac{7\sqrt{3}}{3}\)? Wait, no, wait, maybe I got the angle wrong. Wait, the side of length 7: is it adjacent to 60°? Wait, let's check angle 60°: \(\tan(60^\circ) = \frac{7}{x}\), since \(\tan(60^\circ) = \sqrt{3}\), so \(\sqrt{3} = \frac{7}{x}\), so \(x = \frac{7}{\sqrt{3}}\). Rationalize denominator: \(x = \frac{7\sqrt{3}}{3}\)? Wait, no, wait, maybe the side of length 7 is opposite the 30° angle? No, 30° angle's opposite side would be shorter. Wait, no, let's re-express. In a 30-60-90 triangle, the sides are: short leg (opposite 30°): \(a\), long leg (opposite 60°): \(a\sqrt{3}\), hypotenuse: \(2a\). So if the long leg (opposite 60°) is 7, then \(a\sqrt{3} = 7\), so \(a = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3}\). But wait, the side \(x\): is \(x\) the short leg (opposite 30°)? Then yes, because the 30° angle's opposite side is the short leg. Wait, the angle of 30°: opposite side is \(x\), adjacent is 7 (long leg). So yes, so \(x = \frac{7\sqrt{3}}{3}\)? Wait, no, wait, maybe I made a mistake. Wait, let's use sine. \(\sin(30^\circ) = \frac{x}{\text{hypotenuse}}\), and \(\sin(60^\circ) = \frac{7}{\text{hypotenuse}}\). Since \(\sin(30^\circ) = \frac{1}{2}\), \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). So \(\frac{x}{h} = \frac{1}{2}\), \(\frac{7}{h} = \frac{\sqrt{3}}{2}\). From the second equation, \(h = \frac{14}{\sqrt{3}}\). Then from first equation, \(x = \frac{h}{2} = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3}\). Wait, that's the same as before. So the length of \(x\) is \(\frac{7\sqrt{3}}{3}\)? Wait, no, wait, maybe I mixed up the sides. Wait, the side labeled 7: is it adjacent to the 30° angle? Let's look at the triangle: the right angle is between \(x\) and 7, so \(x\) and 7 are the legs, hypotenuse is the other side. The angle of 30° is at the bottom, so the leg adjacent to 30° is 7, and the leg opposite is \(x\). So \(\tan(30^\circ) = \frac{x}{7}\), so \(x = 7 \tan(30^\circ) = 7 \times \frac{1}{\sqrt{3}} = \frac{7\sqrt{3}}{3}\). Yes, that's correct.

Step2: Rationalize the denominator

We have \(x = \frac{7}{\sqrt{3}}\). To rationalize, multiply numerator and denominator by \(\sqrt{3}\): \(x = \frac{7\sqrt{3}}{3}\).

Answer:

\(\boxed{\dfrac{7\sqrt{3}}{3}}\)