Question

find the length of side x in simplest radical form with a rational denominator.
the number of attempts on this specific problem before you are marked incorrect
answer attempt 1 out of 2
x =

Explanation:

Step1: Identify triangle type

It's 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 6 is opposite the 30° angle? Wait, no: wait, the angle of 30° is at the bottom, 60° at the left, right angle at the top. So the side of length 6 is opposite the 30° angle? Wait, no: let's label the triangle. Let’s call the right angle vertex \(A\), the 60° vertex \(B\), and the 30° vertex \(C\). So side \(AB = 6\) (opposite angle \(C\) (30°)), side \(AC = x\) (opposite angle \(B\) (60°)), and side \(BC\) is the hypotenuse. In a 30-60-90 triangle, the side opposite 30° is the shortest leg, opposite 60° is the longer leg, hypotenuse is twice the shortest leg. Wait, angle at \(C\) is 30°, so side opposite \(C\) is \(AB = 6\) (shortest leg). Then the longer leg (opposite 60°) is \(AB \times \sqrt{3}\)? Wait no, wait: shortest leg (opposite 30°) is \(a\), longer leg (opposite 60°) is \(a\sqrt{3}\), hypotenuse is \(2a\). Wait, in our case, angle at \(C\) is 30°, so side opposite \(C\) is \(AB = 6\) (so \(a = 6\)? Wait no, wait: angle at \(B\) is 60°, so side opposite \(B\) is \(AC = x\) (longer leg), side opposite \(C\) (30°) is \(AB = 6\) (shortest leg). So longer leg \(x = a\sqrt{3}\), where \(a\) is the shortest leg. Wait, no: shortest leg (opposite 30°) is \(a\), longer leg (opposite 60°) is \(a\sqrt{3}\), hypotenuse \(2a\). So if \(AB = 6\) is opposite 30° (angle \(C\)), then \(a = 6\)? Wait no, that can't be, because then longer leg would be \(6\sqrt{3}\), but wait, maybe I mixed up the angles. Wait, the side of length 6 is adjacent to the 30° angle? Wait, maybe I made a mistake. Let's use trigonometry. Let's use sine or cosine. Let's take angle \(C = 30°\), right angle at \(A\). So in triangle \(ABC\), right-angled at \(A\), angle at \(C = 30°\), angle at \(B = 60°\), side \(AB = 6\) (vertical side), side \(AC = x\) (horizontal side), side \(BC\) (hypotenuse). So for angle \(C = 30°\), the adjacent side is \(AC = x\), opposite side is \(AB = 6\), hypotenuse is \(BC\). So \(\tan(30°) = \frac{opposite}{adjacent} = \frac{AB}{AC} = \frac{6}{x}\). \(\tan(30°) = \frac{1}{\sqrt{3}}\), so \(\frac{1}{\sqrt{3}} = \frac{6}{x}\), so \(x = 6\sqrt{3}\)? Wait, no, that would be if angle \(C\) is 30°, adjacent is \(x\), opposite is 6. Wait, \(\tan(\theta) = \frac{opposite}{adjacent}\). So \(\theta = 30°\), opposite is \(AB = 6\), adjacent is \(AC = x\). So \(\tan(30°) = \frac{6}{x}\) → \(x = \frac{6}{\tan(30°)}\). Since \(\tan(30°) = \frac{1}{\sqrt{3}}\), so \(x = 6 \times \sqrt{3} = 6\sqrt{3}\)? Wait, but let's check with 30-60-90 ratios. Alternatively, if angle at \(B\) is 60°, then for angle \(B = 60°\), opposite side is \(AC = x\), adjacent side is \(AB = 6\). So \(\tan(60°) = \frac{x}{6}\), and \(\tan(60°) = \sqrt{3}\), so \(x = 6\sqrt{3}\). Yes, that makes sense. So using trigonometry: \(\tan(60°) = \frac{opposite}{adjacent} = \frac{x}{6}\), so \(x = 6 \times \tan(60°) = 6\sqrt{3}\). Alternatively, in 30-60-90 triangle, the sides are in ratio \(1 : \sqrt{3} : 2\) (short leg : long leg : hypotenuse). The short leg is opposite 30°, long leg opposite 60°, hypotenuse twice short leg. Wait, if angle at \(C\) is 30°, then short leg is opposite 30° (which is \(AB = 6\)), so long leg (opposite 60°) is \(AB \times \sqrt{3} = 6\sqrt{3}\), which is \(x\). Yes, that matches. So \(x = 6\sqrt{3}\).

Step2: Confirm the ratio

In a 30-60-90 triangle, the sides are related as:
- Short leg (opposite 30°): \(a\)
- Long leg (opposite 60°): \(a\sqrt{3}\)
- Hypotenuse: \(2a\)

Here, the side of length 6 is the short leg (opposite 30°), so the long leg (opposite 60°) is \(6\sqrt{3}\), which is \(x\). So that's consistent.

Answer:

\(x = 6\sqrt{3}\)