Question

find the length of side x in simplest radical form with a rational denominator. 9 30° 60° x answer attempt 1 out of 2 x = submit answer √

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 9 is opposite the 30° angle? Wait, no: wait, the right angle, 30°, 60°. Let's check the angles. The right angle is between the side of length 9 and the other leg. Wait, the side of length 9: let's see, the angle of 60°: the side opposite 30° is the shorter leg, opposite 60° is the longer leg, hypotenuse is opposite 90°. Wait, the side labeled 9: let's see, the angle at the bottom left is 60°, so the side adjacent to 60° is the shorter leg (opposite 30°), and the side opposite 60° is the longer leg, and hypotenuse is \(x\). Wait, no: in a 30-60-90 triangle, the side opposite 30° is the shortest leg, let's call it \(a\), then the side opposite 60° is \(a\sqrt{3}\), and hypotenuse is \(2a\). Wait, looking at the triangle: the right angle is at the top, so the two legs are: one leg is 9 (let's say adjacent to 60° angle), and the other leg is opposite 60°? Wait, no, let's label the triangle. Let's denote the vertices: right angle at \(A\), 30° at \(B\), 60° at \(C\). So side \(AC = 9\) (leg), side \(AB\) is the other leg, side \(BC = x\) (hypotenuse). Angle at \(B\) is 30°, so side opposite 30° is \(AC = 9\). So in 30-60-90 triangle, side opposite 30° is \(a = 9\), then hypotenuse \(x = 2a\)? Wait, no, wait: side opposite 30° is the shorter leg, so if angle at \(B\) is 30°, then side \(AC\) (opposite \(B\)) is the shorter leg, so \(AC = a = 9\), then hypotenuse \(BC = x = 2a\)? Wait, no, that can't be, because then hypotenuse would be 18, but wait, maybe I got the angles wrong. Wait, the angle at the bottom is 60°, so angle at \(C\) is 60°, angle at \(B\) is 30°, right angle at \(A\). So side \(AB\) is adjacent to 60° (angle at \(C\)), side \(AC\) is opposite 60° (angle at \(C\)). Wait, angle at \(C\) is 60°, so side opposite angle \(C\) (60°) is \(AB\), and side opposite angle \(B\) (30°) is \(AC = 9\). So in 30-60-90 triangle, side opposite 30° (angle \(B\)) is \(AC = 9\) (shorter leg), then side opposite 60° (angle \(C\)) is \(AB = 9\sqrt{3}\), and hypotenuse \(BC = x = 2 \times 9 = 18\)? Wait, no, that's not right. Wait, no: wait, the side of length 9: is it the shorter leg (opposite 30°) or the longer leg (opposite 60°)? Let's use trigonometry. Let's use cosine or sine. Let's take angle 30°: \(\sin(30°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{x}\). Since \(\sin(30°) = \frac{1}{2}\), so \(\frac{1}{2} = \frac{9}{x}\), so \(x = 18\)? Wait, that can't be, because then the other leg would be \(x \cos(30°) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3}\). Wait, but maybe I made a mistake. Wait, let's check again. The triangle has a right angle, 30°, 60°. The side of length 9: let's see, the angle of 30°: the side adjacent to 30° is the longer leg, and the side opposite 30° is the shorter leg. Wait, \(\cos(30°) = \frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin(30°) = \frac{\text{opposite}}{\text{hypotenuse}}\). Let's take the 30° angle: the side opposite 30° is 9, so \(\sin(30°) = \frac{9}{x}\), so \(x = \frac{9}{\sin(30°)} = \frac{9}{0.5} = 18\). Wait, but that seems too simple. Wait, but maybe the side of length 9 is the longer leg (opposite 60°). Let's try that. If the side of length 9 is opposite 60°, then \(\sin(60°) = \frac{9}{x}\), so \(x = \frac{9}{\sin(60°)} = \frac{9}{\frac{\sqrt{3}}{2}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}\). Wait, now I'm confused. Let's clarify the triangle. Let's look at the angles: right angle, 30°, 60°. So the two legs: one leg is opposite 30°, one opposite 60°, hypotenuse. Let's see the side labeled 9: is it adjacent to 60° or 30°? The angle at the bottom is 60°, so the side adjacent to 60° is the shorter leg (opposite 30°), and the side opposite 60° is the longer leg. Wait, no: in a right triangle, the side adjacent to an angle is the leg that's not opposite. So angle of 60°: adjacent leg is the shorter leg (opposite 30°), opposite leg is the longer leg. So if the side of length 9 is the adjacent leg to 60° (i.e., opposite 30°), then hypotenuse is \(2 \times 9 = 18\). But if the side of length 9 is the opposite leg to 60° (longer leg), then hypotenuse is \(\frac{9}{\sin(60°)} = \frac{9}{\frac{\sqrt{3}}{2}} = 6\sqrt{3}\). Wait, let's check the angles again. The triangle has angles 90°, 30°, 60°. So the side opposite 30° is the shortest leg, side opposite 60° is longer leg, hypotenuse is longest. So if the side of length 9 is the shortest leg (opposite 30°), then hypotenuse is 18. If it's the longer leg (opposite 60°), hypotenuse is \( \frac{9}{\sin(60°)} \). Wait, let's use trigonometry with angle 30°: \(\cos(30°) = \frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin(30°) = \frac{\text{opposite}}{\text{hypotenuse}}\). Let's suppose the side of length 9 is adjacent to 30° angle. Then \(\cos(30°) = \frac{9}{x}\), so \(x = \frac{9}{\cos(30°)} = \frac{9}{\frac{\sqrt{3}}{2}} = 6\sqrt{3}\). Wait, now I'm confused. Let's draw the triangle mentally: right angle at top, 30° at bottom right, 60° at bottom left. So the two legs: left leg (vertical) is 9, bottom leg (horizontal) is the other leg, hypotenuse is \(x\) (bottom side). So angle at bottom left is 60°, so the left leg (vertical) is opposite 60° angle (since angle at bottom left: the vertical leg is opposite to it). So vertical leg (9) is opposite 60° angle, so it's the longer leg. Then, in 30-60-90 triangle, longer leg (opposite 60°) is \(a\sqrt{3}\), where \(a\) is shorter leg (opposite 30°). So \(a\sqrt{3} = 9\), so \(a = \frac{9}{\sqrt{3}} = 3\sqrt{3}\). Then hypotenuse \(x = 2a = 2 \times 3\sqrt{3} = 6\sqrt{3}\). Wait, that makes sense. So let's confirm: shorter leg (opposite 30°) is \(a = 3\sqrt{3}\), longer leg (opposite 60°) is \(a\sqrt{3} = 3\sqrt{3} \times \sqrt{3} = 9\), which matches the given side. Then hypotenuse is \(2a = 6\sqrt{3}\). Yes, that works. So the key was realizing that the side of length 9 is the longer leg (opposite 60°), so we can find the shorter leg first, then hypotenuse.

So step 1: Identify the triangle as 30-60-90, so sides are in ratio \(a : a\sqrt{3} : 2a\) (shorter leg : longer leg : hypotenuse).

Step 2: The longer leg (opposite 60°) is \(a\sqrt{3} = 9\). So solve for \(a\): \(a = \frac{9}{\sqrt{3}} = 3\sqrt{3}\) (rationalizing the denominator: \(\frac{9}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{9\sqrt{3}}{3} = 3\sqrt{3}\)).

Step 3: Hypotenuse \(x = 2a = 2 \times 3\sqrt{3} = 6\sqrt{3}\). Wait, but let's check with sine. \(\sin(60°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{x}\), so \(x = \frac{9}{\sin(60°)} = \frac{9}{\frac{\sqrt{3}}{2}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}\) (rationalizing: \(\frac{18}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{18\sqrt{3}}{3} = 6\sqrt{3}\)). Yes, that's correct.

So the length of \(x\) is \(6\sqrt{3}\)? Wait, no, wait: wait, if the side of length 9 is the longer leg (opposite 60°), then hypotenuse is \(6\sqrt{3}\)? Wait, no, wait, 30-60-90 triangle: shorter leg \(a\), longer leg \(a\sqrt{3}\), hypotenuse \(2a\). So if longer leg is 9, then \(a\sqrt{3} = 9\) => \(a = 9/\sqrt{3} = 3\sqrt{3}\), then hypotenuse is \(2a = 6\sqrt{3}\). Yes, that's correct.

Answer:

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