Question

find the length of side x in simplest radical form with a rational denominator.
(image of a right triangle with angles 60°, 30°, right angle, one leg 9, hypotenuse x)
answer attempt 2 out of 2
x = blank 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 \(1 : \sqrt{3} : 2\), where the side opposite \(30^\circ\) is the shortest, opposite \(60^\circ\) is \(\sqrt{3}\) times the shortest, and the hypotenuse is twice the shortest. Here, the side of length 9 is opposite \(30^\circ\)? Wait, no: wait, the right angle, \(60^\circ\), \(30^\circ\). So the side adjacent to \(60^\circ\) (opposite \(30^\circ\)): wait, let's label the triangle. Let the right angle be \(C\), \(60^\circ\) at \(A\), \(30^\circ\) at \(B\). Then side \(AC = 9\), angle at \(A\) is \(60^\circ\), so side \(AC\) is adjacent to \(60^\circ\), opposite to \(30^\circ\)? Wait, no, in a right triangle, the side opposite \(30^\circ\) is the shorter leg. Wait, angle at \(B\) is \(30^\circ\), so side opposite \(B\) (which is \(AC\)) is the shorter leg. Then the hypotenuse \(AB = x\), and the longer leg (opposite \(60^\circ\)) would be \(BC\). Wait, no, let's correct: in 30-60-90, hypotenuse is twice the shorter leg (opposite 30°). So if the shorter leg (opposite 30°) is \(a\), hypotenuse is \(2a\), longer leg (opposite 60°) is \(a\sqrt{3}\). Here, the side of length 9: let's see, angle at \(B\) is 30°, so side opposite \(B\) is \(AC = 9\) (shorter leg). Then hypotenuse \(x\) (opposite right angle) should be \(2 \times 9 = 18\)? Wait, no, that can't be, because 9 is opposite 30°, so hypotenuse is twice that. Wait, but wait, the side with length 9: is it the shorter leg? Let's check the angles. The right angle, 60°, 30°. So the side opposite 30° is the shorter leg. So if angle at \(B\) is 30°, then side \(AC\) (opposite \(B\)) is shorter leg, length 9. Then hypotenuse \(AB = x\) is \(2 \times 9 = 18\)? Wait, no, that seems wrong. Wait, maybe I mixed up the angles. Wait, the side of length 9: let's see, the angle at \(A\) is 60°, so side \(BC\) (opposite \(A\)) is longer leg, and side \(AC\) (opposite \(B\), 30°) is shorter leg. Wait, no, let's use trigonometry. In right triangle, \(\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{x}\). Since \(\sin(30^\circ) = \frac{1}{2}\), so \(\frac{1}{2} = \frac{9}{x}\), so \(x = 18\)? Wait, no, that can't be, because if 30° is at \(B\), then opposite side is \(AC = 9\), so \(\sin(30^\circ) = \frac{AC}{AB} = \frac{9}{x}\), so \(x = 9 / \sin(30^\circ) = 9 / (1/2) = 18\). Wait, but wait, maybe the side of length 9 is the longer leg. Wait, let's re-express. If angle at \(A\) is 60°, then \(\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{x}\)? No, that would be if 9 is opposite 60°, but then \(\sin(60^\circ) = \frac{\sqrt{3}}{2} = \frac{9}{x}\), so \(x = \frac{9 \times 2}{\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}\)? Wait, now I'm confused. Let's look at the triangle again. The right angle, 60°, 30°. So the sides: the side adjacent to 60° is the shorter leg (opposite 30°), the side opposite 60° is longer leg, hypotenuse is longest. So let's label the triangle: vertices \(A\) (60°), \(B\) (30°), \(C\) (right angle). Then side \(AC\) is between \(A\) and \(C\), side \(BC\) between \(B\) and \(C\), side \(AB\) (hypotenuse) between \(A\) and \(B\). So angle at \(A\) is 60°, so side \(BC\) (opposite \(A\)) is longer leg. Angle at \(B\) is 30°, so side \(AC\) (opposite \(B\)) is shorter leg. The right angle is at \(C\), so \(AC\) and \(BC\) are legs, \(AB\) is hypotenuse. Given that \(AC = 9\) (shorter leg, opposite 30°), then hypotenuse \(AB = x = 2 \times AC = 2 \times 9 = 18\)? But that contradicts the trigonometry. Wait, no, maybe the side of length 9 is the longer leg. Let's check: if \(BC\) (longer leg, opposite 60°) is 9, then shorter leg \(AC = \frac{9}{\sqrt{3}} = 3\sqrt{3}\), and hypotenuse \(x = 2 \times 3\sqrt{3} = 6\sqrt{3}\)? Wait, now I'm really confused. Wait, let's use trigonometric ratios. In right triangle, \(\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}\). Angle at \(A\) is 60°, adjacent side is \(AC = 9\), hypotenuse is \(x\). So \(\cos(60^\circ) = \frac{9}{x}\). Since \(\cos(60^\circ) = \frac{1}{2}\), then \(\frac{1}{2} = \frac{9}{x}\), so \(x = 18\). Wait, that makes sense. Because \(\cos(60^\circ) = 0.5\), so adjacent over hypotenuse is 0.5, so hypotenuse is 18. Alternatively, \(\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{x}\), and \(\sin(30^\circ) = 0.5\), so \(x = 18\). Yes, that's correct. So the hypotenuse \(x\) is 18. Wait, but wait, the side of length 9: is it adjacent to 60° or opposite to 30°? Yes, in the triangle, angle at \(A\) is 60°, so side \(AC\) (length 9) is adjacent to 60°, and opposite to 30° (angle at \(B\)). So \(\sin(30^\circ) = \frac{9}{x}\), so \(x = 9 / \sin(30^\circ) = 9 / 0.5 = 18\). Alternatively, \(\cos(60^\circ) = \frac{9}{x}\), \(\cos(60^\circ) = 0.5\), so \(x = 18\). So that's the hypotenuse.

Step2: Confirm with 30-60-90 ratios

In 30-60-90 triangle, hypotenuse is twice the shorter leg (opposite 30°). Here, shorter leg (opposite 30°) is 9, so hypotenuse \(x = 2 \times 9 = 18\).

Answer:

\(x = 18\)