Question

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

Explanation:

Step1: Identify the triangle type

This is a right - angled triangle with angles \(30^{\circ}\), \(60^{\circ}\) and \(90^{\circ}\). In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest side (let's call it \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\).

We know that the side with length 9 is opposite the \(30^{\circ}\) angle? Wait, no. Wait, the angle of \(30^{\circ}\) and \(60^{\circ}\): the side adjacent to \(30^{\circ}\) and opposite to \(60^{\circ}\) is 9? Wait, let's re - examine. The right - angled triangle: the angle of \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). The side of length 9: let's see the angles. The angle of \(60^{\circ}\): the side opposite to \(60^{\circ}\)? Wait, no. Wait, in a right - angled triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).

We have a right - angled triangle, with one angle \(30^{\circ}\), one angle \(60^{\circ}\), right angle \(90^{\circ}\). The side of length 9: let's check which angle it is opposite to. The angle of \(60^{\circ}\): the side opposite to \(60^{\circ}\) is 9? Wait, no. Wait, the side with length 9: let's see, the angle of \(30^{\circ}\) has opposite side? Wait, no, let's use trigonometric ratios.

We know that in a right - angled triangle, \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\).

Wait, the side of length 9: let's see, the angle of \(60^{\circ}\): the side adjacent to \(60^{\circ}\) is 9? No, wait, the angle of \(30^{\circ}\): the side opposite to \(30^{\circ}\) is the shorter leg. Wait, the side with length 9: let's check the angles. The angle of \(60^{\circ}\): the side opposite to \(60^{\circ}\) is 9? Wait, no, let's use the fact that in a \(30 - 60 - 90\) triangle, the hypotenuse is twice the shorter leg (the leg opposite \(30^{\circ}\)). Wait, the side of length 9: is it the leg opposite \(60^{\circ}\)?

Wait, let's label the triangle. Let the right angle be \(C\), angle \(A = 60^{\circ}\), angle \(B = 30^{\circ}\), side \(AC = 9\), side \(BC\) is the other leg, and side \(AB=x\) (hypotenuse).

In triangle \(ABC\), right - angled at \(C\), \(\sin(A)=\sin(60^{\circ})=\frac{BC}{AB}\), \(\cos(A)=\cos(60^{\circ})=\frac{AC}{AB}\)

We know that \(\cos(60^{\circ})=\frac{1}{2}\), and \(AC = 9\), \(AB=x\). So \(\cos(60^{\circ})=\frac{AC}{AB}\)

Step2: Apply the cosine ratio

\(\cos(60^{\circ})=\frac{AC}{AB}\)

We know that \(\cos(60^{\circ})=\frac{1}{2}\), \(AC = 9\), and \(AB=x\). So:

\(\frac{1}{2}=\frac{9}{x}\)

Cross - multiply: \(x\times1=9\times2\)

\(x = 18\)? Wait, no, that can't be right. Wait, maybe I mixed up the sides. Wait, the side of length 9: is it the leg opposite \(30^{\circ}\)? Wait, no, the angle of \(30^{\circ}\) has opposite side. Wait, let's use the sine ratio.

Wait, the angle of \(30^{\circ}\): \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). If the side opposite \(30^{\circ}\) is 9, then \(\sin(30^{\circ})=\frac{9}{x}\), and since \(\sin(30^{\circ})=\frac{1}{2}\), then \(\frac{1}{2}=\frac{9}{x}\), so \(x = 18\). Wait, but let's check the other angle. The angle of \(60^{\circ}\): \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite \(60^{\circ}\) would be \(9\sqrt{3}\)? Wait, no, if the hypotenuse is 18, and the side opposite \(60^{\circ}\) is \(\sin(60^{\circ})\times18=\frac{\sqrt{3}}{2}\times18 = 9\sqrt{3}\), but in the diagram, the side of length 9 is one of the legs. Wait, maybe I made a mistake in identifying the sides.

Wait, let's re - examine the triangle. The triangle has angles \(60^{\circ}\), \(30^{\circ}\), and \(90^{\circ}\). The side labeled 9 is adjacent to the \(60^{\circ}\) angle and opposite to the \(30^{\circ}\) angle? No, wait, the side of length 9: let's see, the angle of \(60^{\circ}\): the side adjacent to \(60^{\circ}\) is 9, and the hypotenuse is \(x\). Then \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{9}{x}\)

Since \(\cos(60^{\circ})=\frac{1}{2}\), then \(\frac{1}{2}=\frac{9}{x}\), so \(x = 18\). Wait, but that seems too long. Wait, maybe the side of length 9 is the leg opposite \(60^{\circ}\). Let's try that.

If the side opposite \(60^{\circ}\) is 9, then in a \(30 - 60 - 90\) triangle, the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), where \(a\) is the side opposite \(30^{\circ}\), and the hypotenuse is \(2a\).

So if \(a\sqrt{3}=9\), then \(a=\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\). Then the hypotenuse \(x = 2a=6\sqrt{3}\). Wait, now I'm confused. Which is correct?

Wait, let's look at the angles again. The triangle has angles \(60^{\circ}\), \(30^{\circ}\), and \(90^{\circ}\). The side of length 9: let's see, the angle of \(60^{\circ}\): the side adjacent to \(60^{\circ}\) is the leg that is not the hypotenuse and is next to the \(60^{\circ}\) angle. The side opposite \(60^{\circ}\) is the leg opposite the \(60^{\circ}\) angle.

Wait, let's use the fact that the sum of angles in a triangle is \(180^{\circ}\), so it is a right - angled triangle with angles \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\). So the sides are in the ratio \(1:\sqrt{3}:2\) (shorter leg : longer leg : hypotenuse). The shorter leg is opposite \(30^{\circ}\), the longer leg is opposite \(60^{\circ}\), and the hypotenuse is opposite \(90^{\circ}\).

Now, the side of length 9: is it the shorter leg or the longer leg?

If the side of length 9 is the longer leg (opposite \(60^{\circ}\)), then longer leg \(=a\sqrt{3}\), where \(a\) is the shorter leg (opposite \(30^{\circ}\)). So \(a\sqrt{3}=9\), so \(a = \frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\). Then the hypotenuse \(x = 2a=2\times3\sqrt{3}=6\sqrt{3}\)? No, that can't be. Wait, no, the hypotenuse is twice the shorter leg. Wait, no, in a \(30 - 60 - 90\) triangle, the ratios are:

- Shorter leg (opposite \(30^{\circ}\)): \(a\)
- Longer leg (opposite \(60^{\circ}\)): \(a\sqrt{3}\)
- Hypotenuse: \(2a\)

So if the side of length 9 is the longer leg (opposite \(60^{\circ}\)), then \(a\sqrt{3}=9\), so \(a = \frac{9}{\sqrt{3}} = 3\sqrt{3}\), and the hypotenuse \(x = 2a=6\sqrt{3}\). But if the side of length 9 is the shorter leg (opposite \(30^{\circ}\)), then the hypotenuse \(x = 2\times9 = 18\), and the longer leg is \(9\sqrt{3}\).

Wait, now I need to figure out which side is which. The side of length 9: looking at the diagram, the side labeled 9 is between the \(60^{\circ}\) angle and the right angle. So that side is adjacent to the \(60^{\circ}\) angle and opposite to the \(30^{\circ}\) angle. So it is the shorter leg (opposite \(30^{\circ}\))? Wait, no, the side opposite \(30^{\circ}\) is the shorter leg. So if the side of length 9 is opposite \(30^{\circ}\), then the hypotenuse \(x = 2\times9=18\), and the longer leg (opposite \(60^{\circ}\)) is \(9\sqrt{3}\). But in the diagram, the side \(x\) is the hypotenuse (the side opposite the right angle).

Wait, the right angle is between the side of length 9 and the other leg. So the hypotenuse is \(x\), which is opposite the right angle. So the side of length 9 is one leg, and the other leg is the one opposite \(30^{\circ}\) or \(60^{\circ}\).

Wait, let's use trigonometry. Let's take angle \(B = 30^{\circ}\), angle \(A = 60^{\circ}\), right angle at \(C\). So side \(AC = 9\) (adjacent to angle \(A\)), side \(BC\) (adjacent to angle \(B\)), and side \(AB=x\) (hypotenuse).

\(\cos(A)=\cos(60^{\circ})=\frac{AC}{AB}\)

\(\cos(60^{\circ})=\frac{1}{2}=\frac{9}{x}\)

So \(x = 18\). But that seems like the hypotenuse is 18, and the other leg \(BC\) can be found by \(\sin(60^{\circ})=\frac{BC}{AB}\), \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}=\frac{BC}{18}\), so \(BC = 9\sqrt{3}\).

Alternatively, if we take angle \(B = 30^{\circ}\), \(\cos(B)=\cos(30^{\circ})=\frac{BC}{AB}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}=\frac{BC}{x}\), and \(\sin(B)=\sin(30^{\circ})=\frac{AC}{AB}=\frac{9}{x}\), since \(\sin(30^{\circ})=\frac{1}{2}\), then \(\frac{9}{x}=\frac{1}{2}\), so \(x = 18\). Ah, here we go. So \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{AC}{AB}\), and \(AC = 9\), \(AB=x\), \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{9}{x}\), so \(x = 18\). Wait, but that contradicts the earlier thought. But according to the sine of \(30^{\circ}\), which is \(\frac{1}{2}\), the side opposite \(30^{\circ}\) is half the hypotenuse. So if the side opposite \(30^{\circ}\) is 9, then the hypotenuse is \(2\times9 = 18\). That makes sense.

So the correct approach is:

In the right - angled triangle, angle \(B = 30^{\circ}\), so the side opposite angle \(B\) (which is \(AC\)) is equal to \(\frac{1}{2}\) of the hypotenuse \(AB\) (which is \(x\)).

So \(AC=\frac{1}{2}AB\)

We know \(AC = 9\), so \(9=\frac{1}{2}x\)

Multiply both sides by 2: \(x = 18\)

Wait, but now I'm confused because when I thought of the longer leg, I got a different answer. But according to the sine of \(30^{\circ}\), the side opposite \(30^{\circ}\) is half the hypotenuse. So if the side of length 9 is opposite \(30^{\circ}\), then hypotenuse is 18.

Yes, that's correct. Because in a \(30 - 60 - 90\) triangle, the side opposite \(30^{\circ}\) is the shortest side (shorter leg) and is equal to \(\frac{1}{2}\) of the hypotenuse. The side opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shorter leg.

So in this case, the side of length 9 is the shorter leg (opposite \(30^{\circ}\)), so the hypotenuse \(x = 2\times9=18\).

Wait, but let's check with the other angle. The angle \(A = 60^{\circ}\), the side opposite \(A\) is the longer leg, which should be \(9\sqrt{3}\). Let's check with Pythagoras: \(9^{2}+(9\sqrt{3})^{2}=81 + 243=324=18^{2}\), which is correct. So \(9^{2}+(9\sqrt{3})^{2}=81 + 243 = 324=18^{2}\), so it satisfies the Pythagorean theorem.

So the length of \(x\) is 18.

Answer:

\(18\)