Question

consider the work shown to determine the value of x in the following triangle.

image of a right triangle with a 30° angle, adjacent side 4√3, hypotenuse x

since this right triangle has a 30° angle, it is a 30°–60°–90° special triangle.
the hypotenuse is √3 times larger than the longer leg.

x = 4√3 × √3
x = 12

which statement best describes the work shown?
○ the value for x was not simplified properly.
● the hypotenuse is not √3 times longer than the longer leg.
○ this is not a 30°–60°–90° special triangle.
○ there was no error made.

correct. good job!

correctly solve for x. if there was no error, enter the value for x from the work above.

Explanation:

Step1: Recall 30-60-90 triangle ratios

In a \(30^\circ - 60^\circ - 90^\circ\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where:
- Shorter leg (opposite \(30^\circ\)): \(s\)
- Longer leg (opposite \(60^\circ\)): \(s\sqrt{3}\)
- Hypotenuse: \(2s\)

Step2: Identify legs and hypotenuse

In the given triangle:
- The angle is \(30^\circ\), so the side opposite \(30^\circ\) is the shorter leg.
- The given leg is \(4\sqrt{3}\), which is opposite \(60^\circ\) (longer leg), so \(s\sqrt{3}=4\sqrt{3}\), thus \(s = 4\) (shorter leg).
- Hypotenuse \(x\) should be \(2s=2\times4 = 8\)? Wait, no—wait, the error in the original work: The original work said hypotenuse is \(\sqrt{3}\) times longer leg, but actually, hypotenuse is \(\frac{2}{\sqrt{3}}\) times the longer leg (or longer leg is \(\frac{\sqrt{3}}{2}\) hypotenuse). Wait, let's do it correctly.

Wait, let's re - express:
Longer leg \(= s\sqrt{3}\), hypotenuse \(= 2s\). So from longer leg \(L = s\sqrt{3}\), hypotenuse \(H=2s=\frac{2L}{\sqrt{3}}\).

Given \(L = 4\sqrt{3}\), then \(H=\frac{2\times4\sqrt{3}}{\sqrt{3}}=8\). Wait, but the original work used \(H = L\times\sqrt{3}\), which is wrong. The correct relation: hypotenuse is \(\frac{2}{\sqrt{3}}\) times the longer leg (or longer leg is \(\frac{\sqrt{3}}{2}\) hypotenuse). So the mistake was in the ratio: hypotenuse is not \(\sqrt{3}\) times longer leg. But wait, let's solve for \(x\) correctly.

Let's use trigonometry. In a right triangle, \(\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(30^\circ\) is \(4\sqrt{3}\), hypotenuse is \(x\). \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), so:
\[ \frac{\sqrt{3}}{2}=\frac{4\sqrt{3}}{x} \]
Cross - multiply: \(x\sqrt{3}=2\times4\sqrt{3}\)
Divide both sides by \(\sqrt{3}\): \(x = 8\)

Wait, but the original work had an error in the ratio. But let's check again. Wait, maybe I mixed up shorter and longer legs. Wait, the angle is \(30^\circ\), so the side adjacent to \(30^\circ\) is the longer leg (since adjacent to \(30^\circ\) is opposite \(60^\circ\)). So:

\(\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\), adjacent \(= 4\sqrt{3}\), hypotenuse \(= x\)

\(\cos(30^\circ)=\frac{\sqrt{3}}{2}=\frac{4\sqrt{3}}{x}\)

Solving for \(x\):

\(x=\frac{4\sqrt{3}\times2}{\sqrt{3}} = 8\)

Wait, but the original work's mistake was in the ratio. But the question now is to correctly solve for \(x\). Wait, maybe I made a mistake earlier. Let's re - express the 30 - 60 - 90 ratios:

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

Given that the leg with length \(4\sqrt{3}\) is the longer leg (opposite \(60^\circ\)), so \(s\sqrt{3}=4\sqrt{3}\implies s = 4\) (shorter leg). Then hypotenuse \(x = 2s=8\)? Wait, no—wait, if the angle is \(30^\circ\), the side adjacent to \(30^\circ\) is the longer leg (because adjacent to \(30^\circ\) is opposite \(60^\circ\)). So the shorter leg is opposite \(30^\circ\), so shorter leg \(= s\), longer leg \(= s\sqrt{3}\), hypotenuse \(= 2s\). So if longer leg \(= 4\sqrt{3}\), then \(s\sqrt{3}=4\sqrt{3}\implies s = 4\), so hypotenuse \(= 2\times4 = 8\). Wait, but the original work said \(x = 4\sqrt{3}\times\sqrt{3}=12\), which is wrong. The correct hypotenuse is 8? Wait, no, wait—maybe I got the angle wrong. Wait, the right angle, \(30^\circ\), so the sides:

Wait, let's use cosine: \(\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\). Adjacent to \(30^\circ\) is \(4\sqrt{3}\), hypotenuse is \(x\). \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), so:

\(\frac{\sqrt{3}}{2}=\frac{4\sqrt{3}}{x}\)

Multiply both sides by \(x\): \(\frac{\sqrt{3}}{2}x=4\sqrt{3}\)

Divide both sides by \(\sqrt{3}\): \(\frac{1}{2}x = 4\)

Multiply both sides by 2: \(x = 8\)

Yes, so the correct value of \(x\) is 8.

Answer:

\(8\)