Question

the longer leg of a 30°-60°-90° triangle is \\(\sqrt{6}\\) yards long. how long is the hypotenuse?
write your answer in simplest radical form.
\boxed{} yards

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 the side opposite \(30^\circ\) (shorter leg) is \(x\), the side opposite \(60^\circ\) (longer leg) is \(x\sqrt{3}\), and the hypotenuse is \(2x\).

Step2: Relate longer leg to \(x\)

Given the longer leg is \(\sqrt{6}\) yards. The longer leg is \(x\sqrt{3}\), so we set \(x\sqrt{3}=\sqrt{6}\). Solve for \(x\):
\(x = \frac{\sqrt{6}}{\sqrt{3}}\)
Simplify \(\frac{\sqrt{6}}{\sqrt{3}}\) by rationalizing or using radical properties: \(\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{\frac{6}{3}}=\sqrt{2}\).

Step3: Find hypotenuse

The hypotenuse is \(2x\). Substitute \(x = \sqrt{2}\):
Hypotenuse \(= 2\sqrt{2}\)? Wait, no—wait, let's recheck. Wait, longer leg is \(x\sqrt{3}\), so \(x\sqrt{3}=\sqrt{6}\) ⇒ \(x=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}\). Then hypotenuse is \(2x = 2\sqrt{2}\)? Wait, no, wait: Wait, maybe I mixed up. Wait, in 30-60-90, shorter leg is \(x\), longer leg (opposite 60) is \(x\sqrt{3}\), hypotenuse \(2x\). So if longer leg is \(x\sqrt{3}=\sqrt{6}\), then \(x=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}\). Then hypotenuse is \(2x = 2\sqrt{2}\)? Wait, no, wait, let's compute again. Wait, \(\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{\frac{6}{3}}=\sqrt{2}\), so \(x=\sqrt{2}\). Then hypotenuse is \(2x = 2\sqrt{2}\)? Wait, but let's check with another approach. Alternatively, longer leg is \(\sqrt{3}\) times shorter leg, hypotenuse is \(\frac{2}{\sqrt{3}}\) times longer leg. Let's try that: hypotenuse \(= \frac{2}{\sqrt{3}} \times \) longer leg. Longer leg is \(\sqrt{6}\), so hypotenuse \(= \frac{2\sqrt{6}}{\sqrt{3}} = 2\sqrt{\frac{6}{3}} = 2\sqrt{2}\). Wait, that's the same. Wait, but let's verify: If hypotenuse is \(2\sqrt{2}\), then shorter leg is \(\sqrt{2}\), longer leg is \(\sqrt{2} \times \sqrt{3} = \sqrt{6}\), which matches. So yes, hypotenuse is \(2\sqrt{2}\)? Wait, no, wait—wait, no, wait, I think I made a mistake. Wait, no: Wait, \(x\sqrt{3}=\sqrt{6}\) ⇒ \(x = \sqrt{6}/\sqrt{3} = \sqrt{2}\), then hypotenuse is \(2x = 2\sqrt{2}\). Wait, but let's check with actual values. Let's compute \(2\sqrt{2}\approx2.828\), longer leg \(\sqrt{6}\approx2.449\). Wait, but in 30-60-90, hypotenuse should be longer than longer leg. \(2\sqrt{2}\approx2.828\) is longer than \(\sqrt{6}\approx2.449\), so that works. Wait, but let's do it again. Alternatively, use trigonometry: in a right triangle, \(\sin(60^\circ)=\frac{\text{longer leg}}{\text{hypotenuse}}\). \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{\text{hypotenuse}}\). Solve for hypotenuse: \(\text{hypotenuse}=\frac{2\sqrt{6}}{\sqrt{3}} = 2\sqrt{2}\). Yes, same result.

Answer:

\(2\sqrt{2}\)