Question

find the length of side ( x ) in simplest radical form with a rational denom
answer
( x = )

Explanation:

Step1: Identify the triangle type

This is a 30 - 60 - 90 right triangle. 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 \(2a\). Here, the hypotenuse is \(\sqrt{8}\), and \(x\) is the side opposite \(30^{\circ}\)? Wait, no. Wait, the angle of \(30^{\circ}\): the side opposite \(30^{\circ}\) is the shorter leg, the side opposite \(60^{\circ}\) is the longer leg, and hypotenuse is opposite \(90^{\circ}\). Wait, in the triangle, the hypotenuse is \(\sqrt{8}\), and we need to find \(x\), which is adjacent to \(60^{\circ}\) and opposite to \(30^{\circ}\)? Wait, let's label the triangle. The right angle is between \(x\) and the other leg. The angle of \(30^{\circ}\) is at the bottom right, so the side opposite \(30^{\circ}\) is \(x\), and the hypotenuse is \(\sqrt{8}\). Wait, in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) (shorter leg) is \(a\), hypotenuse is \(2a\), and the side opposite \(60^{\circ}\) (longer leg) is \(a\sqrt{3}\). So if hypotenuse \(c = 2a=\sqrt{8}\), then \(a=\frac{\sqrt{8}}{2}\). But wait, \(x\) is the side opposite \(30^{\circ}\), so \(x = a\). Wait, but let's check again. Wait, the angle of \(30^{\circ}\): the side opposite \(30^{\circ}\) is the shorter leg, so \(x\) is the shorter leg, hypotenuse is \(\sqrt{8}\). So hypotenuse \(= 2\times\) shorter leg. So \(x=\frac{\text{hypotenuse}}{2}\)? Wait, no, wait: in 30 - 60 - 90 triangle, the ratio is shorter leg : longer leg : hypotenuse \(=1:\sqrt{3}:2\). So shorter leg (opposite \(30^{\circ}\)) \(= a\), longer leg (opposite \(60^{\circ}\)) \(= a\sqrt{3}\), hypotenuse \(= 2a\). So here, hypotenuse is \(\sqrt{8}\), so \(2a=\sqrt{8}\), so \(a = \frac{\sqrt{8}}{2}\). Simplify \(\frac{\sqrt{8}}{2}\): \(\sqrt{8}=2\sqrt{2}\), so \(\frac{2\sqrt{2}}{2}=\sqrt{2}\). Wait, but wait, maybe I mixed up the angles. Wait, the angle of \(60^{\circ}\) is at the bottom left, \(30^{\circ}\) at bottom right, right angle at top. So the side opposite \(30^{\circ}\) is \(x\) (top left to bottom left), and the side opposite \(60^{\circ}\) is the other leg (top left to bottom right), and hypotenuse is bottom left to bottom right (\(\sqrt{8}\)). So yes, hypotenuse \(=\sqrt{8}\), so shorter leg (opposite \(30^{\circ}\)) \(x=\frac{\text{hypotenuse}}{2}=\frac{\sqrt{8}}{2}\). Simplify \(\sqrt{8}\): \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\), so \(\frac{2\sqrt{2}}{2}=\sqrt{2}\). Wait, but let's verify with trigonometry. Let's use sine: \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{\sqrt{8}}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{x}{\sqrt{8}}\), so \(x=\frac{\sqrt{8}}{2}\). Simplify \(\sqrt{8}=2\sqrt{2}\), so \(x = \frac{2\sqrt{2}}{2}=\sqrt{2}\). Wait, but wait, maybe I made a mistake. Wait, alternatively, use cosine: \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{\sqrt{8}}\). Since \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{x}{\sqrt{8}}\), so \(x=\frac{\sqrt{8}}{2}=\sqrt{2}\). Yes, that's correct.

Step2: Simplify the expression

We have \(x=\frac{\sqrt{8}}{2}\). Simplify \(\sqrt{8}\): \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\). So \(\frac{2\sqrt{2}}{2}=\sqrt{2}\).

Answer:

\(\sqrt{2}\)