Question

38 use special right triangles to find the missing length. find the length of side x in simplest radical form with a rational denominator. triangle image: right triangle with right angle, 60° angle, 30° angle, one leg (opposite 30°) is 2, hypotenuse is 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 \(2a\).

Step2: Determine which side is given

The side with length 2 is opposite the \(30^{\circ}\) angle? Wait, no. Wait, the right angle is between the side of length 2 and the base. The angle at the bottom is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the side with length 2? Wait, no. Wait, in a right triangle, the hypotenuse is opposite the right angle. The angle at the top is \(60^{\circ}\), the angle at the bottom is \(30^{\circ}\), and the right angle is between the vertical side (length 2) and the horizontal side. So the side of length 2 is adjacent to the \(60^{\circ}\) angle and opposite the \(30^{\circ}\) angle? Wait, no. Let's recall: in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is the shortest side. Let's denote:
- Let the side opposite \(30^{\circ}\) be \(a\).
- Side opposite \(60^{\circ}\) be \(a\sqrt{3}\).
- Hypotenuse be \(2a\).

Looking at the triangle, the side with length 2: let's see the angles. The angle at the bottom is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the vertical side? Wait, no. Wait, the vertical side (length 2) is adjacent to the \(30^{\circ}\) angle? Wait, maybe I got it wrong. Wait, the right angle is between the vertical side (length 2) and the horizontal side. The angle at the top is \(60^{\circ}\), so the vertical side (length 2) is adjacent to the \(60^{\circ}\) angle. Wait, let's use trigonometry. Let's take the angle \(30^{\circ}\). The sine of \(30^{\circ}\) is \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(30^{\circ}\) is the vertical side (length 2). So \(\sin(30^{\circ})=\frac{2}{x}\), since \(x\) is the hypotenuse. We know that \(\sin(30^{\circ})=\frac{1}{2}\). So \(\frac{1}{2}=\frac{2}{x}\).

Step3: Solve for \(x\)

Cross - multiply: \(x\times1 = 2\times2\), so \(x = 4\)? Wait, no, that can't be. Wait, maybe I mixed up the opposite and adjacent. Wait, no. Wait, \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\). The angle at the bottom is \(30^{\circ}\), so the opposite side to \(30^{\circ}\) is the vertical side (length 2), and the hypotenuse is \(x\). So \(\sin(30^{\circ})=\frac{2}{x}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), then \(\frac{1}{2}=\frac{2}{x}\). Solving for \(x\): \(x = 2\times2=4\)? Wait, but let's check with cosine. The adjacent side to \(30^{\circ}\) is the horizontal side, and the hypotenuse is \(x\). \(\cos(30^{\circ})=\frac{\text{adjacent}}{x}\). But maybe a better way: in a 30 - 60 - 90 triangle, the hypotenuse is twice the length of the side opposite \(30^{\circ}\). The side opposite \(30^{\circ}\) is 2, so hypotenuse \(x = 2\times2 = 4\)? Wait, no, that seems off. Wait, no, wait. Wait, if the side opposite \(30^{\circ}\) is \(a\), then hypotenuse is \(2a\). If \(a = 2\), then hypotenuse is \(4\). Let's verify with the other angle. The angle at the top is \(60^{\circ}\), the side opposite \(60^{\circ}\) would be \(a\sqrt{3}=2\sqrt{3}\), which would be the horizontal side. Let's check with sine of \(60^{\circ}\): \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\), which is correct. So yes, the hypotenuse \(x = 4\). Wait, but let's re - examine the triangle. The vertical side is length 2, angle at the bottom is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the vertical side (length 2), so hypotenuse is \(2\times2 = 4\).

Answer:

\(4\)