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

Answer:

$2\sqrt{3}$

Explanation:

Step1: Identify triangle type

This is a 30 - 60 - 90 right triangle. In a 30 - 60 - 90 triangle, the sides are in the ratio $1:\sqrt{3}:2$ (opposite 30°, 60°, 90° respectively).

Step2: Determine side relationships

The side of length 4 is opposite the 90° angle? Wait, no. Wait, the right angle is at the bottom left, so the side opposite 30° is the shorter leg, opposite 60° is the longer leg, and hypotenuse is opposite 90°. Wait, the side with length 4: let's check angles. The angle at the top is 30°, bottom right is 60°, bottom left is 90°. So the side opposite 60° is $x$? Wait, no. Wait, the side labeled 4: let's see, the side adjacent to 60°? Wait, no. Wait, in a 30 - 60 - 90 triangle, the hypotenuse is twice the shorter leg. Wait, maybe I mixed up. Let's use trigonometry. Let's take the angle of 60°. The side adjacent to 60° is the shorter leg (opposite 30°), and the side opposite 60° is the longer leg. Wait, the right angle is at the bottom left, so the sides: the leg opposite 30° (top angle) is the shorter leg, let's call it $a$, the leg opposite 60° is $b$, and hypotenuse $c$. We know that in a 30 - 60 - 90 triangle, $c = 2a$, and $b=a\sqrt{3}$. Wait, but in the diagram, the side with length 4: let's see, the angle at the bottom right is 60°, so the side adjacent to 60° is the shorter leg (opposite 30°), and the side opposite 60° is $x$. Wait, no, the side labeled 4: let's check the angles. The top angle is 30°, so the side opposite 30° is the shorter leg (let's say length $a$), the side opposite 60° is $x$ (longer leg), and hypotenuse is $2a$. Wait, but in the diagram, the side with length 4: is it the hypotenuse? Wait, the right angle is at the bottom left, so the two legs are the bottom side and the left side ($x$), and the hypotenuse is the right side (length 4)? Wait, that makes sense. So hypotenuse $c = 4$, shorter leg (opposite 30°) is $a=\frac{c}{2}=2$, and longer leg (opposite 60°) is $a\sqrt{3}=2\sqrt{3}$. Wait, the left leg ($x$) is opposite 60°, so $x = 2\sqrt{3}$. Let's verify with trigonometry. Using sine: $\sin(60^{\circ})=\frac{x}{4}$. Since $\sin(60^{\circ})=\frac{\sqrt{3}}{2}$, we have $\frac{\sqrt{3}}{2}=\frac{x}{4}$. Solving for $x$, multiply both sides by 4: $x = 4\times\frac{\sqrt{3}}{2}=2\sqrt{3}$.