Question

the triangle below is equilateral. find the length of side ( x ) in simplest radical form with a rational denominator.
answer attempt 1 out of 2
( x = )

Explanation:

Step1: Analyze the angles

Since the triangle is equilateral, each angle is \(60^{\circ}\). The smaller right - triangle has an angle of \(60^{\circ}\), and the side opposite to the \(30^{\circ}\) angle (in the right - triangle formed by the altitude of the equilateral triangle) is not directly given, but we know the side adjacent to the \(60^{\circ}\) angle is \(5\). We use the trigonometric ratio \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\theta = 60^{\circ}\), \(\cos60^{\circ}=\frac{1}{2}\), but another way is to use the relationship in a \(30 - 60-90\) triangle. In a \(30 - 60 - 90\) triangle, if the side adjacent to the \(60^{\circ}\) angle is \(a\) and the hypotenuse is \(x\), and we know that \(\cos60^{\circ}=\frac{5}{x}\). But a better approach is to use the fact that in a \(30 - 60-90\) triangle, if the side opposite to \(30^{\circ}\) is \(y\), the side opposite to \(60^{\circ}\) is \(y\sqrt{3}\), and the hypotenuse is \(2y\). Here, if we consider the right - triangle, the side adjacent to \(60^{\circ}\) (let's use the formula \(\cos60^{\circ}=\frac{5}{x}\) is incorrect. We should use \(\sin60^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}\). Wait, no, let's use the property of \(30 - 60-90\) triangle. The altitude of an equilateral triangle divides it into two \(30 - 60-90\) triangles. Let the side of the equilateral triangle be \(s\). The altitude \(h\) of an equilateral triangle with side \(s\) is \(h = \frac{\sqrt{3}}{2}s\). But in our right - triangle, if we assume the side \(x\) is the side of the equilateral triangle, and the altitude (one of the legs of the right - triangle) is related. Wait, no, in the right - triangle, if the angle is \(60^{\circ}\), and the side adjacent to \(60^{\circ}\) is \(5\). We know that \(\tan60^{\circ}=\frac{\text{opposite}}{\text{adjacent}}\), but no. Wait, using the reciprocal. We know that \(\cos60^{\circ}=\frac{1}{2}\), but actually, if we consider the right - triangle with hypotenuse \(x\) (side of equilateral triangle) and one leg \(5\). Since the triangle is equilateral, the angle in the right - triangle (other than \(90^{\circ}\)) is \(60^{\circ}\). Using the formula \(\cos60^{\circ}=\frac{5}{x}\) is wrong. Wait, no! Wait, the side of length \(5\) is adjacent to the \(30^{\circ}\) angle. Wait, no. Let's start over.
In a \(30 - 60-90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\). Let the side opposite to \(30^{\circ}\) be \(a\), the side opposite to \(60^{\circ}\) be \(a\sqrt{3}\), and the hypotenuse be \(2a\). Here, if we assume the side of length \(5\) is opposite to \(30^{\circ}\), that's wrong. Wait, no. The altitude of an equilateral triangle of side \(x\) is \(\frac{\sqrt{3}}{2}x\). But in our right - triangle (formed by the altitude of the equilateral triangle), if we use the trigonometric ratio \(\sin60^{\circ}=\frac{5}{x}\) is wrong. Wait, no! Wait, the side of length \(5\) is the altitude of the equilateral triangle. For an equilateral triangle of side \(x\), the altitude \(h\) is given by \(h=\frac{\sqrt{3}}{2}x\). But we know \(h = 5\). So, \(\frac{\sqrt{3}}{2}x=5\).

Step2: Solve for \(x\)

Cross - multiply the equation \(\frac{\sqrt{3}}{2}x = 5\) to get \(x=\frac{10}{\sqrt{3}}\). Rationalize the denominator: \(x=\frac{10\sqrt{3}}{3}\).

Answer:

\(x=\frac{10\sqrt{3}}{3}\)