Question

question find the length of side x in simplest radical form with a rational denominator. answer attempt 1 out of 2 x=

Explanation:

Step1: Identify the triangle type

This is a \(30^{\circ}-60^{\circ}-90^{\circ}\) right - triangle. In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\). If the side opposite the \(30^{\circ}\) angle is \(a\), the side opposite the \(60^{\circ}\) angle is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, the side opposite the \(30^{\circ}\) angle is not given, but we know the side opposite the \(60^{\circ}\) angle is \(8\). Let the side opposite the \(30^{\circ}\) angle be \(y\). Then \(y\sqrt{3}=8\), so \(y = \frac{8}{\sqrt{3}}\).

Step2: Use the Pythagorean theorem or the side - ratio relationship

We can also use the tangent function. \(\tan60^{\circ}=\frac{x}{8}\). Since \(\tan60^{\circ}=\sqrt{3}\), we have \(x = 8\sqrt{3}\).

Another way: In a \(30 - 60-90\) triangle, if the side opposite \(60^{\circ}\) is \(s\), and the side opposite \(30^{\circ}\) is \(t\), and the side opposite \(90^{\circ}\) is \(h\). We know that \(\sin60^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{8}{h}\), \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), so \(h=\frac{16}{\sqrt{3}}\). Also, \(\cos60^{\circ}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{h}\), and \(\cos60^{\circ}=\frac{1}{2}\), then \(x=\frac{h}{2}\). Substitute \(h = \frac{16}{\sqrt{3}}\) into \(x=\frac{h}{2}\), we get \(x=\frac{8}{\sqrt{3}}\times\sqrt{3}=8\sqrt{3}\) (by rationalizing the denominator of the side - ratio based calculation in Step1: \(y=\frac{8}{\sqrt{3}}\), and \(x = y\sqrt{3}\times\sqrt{3}\) is wrong, correct using the ratio: if the side opposite \(60^{\circ}\) is \(8\) (let the side opposite \(30^{\circ}\) be \(a\), then \(a\sqrt{3}=8\), \(a=\frac{8}{\sqrt{3}}\), and \(x\) (side opposite \(90^{\circ}\) is \(2a\), no \(x\) is side opposite \(60^{\circ}\) wrong, correct: in a \(30 - 60-90\) triangle, if the side opposite \(30^{\circ}\) is \(a\), side opposite \(60^{\circ}\) is \(a\sqrt{3}\), here \(a\sqrt{3}=8\) (wrong, no \(x\) is adjacent to \(60^{\circ}\), correct: \(\tan60^{\circ}=\frac{x}{8}\), \(\tan60^{\circ}=\sqrt{3}\))

Answer:

\(8\sqrt{3}\)