Question

find the values of x and y. write your answers in simplest form.

x =

y =

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 \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest (let's call it \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, the hypotenuse is \(12\sqrt{3}\), and \(x\) is adjacent to \(30^{\circ}\) (opposite \(60^{\circ}\)), \(y\) is opposite \(30^{\circ}\).

Step2: Find \(y\) (opposite \(30^{\circ}\))

The hypotenuse \(c = 2a\), where \(a=y\). So \(2y=12\sqrt{3}\), then \(y=\frac{12\sqrt{3}}{2}=6\sqrt{3}\)? Wait, no, wait. Wait, in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is half the hypotenuse? Wait, no, I mixed up. Wait, the hypotenuse is opposite the right angle. Let's label the triangle: right angle at the corner with \(x\) and \(y\), angle \(30^{\circ}\) at the corner with \(x\) and hypotenuse. So the side opposite \(30^{\circ}\) is \(y\), the side opposite \(60^{\circ}\) is \(x\), and hypotenuse is \(12\sqrt{3}\).

In a 30 - 60 - 90 triangle, the ratio of sides (opposite \(30^{\circ}\), opposite \(60^{\circ}\), hypotenuse) is \(a:a\sqrt{3}:2a\). So hypotenuse \(= 2a=12\sqrt{3}\), so \(a = \frac{12\sqrt{3}}{2}=6\sqrt{3}\). Then the side opposite \(30^{\circ}\) (which is \(y\)) is \(a = 6\sqrt{3}\)? Wait, no, wait, no. Wait, no, the side opposite \(30^{\circ}\) is the shortest side. Wait, maybe I got the labels wrong. Let's re - label: let the right - angled triangle have angles \(90^{\circ}\), \(30^{\circ}\), and \(60^{\circ}\). The side opposite \(30^{\circ}\) is the shortest side, let's call it \(s\), the side opposite \(60^{\circ}\) is \(s\sqrt{3}\), and the hypotenuse is \(2s\).

In the given triangle, the hypotenuse is \(12\sqrt{3}\), and the angle of \(30^{\circ}\) is at the vertex where the sides \(x\) and hypotenuse meet. So the side opposite \(30^{\circ}\) is \(y\), so \(y\) is the shortest side. So hypotenuse \(= 2y\), so \(2y = 12\sqrt{3}\), so \(y=\frac{12\sqrt{3}}{2}=6\sqrt{3}\)? Wait, no, that can't be. Wait, no, maybe the side adjacent to \(30^{\circ}\) is \(x\), and opposite is \(y\). Wait, let's use trigonometry.

Using cosine and sine. \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{12\sqrt{3}}\), and \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{y}{12\sqrt{3}}\).

We know that \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{x}{12\sqrt{3}}=\frac{\sqrt{3}}{2}\). Cross - multiply: \(x=\frac{\sqrt{3}}{2}\times12\sqrt{3}=\frac{12\times3}{2}=18\).

And \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{y}{12\sqrt{3}}=\frac{1}{2}\), so \(y = \frac{12\sqrt{3}}{2}=6\sqrt{3}\). Ah, there we go. I made a mistake earlier in the ratio. Let's verify with trigonometry:

\(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{12\sqrt{3}}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(x = 12\sqrt{3}\times\frac{\sqrt{3}}{2}=\frac{12\times3}{2}=18\).

\(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{y}{12\sqrt{3}}\), \(\sin(30^{\circ})=\frac{1}{2}\), so \(y=\frac{12\sqrt{3}}{2}=6\sqrt{3}\).

Answer:

\(x = 18\), \(y = 6\sqrt{3}\)