Question

find x and y using special right triangles. write your answer in simplified radical form.

Explanation:

Step1: Recall the properties of a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle

In a \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle, if the hypotenuse is \(c\), the side opposite \(30^{\circ}\) is \(a=\frac{c}{2}\), and the side opposite \(60^{\circ}\) is \(b = \frac{\sqrt{3}}{2}c\). Here, we can also use the sine formula. \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\)

Step2: Find \(x\)

Since \(x\) is opposite the \(30^{\circ}\) angle. Using \(\sin30^{\circ}=\frac{x}{12}\) (because \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\theta = 30^{\circ}\), hypotenuse \( = 12\)).
We know that \(\sin30^{\circ}=\frac{1}{2}\), so \(x = 12\times\sin30^{\circ}=12\times\frac{1}{2}=6\)

Step3: Find \(y\)

Using \(\cos30^{\circ}=\frac{y}{12}\) (because \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\theta = 30^{\circ}\), hypotenuse \(=12\)).
We know that \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), so \(y = 12\times\cos30^{\circ}=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}\)

Answer:

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