Question

a right triangle with a 60-degree angle, hypotenuse labeled x, vertical leg labeled 5√3, horizontal leg labeled y.

Explanation:

Step1: Use cosine function to find \(x\)

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here \(\theta = 30^{\circ}\), adjacent side to \(30^{\circ}\) is \(5\sqrt{3}\), and hypotenuse is \(x\).
\(\cos30^{\circ}=\frac{5\sqrt{3}}{x}\)
Since \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{5\sqrt{3}}{x}\).
Cross - multiply: \(\sqrt{3}x = 10\sqrt{3}\), then \(x = 10\).

Step2: Use sine function to find \(y\)

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here \(\theta = 30^{\circ}\), opposite side to \(30^{\circ}\) is \(y\), and hypotenuse \(x = 10\).
\(\sin30^{\circ}=\frac{y}{x}\)
Since \(\sin30^{\circ}=\frac{1}{2}\) and \(x = 10\), then \(y=\frac{1}{2}\times10 = 5\).

Answer:

\(x = 10,y = 5\)