Question

1. right triangle with right angle, 60° angle, leg labeled (5sqrt{3}), hypotenuse labeled (x), other leg labeled (y)

Explanation:

Step1: Identify the triangle type

This is a right - triangle with one angle \(60^{\circ}\), so the other non - right angle is \(30^{\circ}\) (since the sum of angles in a triangle is \(180^{\circ}\) and one angle is \(90^{\circ}\)). In a \(30 - 60 - 90\) right - triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest side (let's call it \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\).

The side opposite \(60^{\circ}\) is \(5\sqrt{3}\). Let the side opposite \(30^{\circ}\) be \(y\) and the hypotenuse be \(x\).

Step2: Find the length of \(y\)

We know that the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and here \(a\sqrt{3}=5\sqrt{3}\). Dividing both sides by \(\sqrt{3}\), we get \(a = 5\). The side opposite \(30^{\circ}\) (which is \(y\)) is \(a\), so \(y = 5\).

Step3: Find the length of \(x\) (the hypotenuse)

In a \(30 - 60 - 90\) triangle, the hypotenuse \(x = 2a\). Since \(a = 5\), then \(x=2\times5 = 10\). We can also use trigonometric ratios. Using \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{5\sqrt{3}}{x}\). We know that \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{5\sqrt{3}}{x}\). Cross - multiplying gives \(x\times\sqrt{3}=2\times5\sqrt{3}\). Dividing both sides by \(\sqrt{3}\), we get \(x = 10\). Also, using \(\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{5\sqrt{3}}{y}\), and \(\tan(60^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{5\sqrt{3}}{y}\), which gives \(y = 5\).

Answer:

If we are finding \(x\), the answer is \(10\); if we are finding \(y\), the answer is \(5\). (Assuming we are asked to find \(x\) and \(y\) for the right - triangle with angle \(60^{\circ}\), right - angle, and opposite side \(5\sqrt{3}\) to \(60^{\circ}\) angle)