Question

find the values of x and y. x 60° 8√3 y x = 4, y = 4√3 x = 4√3, y = 12 x = 4√3, y = 36 x = 16√3, y = 48

Explanation:

Step1: Identify Triangle Type

This is a right - triangle with one angle \(60^{\circ}\), so the other non - right angle is \(30^{\circ}\). In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (opposite to \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\) angles respectively). Let the side opposite \(30^{\circ}\) be \(a\), opposite \(60^{\circ}\) be \(a\sqrt{3}\), and hypotenuse be \(2a\).

Step2: Analyze the Given Side

The given side \(8\sqrt{3}\) is the hypotenuse? Wait, no. Wait, the side with length \(8\sqrt{3}\): let's see the angles. The angle of \(60^{\circ}\) is adjacent to side \(x\) and opposite to side \(y\), and the hypotenuse is the side with length \(8\sqrt{3}\)? Wait, no, in the right - triangle, the hypotenuse is the side opposite the right angle. Wait, the right angle is between \(x\) and \(y\), so the hypotenuse is the side with length \(8\sqrt{3}\). Wait, no, let's re - check. The angle of \(60^{\circ}\): so \(\cos(60^{\circ})=\frac{x}{8\sqrt{3}}\) and \(\sin(60^{\circ})=\frac{y}{8\sqrt{3}}\)

We know that \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{x}{8\sqrt{3}}=\frac{1}{2}\), then \(x = \frac{8\sqrt{3}}{2}=4\sqrt{3}\)

We know that \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{y}{8\sqrt{3}}=\frac{\sqrt{3}}{2}\), then \(y=\frac{8\sqrt{3}\times\sqrt{3}}{2}=\frac{8\times3}{2}=12\)

Answer:

\(x = 4\sqrt{3},y = 12\) (corresponding to the option \(x = 4\sqrt{3},y = 12\))