Question

1. a right - angled triangle with a 60° angle, hypotenuse (8sqrt{5}), one leg labeled (y) (adjacent to the 60° angle), another leg labeled (x) (opposite to the 60° angle), and a right angle (the right angle is between legs (x) and (y)).

Explanation:

This is a right - triangle problem. In a right - triangle, we can use trigonometric ratios (sine, cosine) and the Pythagorean theorem. Given that one of the non - right angles is \(60^{\circ}\), the hypotenuse \(c = 8\sqrt{5}\), and we need to find the lengths of the legs \(x\) and \(y\).

Step 1: Recall the trigonometric ratios for a \(60^{\circ}\) angle in a right - triangle

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), where \(\theta\) is one of the non - right angles. Also, for a \(60^{\circ}\) angle, \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\) and \(\cos60^{\circ}=\frac{1}{2}\)

Let's first find the length of \(x\) (opposite to the \(60^{\circ}\) angle) and \(y\) (adjacent to the \(60^{\circ}\) angle)

For \(x\):
We know that \(\sin60^{\circ}=\frac{x}{8\sqrt{5}}\)
So, \(x = 8\sqrt{5}\times\sin60^{\circ}\)
Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), then \(x = 8\sqrt{5}\times\frac{\sqrt{3}}{2}=4\sqrt{15}\)

For \(y\):
We know that \(\cos60^{\circ}=\frac{y}{8\sqrt{5}}\)
Since \(\cos60^{\circ}=\frac{1}{2}\), then \(y = 8\sqrt{5}\times\frac{1}{2}=4\sqrt{5}\)

We can also verify using the Pythagorean theorem \(x^{2}+y^{2}=c^{2}\)

Left - hand side: \((4\sqrt{15})^{2}+(4\sqrt{5})^{2}=16\times15 + 16\times5=240 + 80 = 320\)

Right - hand side: \((8\sqrt{5})^{2}=64\times5 = 320\)

Answer:

If we want to find \(x\), \(x = 4\sqrt{15}\); if we want to find \(y\), \(y = 4\sqrt{5}\)