Question

1. solve for x triangle with right angle, angle 60°, side x, hypotenuse 2√2

Explanation:

Step1: Identify the triangle type

The triangle is a right - triangle with one angle of \(60^{\circ}\), so it is a \(30 - 60 - 90\) triangle? Wait, no, let's check the sides. Wait, the hypotenuse? Wait, no, the side adjacent to the \(60^{\circ}\) angle? Wait, no, let's use trigonometry. In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\) and \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Wait, the angle given is \(60^{\circ}\), the side adjacent to \(60^{\circ}\) is \(x\), and the hypotenuse? Wait, no, the side labeled \(2\sqrt{2}\) is the hypotenuse? Wait, no, in the right - triangle, the right angle is at the top - left, the angle at the bottom - left is \(60^{\circ}\), so the side opposite to \(60^{\circ}\) is the top - right side, the side adjacent to \(60^{\circ}\) is \(x\), and the hypotenuse is the side opposite the right angle, which is the side with length \(2\sqrt{2}\)? Wait, no, let's correct. The right angle is between the side of length \(x\) and the horizontal side. The angle at the bottom is \(60^{\circ}\), so the side adjacent to \(60^{\circ}\) is \(x\), and the hypotenuse is the side with length \(2\sqrt{2}\). Wait, in a right - triangle, \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). We know that \(\cos(60^{\circ})=\frac{1}{2}\). So \(\cos(60^{\circ})=\frac{x}{2\sqrt{2}}\)? Wait, no, that can't be. Wait, maybe I mixed up adjacent and opposite. Wait, the angle is \(60^{\circ}\), the side opposite to \(30^{\circ}\) would be half the hypotenuse, but here the angle is \(60^{\circ}\). Wait, let's re - examine. The right - triangle has angles \(90^{\circ}\), \(60^{\circ}\), so the third angle is \(30^{\circ}\). Wait, no, \(90 + 60+\alpha=180\), so \(\alpha = 30^{\circ}\). So the side opposite \(30^{\circ}\) is the shortest side. Wait, maybe the side with length \(2\sqrt{2}\) is not the hypotenuse. Wait, no, the hypotenuse is always opposite the right angle. So the hypotenuse is the side opposite the right angle, so the side labeled \(2\sqrt{2}\) is the hypotenuse. Then, the side adjacent to \(60^{\circ}\) is \(x\), and we can use \(\cos(60^{\circ})=\frac{x}{\text{hypotenuse}}\). Since \(\cos(60^{\circ})=\frac{1}{2}\), then \(\frac{1}{2}=\frac{x}{2\sqrt{2}}\), so \(x = \sqrt{2}\)? Wait, no, that seems wrong. Wait, maybe the angle is \(45^{\circ}\)? Wait, the triangle looks like a \(45 - 45 - 90\) triangle? Wait, no, the angle is labeled \(60^{\circ}\). Wait, maybe the user made a typo, or I misread the angle. Wait, if it's a \(45 - 45 - 90\) triangle, then the legs are equal, and hypotenuse \(=x\sqrt{2}\). But here the hypotenuse is \(2\sqrt{2}\), so \(x\sqrt{2}=2\sqrt{2}\), so \(x = 2\). But the angle is labeled \(60^{\circ}\). Wait, maybe the angle is \(45^{\circ}\). Alternatively, maybe the side \(2\sqrt{2}\) is not the hypotenuse. Wait, let's start over.

In a right - triangle, let's denote the right angle as \(C\), angle \(A = 60^{\circ}\), angle \(B=30^{\circ}\). Side \(a\) is opposite angle \(A\), side \(b\) opposite angle \(B\), side \(c\) (hypotenuse) opposite angle \(C\).

We have side \(c = 2\sqrt{2}\), angle \(A = 60^{\circ}\), and side \(b=x\) (opposite angle \(B = 30^{\circ}\)). Wait, in a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is the shortest one (length \(s\)), opposite \(60^{\circ}\) is \(s\sqrt{3}\), hypotenuse is \(2s\).

If angle \(B = 30^{\circ}\), then side \(b=x\) (opposite \(30^{\circ}\)), side \(a\) (opposite \(60^{\circ}\)) is \(x\sqrt{3}\), hypotenuse \(c = 2x\).

We are given that \(c = 2\sqrt{2}\), so \(2x=2\sqrt{2}\), then \(x=\sqrt{2}\). Wait, that makes sense. Because if hypotenuse \(c = 2x\), and \(c = 2\sqrt{2}\), then \(x=\sqrt{2}\). Let's verify with trigonometry. \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(60^{\circ}\) (angle \(A\)) is \(x\) (wait, no, adjacent to \(60^{\circ}\) is side \(b\)? No, adjacent to angle \(A = 60^{\circ}\) is side \(b\) (since angle \(A\) is at vertex \(A\), side \(b\) is between angle \(A\) and angle \(C\)). Wait, maybe my initial ratio was wrong. Let's use \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). Adjacent to \(60^{\circ}\) is \(x\), hypotenuse is \(2\sqrt{2}\). \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{x}{2\sqrt{2}}=\frac{1}{2}\), cross - multiply: \(2x = 2\sqrt{2}\), so \(x=\sqrt{2}\). Yes, that's correct.

Step2: Solve for \(x\)

From the trigonometric relationship \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), where adjacent side to \(60^{\circ}\) is \(x\) and hypotenuse is \(2\sqrt{2}\).

We know that \(\cos(60^{\circ})=\frac{1}{2}\), so:

\[ \frac{x}{2\sqrt{2}}=\frac{1}{2} \]

Multiply both sides by \(2\sqrt{2}\):

\[ x=\frac{1}{2}\times2\sqrt{2}=\sqrt{2} \]

Answer:

\(\sqrt{2}\)