Question

1. solve for x

(right triangle with 60° angle, hypotenuse 16√3, right angle, and side x opposite to 60°? or adjacent? wait, the triangle has a 60° angle, right angle, so the other angle is 30°. wait, the given side is 16√3, maybe the hypotenuse? wait, the image shows a right triangle with one angle 60°, hypotenuse 16√3, and side x (maybe the shorter leg or longer leg). wait, ocr text: 4. solve for x and the triangle with 60°, hypotenuse 16√3, right angle, and side x.

Explanation:

Step1: Identify triangle type

This is a right - triangle with one angle \(30^{\circ}\) (wait, the angle given is \(30^{\circ}\)? Wait, the angle is \(30^{\circ}\), hypotenuse? Wait, no, the side given is \(16\sqrt{3}\), and we have a right - triangle with a \(30^{\circ}\) angle. 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 side (let's call it \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Wait, the angle given is \(30^{\circ}\), the side adjacent to \(30^{\circ}\) is \(x\), and the side opposite \(30^{\circ}\) is? Wait, no, let's re - examine. The triangle has a right angle, and an angle of \(30^{\circ}\). The side labeled \(16\sqrt{3}\) is the side opposite the \(60^{\circ}\) angle? Wait, no, the angle is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is \(x\)? Wait, no, let's use trigonometry. \(\tan(30^{\circ})=\frac{x}{16\sqrt{3}}\)? Wait, no, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). If the angle is \(30^{\circ}\), the opposite side is \(x\), and the adjacent side is \(16\sqrt{3}\)? Wait, no, that can't be. Wait, maybe the angle is \(60^{\circ}\). Wait, the triangle is a right - triangle, so the other non - right angle is \(60^{\circ}\) if one is \(30^{\circ}\). Wait, let's use \(\tan(30^{\circ})=\frac{x}{16\sqrt{3}}\). We know that \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\). So \(\frac{1}{\sqrt{3}}=\frac{x}{16\sqrt{3}}\). Cross - multiply: \(x = \frac{16\sqrt{3}}{\sqrt{3}}=16\). Wait, that makes sense. Alternatively, in a \(30 - 60 - 90\) triangle, the side opposite \(30^{\circ}\) is the shortest side. If the side opposite \(60^{\circ}\) is \(16\sqrt{3}\), then the side opposite \(30^{\circ}\) (which is \(x\)) is \(\frac{16\sqrt{3}}{\sqrt{3}} = 16\), and the hypotenuse would be \(32\). So using trigonometry: \(\tan(30^{\circ})=\frac{x}{16\sqrt{3}}\), so \(x = 16\sqrt{3}\times\tan(30^{\circ})\). Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), then \(x = 16\sqrt{3}\times\frac{1}{\sqrt{3}} = 16\).

Step2: Solve for x

Using the tangent function in the right - triangle. For the angle of \(30^{\circ}\), \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\). Here, the opposite side to \(30^{\circ}\) is \(x\), and the adjacent side is \(16\sqrt{3}\). So \(\tan(30^{\circ})=\frac{x}{16\sqrt{3}}\). We know that \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\). Substituting this value in: \(\frac{1}{\sqrt{3}}=\frac{x}{16\sqrt{3}}\). Multiply both sides by \(16\sqrt{3}\): \(x=\frac{16\sqrt{3}}{\sqrt{3}} = 16\).

Answer:

\(x = 16\)