Question

find the value of x for the right triangle.
(right triangle image with 30° angle, hypotenuse 12, right angle, and leg x)
x = \square

Explanation:

Step1: Identify triangle type

This is a 30 - 60 - 90 right triangle. In a 30 - 60 - 90 triangle, the side opposite the 30° angle is half the hypotenuse, and the side opposite the 60° angle (which is \(x\) here) is \(\frac{\sqrt{3}}{2}\) times the hypotenuse, or alternatively, the side opposite 30° is \(a\), hypotenuse \(2a\), and side opposite 60° is \(a\sqrt{3}\). Wait, actually, looking at the triangle, the hypotenuse is 12? Wait no, wait the angle given is 30°, and the side adjacent to 30°? Wait no, the right angle, and the angle at the top is 30°, so the side opposite the 30° angle? Wait, no, let's re - examine. The right triangle has a right angle, and one angle is 30°, so the sides: in a 30 - 60 - 90 triangle, the side opposite 30° is the shortest side, hypotenuse is twice that, and the other leg (opposite 60°) is \(\sqrt{3}\) times the shortest side. Wait, but here, the side labeled 12: is it the hypotenuse? Wait, the triangle has a right angle, and the angle at the top is 30°, so the side opposite the 30° angle would be the side opposite to that 30° angle. Wait, no, maybe I made a mistake. Wait, in the triangle, the hypotenuse is the side opposite the right angle. So if the angle at the top is 30°, then the side opposite 30° is the side \(x\)? No, wait no. Wait, let's use trigonometry. Sine of an angle in a right triangle is \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), cosine is \(\frac{\text{adjacent}}{\text{hypotenuse}}\), tangent is \(\frac{\text{opposite}}{\text{adjacent}}\).

Wait, the angle given is 30°, the hypotenuse? Wait, no, the side with length 12: is that the hypotenuse? Wait, the right angle is at the bottom left, so the hypotenuse is the side from the bottom right to the top left, which is length 12? Wait, no, the label 12 is on the side that is adjacent to the 30° angle? Wait, no, let's look again. The triangle: right angle at the bottom left, so vertices: bottom left (right angle), bottom right (\(x\) side), top left (30° angle). So the side between top left and bottom right is the hypotenuse, length 12? Wait, no, the side between top left and bottom left is one leg, between bottom left and bottom right is \(x\) (another leg), and between top left and bottom right is hypotenuse. The angle at top left is 30°, so the side opposite 30° is the side opposite to that angle, which is the side from bottom left to bottom right, i.e., \(x\)? No, wait, angle at top left is 30°, so the side opposite to 30° is the side \(x\) (bottom right to bottom left? No, bottom left to bottom right is \(x\), which is opposite the 30° angle? Wait, no, the angle at top left is 30°, so the sides: the side adjacent to 30° is the leg from top left to bottom left, and the side opposite to 30° is the leg from bottom left to bottom right (which is \(x\)), and the hypotenuse is from top left to bottom right (length 12). Wait, no, that can't be. Wait, in a right triangle, \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). If the angle is 30°, opposite side is \(x\), hypotenuse is 12, then \(\sin(30^{\circ})=\frac{x}{12}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), then \(x = 12\times\sin(30^{\circ})=12\times\frac{1}{2}=6\)? Wait, no, that's not right. Wait, no, maybe the hypotenuse is not 12. Wait, maybe the side with length 12 is the leg adjacent to the 30° angle. Wait, let's correct. Let's denote: angle \(A = 30^{\circ}\), right angle at \(C\), so side \(BC=x\) (opposite \(A\)), side \(AC\) (adjacent to \(A\)), and side \(AB\) (hypotenuse). If angle \(A = 30^{\circ}\), then \(\sin(A)=\frac{BC}{AB}\), \(\cos(A)=\frac{AC}{AB}\), \(\tan(A)=\frac{BC}{AC}\). Wait, maybe the side labeled 12 is the hypotenuse? Wait, the problem's diagram: the side with length 12 is the hypotenuse? Then, in a 30 - 60 - 90 triangle, the side opposite 30° is half the hypotenuse. Wait, no, the side opposite 30° is the shortest side. Wait, if the hypotenuse is 12, then the side opposite 30° is 6, and the side opposite 60° is \(6\sqrt{3}\). But wait, maybe I got the angle wrong. Wait, maybe the 30° angle is at the bottom right. Wait, the diagram is a bit unclear, but the standard 30 - 60 - 90 triangle: if the hypotenuse is \(c\), the side opposite 30° is \(c/2\), side opposite 60° is \((c\sqrt{3})/2\). Alternatively, if we have a right triangle with angle 30°, and the leg opposite 30° is \(a\), hypotenuse \(2a\), leg opposite 60° is \(a\sqrt{3}\). Wait, maybe the side with length 12 is the leg adjacent to 30°, and we need to find the leg opposite 30°? No, wait, let's start over.

Wait, the correct approach: in a right triangle, if one angle is 30°, then the side opposite 30° is half the hypotenuse. Wait, suppose the hypotenuse is \(H\), side opposite 30° is \(H/2\), side opposite 60° is \((H\sqrt{3})/2\). Now, looking at the triangle, the side labeled 12: if that is the hypotenuse, then the side opposite 30° is \(12/2 = 6\), but that would be the shorter leg. But maybe the side labeled 12 is the leg adjacent to the 30° angle, and we need to find the leg opposite 30° (which is \(x\)). Wait, no, let's use trigonometry. Let's assume that the angle is 30°, and the hypotenuse is 12. Then the side opposite 30° is \(x\), so \(\sin(30^{\circ})=\frac{x}{12}\), so \(x = 12\times\sin(30^{\circ})=12\times\frac{1}{2}=6\). Wait, but that seems too simple. Wait, maybe the triangle is such that the side with length 12 is the leg opposite 60°, and we need to find the leg opposite 30°. Wait, no, let's check the angles. The sum of angles in a triangle is 180°, so the other angle is 60°. So it's a 30 - 60 - 90 triangle. In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where 1 is the side opposite 30°, \(\sqrt{3}\) opposite 60°, and 2 is the hypotenuse.

Wait, maybe the side with length 12 is the hypotenuse. Then the side opposite 30° is \(12/2 = 6\), and the side opposite 60° is \(6\sqrt{3}\). But the problem is asking for \(x\), which is the side opposite 30°? Wait, no, the diagram: the right angle, and the angle at the top is 30°, so the side \(x\) is opposite the 30° angle? Then \(x = 6\). Wait, but let's confirm with trigonometry. \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{12}\), \(\sin(30^{\circ}) = 0.5\), so \(x=12\times0.5 = 6\).

Step2: Calculate \(x\)

Using the sine function in a right triangle: \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\theta = 30^{\circ}\), opposite side is \(x\), hypotenuse is 12.

We know that \(\sin(30^{\circ})=\frac{1}{2}\).

So, \(\frac{x}{12}=\frac{1}{2}\).

Multiply both sides by 12: \(x = 12\times\frac{1}{2}=6\).

Wait, but that seems too simple. Wait, maybe I misidentified the hypotenuse. Wait, maybe the side with length 12 is the leg adjacent to the 30° angle, and \(x\) is the leg opposite 60°? Wait, no, the angle is 30°, so if the adjacent leg is 12, then \(\tan(30^{\circ})=\frac{x}{12}\), but \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(x=\frac{12}{\sqrt{3}} = 4\sqrt{3}\), which is not 6. So my initial assumption about the hypotenuse must be correct.

Wait, the key is: in a right triangle with a 30° angle, the side opposite the 30° angle is half the hypotenuse. So if the hypotenuse is 12, then the side opposite 30° (which is \(x\)) is \(12\div2 = 6\).

Answer:

\(x = 6\)