Question

find v.

6√2 mi

write your answer in simplest radical form.

miles

Explanation:

Step1: Identify the triangle type

This is a 30 - 60 - 90 right triangle. 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 (let's call it \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Here, the hypotenuse is \(6\sqrt{2}\) mi, and we need to find the side \(v\) which is opposite the \(30^{\circ}\) angle? Wait, no, wait. Wait, the angle at the bottom is \(60^{\circ}\), the right angle is at the bottom - right, so the side \(v\) is adjacent to the \(60^{\circ}\) angle and opposite to the \(30^{\circ}\) angle? Wait, no, let's label the triangle. Let's call the right - angled vertex \(C\), the \(60^{\circ}\) vertex \(A\), and the \(30^{\circ}\) vertex \(B\). So side \(AB\) is the hypotenuse (\(6\sqrt{2}\) mi), side \(AC = v\), side \(BC\) is the other leg. In triangle \(ABC\), angle at \(A\) is \(60^{\circ}\), angle at \(B\) is \(30^{\circ}\), angle at \(C\) is \(90^{\circ}\). So the side opposite \(30^{\circ}\) (angle \(B\)) is \(AC=v\), and the hypotenuse \(AB = 6\sqrt{2}\). In a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is half of the hypotenuse? Wait, no, wait: in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is the shortest side, and it is equal to \(\frac{1}{2}\) of the hypotenuse. Wait, no, let's recall the ratios. Let the side opposite \(30^{\circ}\) be \(x\), then the hypotenuse is \(2x\), and the side opposite \(60^{\circ}\) is \(x\sqrt{3}\). So here, angle \(B\) is \(30^{\circ}\), so the side opposite angle \(B\) (which is \(AC = v\)) should be \(x\), and the hypotenuse \(AB=2x\). We know that \(AB = 6\sqrt{2}\), so \(2x=6\sqrt{2}\), then \(x = 3\sqrt{2}\)? Wait, that can't be right. Wait, maybe I mixed up the angles. Wait, the angle at the bottom is \(60^{\circ}\), so the side \(v\) is adjacent to the \(60^{\circ}\) angle. Wait, let's use trigonometric ratios. \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{v}{6\sqrt{2}}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), then \(\frac{v}{6\sqrt{2}}=\frac{1}{2}\), so \(v=\frac{6\sqrt{2}}{2}=3\sqrt{2}\)? Wait, no, that seems too simple. Wait, no, wait, maybe I got the angle wrong. Wait, the angle inside the triangle at the top is \(30^{\circ}\), the angle at the bottom is \(60^{\circ}\), so the side \(v\) is adjacent to the \(60^{\circ}\) angle and opposite to the \(30^{\circ}\) angle. Wait, let's use cosine. \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{v}{6\sqrt{2}}\). Since \(\cos(60^{\circ})=\frac{1}{2}\), then \(v = 6\sqrt{2}\times\frac{1}{2}=3\sqrt{2}\)? Wait, no, that can't be. Wait, maybe I made a mistake in identifying the angle. Wait, let's look at the triangle again. The hypotenuse is \(6\sqrt{2}\), the angle at the top is \(30^{\circ}\), the angle at the bottom is \(60^{\circ}\), right - angled at the bottom - right. So the side \(v\) is the adjacent side to the \(60^{\circ}\) angle. Wait, \(\cos(60^{\circ})=\frac{v}{\text{hypotenuse}}\), \(\cos(60^{\circ}) = 0.5\), so \(v=6\sqrt{2}\times0.5 = 3\sqrt{2}\)? Wait, but let's check with sine. \(\sin(30^{\circ})=\frac{v}{\text{hypotenuse}}\), \(\sin(30^{\circ}) = 0.5\), so \(v = 6\sqrt{2}\times0.5=3\sqrt{2}\). Alternatively, maybe the side \(v\) is opposite the \(60^{\circ}\) angle? Wait, no, the right angle is at the bottom - right, so the two legs: one is horizontal (\(v\)) and one is vertical. The hypotenuse is the slant side. The angle at the bottom (left) is \(60^{\circ}\), so the horizontal leg \(v\) is adjacent to the \(60^{\circ}\) angle, and the vertical leg is opposite to the \(60^{\circ}\) angle. Wait, I think I messed up the angle - side correspondence. Let's start over. In a right - triangle, \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). Let's take the \(30^{\circ}\) angle. The \(30^{\circ}\) angle: the opposite side is \(v\) (the horizontal leg), and the hypotenuse is \(6\sqrt{2}\). So \(\sin(30^{\circ})=\frac{v}{6\sqrt{2}}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), then \(v = 6\sqrt{2}\times\frac{1}{2}=3\sqrt{2}\). Wait, but let's check with the \(60^{\circ}\) angle. The \(60^{\circ}\) angle: the adjacent side is \(v\), and the hypotenuse is \(6\sqrt{2}\). So \(\cos(60^{\circ})=\frac{v}{6\sqrt{2}}\), and \(\cos(60^{\circ})=\frac{1}{2}\), so \(v = 3\sqrt{2}\). Alternatively, if we consider the side opposite \(60^{\circ}\), it would be \(\sin(60^{\circ})=\frac{\text{opposite}}{6\sqrt{2}}\), \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), but that's the other leg. So in this case, since \(v\) is opposite the \(30^{\circ}\) angle, we use \(\sin(30^{\circ})\) to find \(v\).

Step2: Calculate \(v\)

We know that in a right - triangle, \(\sin(\alpha)=\frac{\text{opposite}}{\text{hypotenuse}}\), where \(\alpha = 30^{\circ}\), the opposite side to \(\alpha\) is \(v\), and the hypotenuse is \(6\sqrt{2}\) mi.
Since \(\sin(30^{\circ})=\frac{1}{2}\), we have the equation:
\(\frac{v}{6\sqrt{2}}=\frac{1}{2}\)
To solve for \(v\), we multiply both sides of the equation by \(6\sqrt{2}\):
\(v=\frac{1}{2}\times6\sqrt{2}\)
\(v = 3\sqrt{2}\)

Answer:

\(3\sqrt{2}\)