Question

find u.
right triangle with right angle, 60° angle, 30° angle, side labeled 6√2 km, side labeled u
write your answer in simplest radical form.
blank kilometers

Explanation:

Step1: Identify triangle type

It's a right - angled triangle with angles \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\). We can use trigonometric ratios. Let's consider the angle of \(30^{\circ}\). The side opposite to \(30^{\circ}\) is related to the hypotenuse, and the other sides can be found using sine and cosine. Alternatively, we can use the fact that in a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\) and \(\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}\). Let's take the angle of \(30^{\circ}\), the side adjacent to \(30^{\circ}\) is \(u\), and the hypotenuse is \(6\sqrt{2}\) km? Wait, no. Wait, the right angle is at the top. Let's re - label: the right angle is between \(u\) and the side opposite to \(60^{\circ}\). Wait, let's use the angle of \(60^{\circ}\). Wait, the angle of \(30^{\circ}\): the side adjacent to \(30^{\circ}\) is \(u\), and the side opposite to \(30^{\circ}\) is, let's say, \(x\), and the hypotenuse is the side of length \(6\sqrt{2}\)? Wait, no. Wait, in a right - triangle, the sum of angles is \(180^{\circ}\), so the right angle is \(90^{\circ}\), one angle is \(30^{\circ}\), another is \(60^{\circ}\). Let's use the cosine of \(30^{\circ}\). The cosine of an angle in a right - triangle is \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, if the angle is \(30^{\circ}\), the adjacent side to \(30^{\circ}\) is \(u\), and the hypotenuse is the side with length \(6\sqrt{2}\) km? Wait, no, maybe I got the sides wrong. Wait, let's look at the triangle: the right angle is at the top left. So the sides: one leg is \(u\) (vertical leg), the other leg is horizontal (let's call it \(v\)), and the hypotenuse is the side with length \(6\sqrt{2}\) km. The angle at the bottom is \(30^{\circ}\), so the angle between the hypotenuse and the vertical leg (\(u\)) is \(30^{\circ}\). So, \(\cos(30^{\circ})=\frac{u}{\text{hypotenuse}}\). The hypotenuse is \(6\sqrt{2}\) km. \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\)? Wait, no, wait, maybe it's the sine of \(60^{\circ}\). Wait, let's use the angle of \(60^{\circ}\). The angle at the top right is \(60^{\circ}\), so the angle between the hypotenuse and the horizontal leg is \(60^{\circ}\), and the vertical leg \(u\) is opposite to \(60^{\circ}\). So \(\sin(60^{\circ})=\frac{u}{\text{hypotenuse}}\). \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), but hypotenuse is \(6\sqrt{2}\)? Wait, no, that can't be. Wait, maybe I made a mistake. 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 (length \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Let's check the angles: the angle at the bottom is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the horizontal leg (let's say \(v\)), the side opposite \(60^{\circ}\) is \(u\), and the hypotenuse is \(6\sqrt{2}\). So, hypotenuse \( = 2x\), where \(x\) is the side opposite \(30^{\circ}\). Then \(u=x\sqrt{3}\), and hypotenuse \( = 2x=6\sqrt{2}\), so \(x = 3\sqrt{2}\), then \(u=x\sqrt{3}=3\sqrt{2}\times\sqrt{3}=3\sqrt{6}\)? Wait, no, that doesn't seem right. Wait, maybe I should use the cosine of \(30^{\circ}\). Wait, \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), adjacent is \(u\), hypotenuse is \(6\sqrt{2}\). \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(u = 6\sqrt{2}\times\cos(30^{\circ})\). \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(u=6\sqrt{2}\times\frac{\sqrt{3}}{2}=3\sqrt{6}\)? Wait, no, maybe I mixed up sine and cosine. Wait, the angle at the bottom is \(30^{\circ}\), so the angle between the hypotenuse and the vertical leg (\(u\)) is \(30^{\circ}\), so the vertical leg \(u\) is adjacent to the \(30^{\circ}\) angle, and the horizontal leg is opposite to the \(30^{\circ}\) angle. So \(\cos(30^{\circ})=\frac{u}{\text{hypotenuse}}\), so \(u=\text{hypotenuse}\times\cos(30^{\circ})\). Hypotenuse is \(6\sqrt{2}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(u = 6\sqrt{2}\times\frac{\sqrt{3}}{2}=3\sqrt{6}\)? Wait, but let's check with sine of \(60^{\circ}\). The angle at the top right is \(60^{\circ}\), the vertical leg \(u\) is opposite to the \(60^{\circ}\) angle. So \(\sin(60^{\circ})=\frac{u}{\text{hypotenuse}}\). \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(u = 6\sqrt{2}\times\frac{\sqrt{3}}{2}=3\sqrt{6}\). Wait, but maybe there's a mistake in the hypotenuse. Wait, no, the side labeled \(6\sqrt{2}\) is the hypotenuse? Wait, maybe the triangle is not a \(30 - 60 - 90\) triangle? Wait, no, the angles are \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\), so it is a \(30 - 60 - 90\) triangle. Wait, but let's re - examine the triangle. The right angle is at the top, so the two legs are \(u\) (vertical) and the horizontal leg, and the hypotenuse is the side connecting the bottom vertex to the top right vertex, with length \(6\sqrt{2}\). The angle at the bottom is \(30^{\circ}\), so the angle between the hypotenuse and the vertical leg is \(30^{\circ}\). So, using cosine: \(\cos(30^{\circ})=\frac{u}{\text{hypotenuse}}\), so \(u = 6\sqrt{2}\times\cos(30^{\circ})\). \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(u=6\sqrt{2}\times\frac{\sqrt{3}}{2}=3\sqrt{6}\). Wait, but let's check with another approach. Let's use the fact that in a right - triangle, \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\). The opposite side to \(30^{\circ}\) is the horizontal leg (let's say \(x\)), and the adjacent side is \(u\). So \(\tan(30^{\circ})=\frac{x}{u}\), and \(x^{2}+u^{2}=(6\sqrt{2})^{2}\). \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(x=\frac{u}{\sqrt{3}}\). Then \(x^{2}+u^{2}=72\) (since \((6\sqrt{2})^{2}=36\times2 = 72\)). Substitute \(x\): \(\frac{u^{2}}{3}+u^{2}=72\), \(\frac{u^{2}+3u^{2}}{3}=72\), \(\frac{4u^{2}}{3}=72\), \(4u^{2}=216\), \(u^{2}=54\), \(u = \sqrt{54}=3\sqrt{6}\). Yes, that matches. So the value of \(u\) is \(3\sqrt{6}\) kilometers.

Step1: Identify the triangle type

The triangle is a right - angled triangle with angles \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). Let the hypotenuse be \(c = 6\sqrt{2}\) km, the side adjacent to \(30^{\circ}\) be \(u\), and the side opposite to \(30^{\circ}\) be \(x\).

Step2: Use the Pythagorean theorem and trigonometric ratios (or direct trigonometric ratio)

We can use the cosine of \(30^{\circ}\): \(\cos(30^{\circ})=\frac{u}{c}\), where \(c = 6\sqrt{2}\) and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\).
So, \(u=c\times\cos(30^{\circ})=6\sqrt{2}\times\frac{\sqrt{3}}{2}\)
Simplify the expression: \(6\sqrt{2}\times\frac{\sqrt{3}}{2}=3\sqrt{2\times3}=3\sqrt{6}\)

Answer:

\(3\sqrt{6}\)