Question

find x.

(image of a right triangle with angles 60°, 30°, and right angle, one leg is \\(\sqrt{3}\\) km, side x is another side. )

write your answer in simplest radical form.

\\(\square\\) kilometers

Explanation:

Step1: Identify the triangle type

This is a right - angled triangle with angles \(30^{\circ}\), \(60^{\circ}\) and \(90^{\circ}\). 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 \(2a\).

We are given the side adjacent to the \(30^{\circ}\) angle (and opposite to the \(60^{\circ}\) angle) as \(\sqrt{3}\) km. Let the side opposite \(30^{\circ}\) be \(x\) (wait, no, wait. Wait, the side with length \(\sqrt{3}\) is adjacent to \(30^{\circ}\) and opposite to \(60^{\circ}\). Let's use trigonometric ratios. Let's consider the angle \(30^{\circ}\). The tangent of an angle in a right - triangle is \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\theta = 30^{\circ}\), \(\tan(30^{\circ})=\frac{x}{\sqrt{3}}\), but \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\). Wait, or we can use the ratio of sides in \(30 - 60 - 90\) triangle.

Wait, the side opposite \(30^{\circ}\) is \(x\), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). We are given that the side opposite \(60^{\circ}\) is \(\sqrt{3}\) km. So if \(x\sqrt{3}=\sqrt{3}\), then \(x = 1\)? No, wait, no. Wait, maybe I mixed up the angles. Wait, the right angle is at the top, the angle at the top - right is \(60^{\circ}\), the angle at the bottom is \(30^{\circ}\). So the side adjacent to \(30^{\circ}\) is \(\sqrt{3}\) km, and the side opposite to \(30^{\circ}\) is \(x\) km.

We know that \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{\sqrt{3}}\). Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), we have \(\frac{1}{\sqrt{3}}=\frac{x}{\sqrt{3}}\), so \(x = 1\)? No, that can't be. Wait, maybe using cosine. \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), but we can also use the ratio of sides. Wait, another approach: in a \(30 - 60 - 90\) triangle, if the side opposite \(30^{\circ}\) is \(a\), side opposite \(60^{\circ}\) is \(a\sqrt{3}\), hypotenuse \(2a\).

Looking at the triangle, the side with length \(\sqrt{3}\) is opposite the \(60^{\circ}\) angle. So if the side opposite \(60^{\circ}\) is \(a\sqrt{3}=\sqrt{3}\), then \(a = 1\). But the side we are looking for \(x\) is opposite the \(30^{\circ}\) angle, so \(x=a = 1\)? No, that seems wrong. Wait, no, wait the side with \(\sqrt{3}\) is adjacent to the \(30^{\circ}\) angle. Wait, let's label the triangle: let the right - angle vertex be \(A\), the vertex with \(60^{\circ}\) be \(B\), and the vertex with \(30^{\circ}\) be \(C\). So \(AB\) is one leg, \(AC\) is the other leg (\(\sqrt{3}\) km), and \(BC\) is the hypotenuse. Angle at \(C\) is \(30^{\circ}\), angle at \(B\) is \(60^{\circ}\), angle at \(A\) is \(90^{\circ}\).

So, \(\tan(\angle C)=\tan(30^{\circ})=\frac{AB}{AC}\). We know \(AC = \sqrt{3}\) km, \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(\frac{1}{\sqrt{3}}=\frac{AB}{\sqrt{3}}\), so \(AB = 1\) km? Wait, but maybe we can use sine. \(\sin(30^{\circ})=\frac{AB}{BC}\), and \(\cos(30^{\circ})=\frac{AC}{BC}\). Since \(AC=\sqrt{3}\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{BC}\), so \(BC = 2\) km. But we need to find \(x\) (assuming \(x\) is \(AB\)). Wait, maybe the problem is that the side with length \(\sqrt{3}\) is the adjacent side to \(30^{\circ}\), and we need to find the opposite side.

Wait, let's use trigonometric ratio: \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For \(\theta = 30^{\circ}\), \(\tan(30^{\circ})=\frac{x}{\sqrt{3}}\). Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), then \(\frac{1}{\sqrt{3}}=\frac{x}{\sqrt{3}}\), so \(x = 1\)? No, that seems too simple. Wait, maybe I made a mistake in identifying the sides. Wait, the triangle has a right angle, a \(30^{\circ}\) angle, and a \(60^{\circ}\) angle. The side labeled \(\sqrt{3}\) is one of the legs. Let's consider the \(30^{\circ}\) angle: the side opposite to \(30^{\circ}\) is the shortest side. Let's call the shortest side (opposite \(30^{\circ}\)) as \(x\), the side opposite \(60^{\circ}\) as \(x\sqrt{3}\), and the hypotenuse as \(2x\).

We are given that the side opposite \(60^{\circ}\) is \(\sqrt{3}\) km. So \(x\sqrt{3}=\sqrt{3}\), then \(x = 1\) km. Wait, that makes sense. Because if \(x = 1\), then the side opposite \(60^{\circ}\) is \(1\times\sqrt{3}=\sqrt{3}\), which matches the given length. So the side opposite \(30^{\circ}\) (which is \(x\)) is \(1\) km? But that seems too easy. Wait, maybe the problem is that the side with \(\sqrt{3}\) is adjacent to \(30^{\circ}\), and we need to find the opposite side. Let's check with cosine: \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). If adjacent is \(\sqrt{3}\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), then hypotenuse \(=\frac{\text{adjacent}}{\cos(30^{\circ})}=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2\). Then \(\sin(30^{\circ})=\frac{\text{opposite}}{2}\), and \(\sin(30^{\circ})=\frac{1}{2}\), so opposite \(= 1\). So yes, \(x = 1\) km? Wait, but the problem says "Write your answer in simplest radical form". But \(1\) is an integer, which is in simplest radical form (since \(\sqrt{1}=1\)).

Wait, maybe I misread the triangle. Wait, maybe the side with \(\sqrt{3}\) is the hypotenuse? No, the right angle is marked, so the hypotenuse is the side opposite the right angle. Wait, the right angle is at the top, so the hypotenuse is the side connecting the \(60^{\circ}\) angle and the \(30^{\circ}\) angle. The side with \(\sqrt{3}\) is one of the legs. Let's re - examine the triangle:

- Right angle at the top.
- Angle at the top - right: \(60^{\circ}\)
- Angle at the bottom: \(30^{\circ}\)
- The leg from the right - angle to the bottom vertex (the one with \(30^{\circ}\)) has length \(\sqrt{3}\) km.
- We need to find the leg from the right - angle to the \(60^{\circ}\) vertex (let's call this length \(x\)).

So, in this case, for the angle of \(30^{\circ}\) (at the bottom), the adjacent side is \(\sqrt{3}\) km, and the opposite side is \(x\) km.

We know that \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{\sqrt{3}}\)

Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), we have:

\(\frac{1}{\sqrt{3}}=\frac{x}{\sqrt{3}}\)

Multiply both sides by \(\sqrt{3}\):

\(x = 1\)

Step2: Confirm the result

We can also use the ratio of sides in a \(30 - 60 - 90\) triangle. The sides are in the ratio \(1:\sqrt{3}:2\) (opposite \(30^{\circ}\), opposite \(60^{\circ}\), hypotenuse). The side opposite \(60^{\circ}\) is \(\sqrt{3}\) times the side opposite \(30^{\circ}\). We found that the side opposite \(30^{\circ}\) (which is \(x\)) is \(1\) km, and the side opposite \(60^{\circ}\) is \(\sqrt{3}\times1=\sqrt{3}\) km, which matches the given length. So our calculation is correct.

Answer:

\(1\)