Question

find q.
right triangle with angles 30°, 60°, right angle; hypotenuse 6√3 mm, side q opposite 30°? wait, no, the right angle is at top left, angle at bottom left is 30°, angle at top right is 60°, hypotenuse is 6√3 mm, side q is top horizontal side. write your answer in simplest radical form.
blank millimeters

Explanation:

Step1: Identify 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 side (let's call it \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\).

Step2: Determine which side is given

The hypotenuse is given as \(6\sqrt{3}\) mm? Wait, no. Wait, the angle of \(30^{\circ}\), the side opposite \(30^{\circ}\) is \(q\)? Wait, no. Wait, let's label the triangle. The right angle, one angle \(30^{\circ}\), one angle \(60^{\circ}\). So the side opposite \(30^{\circ}\) is the shorter leg, opposite \(60^{\circ}\) is the longer leg, and hypotenuse is opposite the right angle. Wait, the side labeled \(q\) is adjacent to \(60^{\circ}\) and opposite to \(30^{\circ}\). The hypotenuse is \(6\sqrt{3}\) mm? Wait, no, wait the side with length \(6\sqrt{3}\) is the hypotenuse? Wait, no, in a 30 - 60 - 90 triangle, the hypotenuse is the longest side. Wait, let's use trigonometry. Let's take the angle \(60^{\circ}\). The cosine of \(60^{\circ}\) is adjacent over hypotenuse. Wait, no, \(q\) is adjacent to \(60^{\circ}\), and the hypotenuse is \(6\sqrt{3}\)? Wait, no, maybe the side with \(6\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)). Wait, let's recall: in 30 - 60 - 90 triangle, if the shorter leg (opposite \(30^{\circ}\)) is \(x\), then longer leg (opposite \(60^{\circ}\)) is \(x\sqrt{3}\), hypotenuse is \(2x\).

Wait, the angle of \(30^{\circ}\): the side opposite \(30^{\circ}\) is \(q\), the side opposite \(60^{\circ}\) is, let's say, \(a\), and hypotenuse \(h\). Then \(q=\frac{h}{2}\), and \(a = q\sqrt{3}\). Wait, but the given side is \(6\sqrt{3}\). Let's see, if the hypotenuse is \(h\), and we have angle \(60^{\circ}\), then \(\cos(60^{\circ})=\frac{q}{h}\), and \(\sin(60^{\circ})=\frac{\text{opposite}}{h}\). Wait, maybe better to use the ratio. Let's assume that the side opposite \(60^{\circ}\) is \(6\sqrt{3}\)? No, wait the side labeled \(q\) is adjacent to \(60^{\circ}\), so let's use cosine of \(60^{\circ}\). \(\cos(60^{\circ})=\frac{q}{\text{hypotenuse}}\), but we need to find which side is which. Wait, alternatively, since it's a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is \(q\), and the side opposite \(60^{\circ}\) is \(q\sqrt{3}\), and hypotenuse is \(2q\). Wait, but the given side is \(6\sqrt{3}\). If the side opposite \(60^{\circ}\) is \(6\sqrt{3}\), then \(q\sqrt{3}=6\sqrt{3}\), so \(q = 6\)? No, that can't be. Wait, no, maybe the hypotenuse is \(6\sqrt{3}\). Then \(q\) is the side opposite \(30^{\circ}\), so \(q=\frac{\text{hypotenuse}}{2}=\frac{6\sqrt{3}}{2}=3\sqrt{3}\)? Wait, no, that doesn't seem right. Wait, let's start over.

Let's label the triangle: right angle at top, \(30^{\circ}\) at bottom, \(60^{\circ}\) at right. So the sides: the vertical side (left) is opposite \(60^{\circ}\), the horizontal side (top, \(q\)) is opposite \(30^{\circ}\), and the hypotenuse (the slant side) is \(6\sqrt{3}\) mm. So in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) (which is \(q\)) is half the hypotenuse? Wait, no, hypotenuse is opposite the right angle. Wait, the side opposite \(30^{\circ}\) is the shorter leg, so length \(x\), hypotenuse is \(2x\), longer leg (opposite \(60^{\circ}\)) is \(x\sqrt{3}\). So if the hypotenuse is \(6\sqrt{3}\), then \(2x = 6\sqrt{3}\), so \(x = 3\sqrt{3}\). But that would be the shorter leg (opposite \(30^{\circ}\)), which is \(q\). Wait, but let's check with trigonometry. Let's take angle \(30^{\circ}\): \(\sin(30^{\circ})=\frac{q}{\text{hypotenuse}}\). \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{q}{6\sqrt{3}}\), then \(q=\frac{6\sqrt{3}}{2}=3\sqrt{3}\). Wait, but that seems conflicting. Wait, maybe the side with \(6\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)). Then, longer leg \(=x\sqrt{3}\), where \(x\) is the shorter leg ( \(q\) ). So \(x\sqrt{3}=6\sqrt{3}\), so \(x = 6\)? Wait, that can't be. Wait, I think I made a mistake in identifying the sides. Let's use cosine of \(60^{\circ}\). The angle \(60^{\circ}\): adjacent side is \(q\), hypotenuse is \(h\). \(\cos(60^{\circ})=\frac{q}{h}\), and \(\cos(60^{\circ}) = 0.5\). Also, the side opposite \(60^{\circ}\) is \(h\sin(60^{\circ})=h\frac{\sqrt{3}}{2}\). But we need to know which side is given. Wait, the problem says "Find \(q\)", and the triangle has a right angle, \(30^{\circ}\), \(60^{\circ}\), and one side is \(6\sqrt{3}\) mm. Let's assume that the side with \(6\sqrt{3}\) is the hypotenuse. Then, \(q\) is the side adjacent to \(60^{\circ}\), so using \(\cos(60^{\circ})=\frac{q}{6\sqrt{3}}\), so \(q = 6\sqrt{3}\times\cos(60^{\circ})=6\sqrt{3}\times\frac{1}{2}=3\sqrt{3}\). Wait, but that seems small. Alternatively, if the side with \(6\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)), then longer leg \(=x\sqrt{3}\), where \(x\) is the shorter leg ( \(q\) ). So \(x\sqrt{3}=6\sqrt{3}\), so \(x = 6\). Wait, now I'm confused. Wait, let's look at the angles. The angle of \(30^{\circ}\): the side opposite \(30^{\circ}\) is the shortest side. So if \(q\) is opposite \(30^{\circ}\), then it's the shortest side. The side opposite \(60^{\circ}\) is longer, and hypotenuse is longest. So if the hypotenuse is \(6\sqrt{3}\), then \(q=\frac{\text{hypotenuse}}{2}=\frac{6\sqrt{3}}{2}=3\sqrt{3}\). But if the side opposite \(60^{\circ}\) is \(6\sqrt{3}\), then \(q=\frac{6\sqrt{3}}{\sqrt{3}} = 6\). Wait, which is it?

Wait, let's use the angle \(30^{\circ}\). The sine of \(30^{\circ}\) is opposite over hypotenuse. So \(\sin(30^{\circ})=\frac{q}{h}\), where \(h\) is hypotenuse. \(\sin(30^{\circ})=\frac{1}{2}\), so \(q=\frac{h}{2}\). The cosine of \(30^{\circ}\) is adjacent over hypotenuse, so adjacent (opposite \(60^{\circ}\)) is \(h\frac{\sqrt{3}}{2}\). Now, if the adjacent side (opposite \(60^{\circ}\)) is \(6\sqrt{3}\), then \(h\frac{\sqrt{3}}{2}=6\sqrt{3}\). Solving for \(h\): multiply both sides by \(\frac{2}{\sqrt{3}}\), \(h = 6\sqrt{3}\times\frac{2}{\sqrt{3}}=12\). Then \(q=\frac{h}{2}=\frac{12}{2}=6\). Wait, that makes sense. Wait, I think I misidentified the given side. The side with \(6\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)), so adjacent to \(30^{\circ}\). So then hypotenuse \(h = \frac{\text{longer leg}}{\sin(60^{\circ})}=\frac{6\sqrt{3}}{\frac{\sqrt{3}}{2}}=12\). Then the shorter leg (opposite \(30^{\circ}\), which is \(q\)) is \(\frac{h}{2}=\frac{12}{2}=6\). Wait, but earlier I thought \(q\) was adjacent to \(60^{\circ}\), which is correct, and adjacent to \(60^{\circ}\) is the shorter leg (opposite \(30^{\circ}\)). So yes, if the longer leg (opposite \(60^{\circ}\)) is \(6\sqrt{3}\), then hypotenuse is \(12\), and shorter leg \(q = 6\). Wait, but let's check with trigonometry using angle \(60^{\circ}\). \(\cos(60^{\circ})=\frac{q}{h}\), \(h = 12\), so \(\cos(60^{\circ})=\frac{q}{12}\), \(q = 12\times0.5 = 6\). Yes, that works. So the mistake earlier was misidentifying which side is given. The given side \(6\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)), so we find hypotenuse first, then shorter leg \(q\).

Step3: Calculate \(q\)

1. Let the longer leg (opposite \(60^{\circ}\)) be \(L = 6\sqrt{3}\) mm.
2. In a 30 - 60 - 90 triangle, \(L=x\sqrt{3}\), where \(x\) is the shorter leg ( \(q\) ), and hypotenuse \(h = 2x\).
3. From \(L=x\sqrt{3}\), we solve for \(x\): \(x=\frac{L}{\sqrt{3}}=\frac{6\sqrt{3}}{\sqrt{3}} = 6\). So \(q = 6\) mm.

Answer:

\(6\)