Question

find p.
write your answer in simplest radical form.
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, opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shortest, and the hypotenuse (opposite \(90^{\circ}\)) is twice the shortest. Here, the side of length \(12\) mm is adjacent to \(30^{\circ}\) and opposite \(60^{\circ}\), and \(p\) is the hypotenuse. Wait, no, let's correct: the right angle is between the \(12\) mm side and the other leg. Wait, the angles are \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\). So the side opposite \(30^{\circ}\) is the shorter leg, opposite \(60^{\circ}\) is the longer leg, and hypotenuse is \(p\). Wait, the side given is \(12\) mm, which is adjacent to \(30^{\circ}\), so it's the longer leg (opposite \(60^{\circ}\)). In a 30 - 60 - 90 triangle, longer leg \(=\) shorter leg \(\times\sqrt{3}\), and hypotenuse \(=\) shorter leg \(\times2\). Let the shorter leg be \(x\). Then \(12 = x\sqrt{3}\), so \(x=\frac{12}{\sqrt{3}} = 4\sqrt{3}\). Then hypotenuse \(p = 2x=2\times4\sqrt{3}=8\sqrt{3}\)? Wait, no, wait: Wait, the angle of \(30^{\circ}\) is at the end of \(p\), so the side opposite \(30^{\circ}\) is the leg adjacent to \(60^{\circ}\)? Wait, maybe better to use trigonometry. Let's use cosine. In the right triangle, \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(30^{\circ}\) is \(12\) mm? No, wait the right angle is between the \(12\) mm side and the other leg. So the angle of \(30^{\circ}\): the sides: the leg opposite \(30^{\circ}\) is, let's say, \(a\), opposite \(60^{\circ}\) is \(12\) mm, and hypotenuse is \(p\). So \(\sin(60^{\circ})=\frac{12}{p}\), since \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\). \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{12}{p}\). Then \(p=\frac{12\times2}{\sqrt{3}}=\frac{24}{\sqrt{3}} = 8\sqrt{3}\)? Wait, no, that can't be. Wait, no, let's re - examine the triangle. The angles are \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\). So the side opposite \(30^{\circ}\) is the shortest leg, opposite \(60^{\circ}\) is the longer leg, hypotenuse is \(p\). The side of length \(12\) mm: which angle is it adjacent to? The \(60^{\circ}\) angle: so adjacent to \(60^{\circ}\) is the shorter leg? No, wait, in a right triangle, the two legs are adjacent to the right angle. Let's label the triangle: let \(C = 90^{\circ}\), \(A = 60^{\circ}\), \(B = 30^{\circ}\). Then side \(AC = 12\) mm (adjacent to \(A = 60^{\circ}\), opposite to \(B = 30^{\circ}\)), side \(BC\) is the other leg, and \(AB = p\) (hypotenuse). So \(\cos(60^{\circ})=\frac{AC}{AB}\), since \(\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\). \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{12}{p}\), so \(p = 24\)? Wait, that's different. Wait, I messed up the angle. Let's look at the triangle: the angle of \(30^{\circ}\) is at the top right, \(60^{\circ}\) at top left, right angle at bottom. So the sides: the bottom leg (adjacent to \(60^{\circ}\)) is \(12\) mm, the other leg (adjacent to \(30^{\circ}\)) is, say, \(x\), and hypotenuse is \(p\). So angle at top right is \(30^{\circ}\), so the leg opposite \(30^{\circ}\) is the bottom leg (length \(12\) mm)? Wait, no: in a right triangle, the side opposite an angle is the one not adjacent to it. So angle \(30^{\circ}\): opposite side is the leg that's perpendicular to the hypotenuse and adjacent to \(60^{\circ}\). Wait, maybe the correct approach is: in a 30 - 60 - 90 triangle, the hypotenuse is twice the shorter leg. The shorter leg is opposite the \(30^{\circ}\) angle. Wait, the side of length \(12\) mm: if the angle of \(30^{\circ}\) is present, then the side opposite \(30^{\circ}\) is the shorter leg. Wait, maybe I had the sides reversed. Let's use trigonometry: \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(30^{\circ}\) is the leg that's not opposite \(30^{\circ}\). Wait, the right angle is between the \(12\) mm side and the other leg. So the angle of \(30^{\circ}\): the adjacent side is the leg with length equal to the longer leg, and the hypotenuse is \(p\). Wait, let's start over. The triangle has angles \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\). So it's a 30 - 60 - 90 triangle. The ratios are: shorter leg (opposite \(30^{\circ}\)): \(x\), longer leg (opposite \(60^{\circ}\)): \(x\sqrt{3}\), hypotenuse: \(2x\). Now, looking at the triangle, the side of length \(12\) mm: is it the longer leg or the shorter leg? The angle of \(60^{\circ}\) is at the vertex where the \(12\) mm side is connected to the right angle. So the side opposite \(60^{\circ}\) is the hypotenuse? No, the right angle is between the \(12\) mm side and the other leg. So the side \(12\) mm is adjacent to \(60^{\circ}\) and opposite to \(30^{\circ}\). Wait, no: in a triangle, the side opposite an angle is the one that doesn't form the angle. So angle \(60^{\circ}\): the sides forming it are the \(12\) mm side and the hypotenuse \(p\). So the side opposite \(60^{\circ}\) is the other leg (let's call it \(y\)). Then \(\sin(60^{\circ})=\frac{y}{p}\), and \(\cos(60^{\circ})=\frac{12}{p}\). Since \(\cos(60^{\circ}) = 0.5=\frac{1}{2}\), so \(\frac{12}{p}=\frac{1}{2}\), so \(p = 24\). Wait, that makes sense. Because \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), adjacent to \(60^{\circ}\) is \(12\) mm, hypotenuse is \(p\). So \(\cos(60^{\circ})=\frac{12}{p}\), \(\cos(60^{\circ})=\frac{1}{2}\), so \(p = 12\times2 = 24\)? Wait, but that contradicts the earlier thought. Wait, no, maybe the \(12\) mm is the shorter leg. Wait, no, the angle of \(30^{\circ}\) is at the end of \(p\), so the side opposite \(30^{\circ}\) is the leg adjacent to \(60^{\circ}\), which would be the shorter leg. Wait, I think I made a mistake in the angle - side correspondence. Let's use the fact that in a 30 - 60 - 90 triangle, the hypotenuse is twice the shorter leg. If the angle of \(30^{\circ}\) is present, the side opposite \(30^{\circ}\) is the shorter leg. Let's assume that the side of length \(12\) mm is the longer leg (opposite \(60^{\circ}\)). Then shorter leg \(=\frac{12}{\sqrt{3}} = 4\sqrt{3}\), and hypotenuse \(= 2\times4\sqrt{3}=8\sqrt{3}\). But that doesn't match the cosine approach. Wait, where is the error? Let's look at the triangle again. The right angle is at the bottom, so the two legs are vertical and horizontal, and hypotenuse is \(p\) at the top. The angle at the top left is \(60^{\circ}\), top right is \(30^{\circ}\). So the leg adjacent to \(60^{\circ}\) is the horizontal leg? No, the legs are the two sides forming the right angle. So one leg is, say, vertical (length \(a\)), one is horizontal (length \(12\) mm), and hypotenuse \(p\). The angle at top left (60°) is between the horizontal leg (12 mm) and hypotenuse \(p\). So the angle between the horizontal leg and hypotenuse is \(60^{\circ}\), so the vertical leg is opposite \(60^{\circ}\), horizontal leg is adjacent to \(60^{\circ}\). Then \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{p}\), so \(p=\frac{12}{\cos(60^{\circ})}=\frac{12}{0.5}=24\). Yes, that's correct. Because \(\cos(60^{\circ}) = 0.5\), so \(p = 24\) mm. Wait, but earlier when I thought of 30 - 60 - 90 ratios, I confused the sides. Let's confirm with ratios. In 30 - 60 - 90, the sides are \(x\) (opposite 30°), \(x\sqrt{3}\) (opposite 60°), \(2x\) (hypotenuse). Here, the angle of 30° is at the top right, so the side opposite 30° is the vertical leg (length \(x\)), the side opposite 60° is the horizontal leg (length \(x\sqrt{3}=12\) mm), and hypotenuse is \(2x\). Wait, so if \(x\sqrt{3}=12\), then \(x = \frac{12}{\sqrt{3}} = 4\sqrt{3}\), and hypotenuse \(2x = 8\sqrt{3}\). But this contradicts the cosine result. So where is the mistake? Ah! The angle at the top right is 30°, so the side opposite 30° is the vertical leg (length \(x\)), the side opposite 60° is the horizontal leg (length \(12\) mm), so \(x\sqrt{3}=12\) (since opposite 60° is \(x\sqrt{3}\)), so \(x = \frac{12}{\sqrt{3}} = 4\sqrt{3}\), and hypotenuse \(p = 2x = 8\sqrt{3}\). But then \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{p}\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{12}{p}=\frac{\sqrt{3}}{2}\), so \(p=\frac{24}{\sqrt{3}} = 8\sqrt{3}\), which matches the ratio method. So my earlier cosine approach was wrong because I misidentified the adjacent side. The angle of 60°: the adjacent side is the vertical leg, not the horizontal leg. Oh! That's the mistake. The angle at the top left is 60°, which is between the vertical leg and the hypotenuse, not the horizontal leg. So the horizontal leg (12 mm) is opposite the 60° angle, and the vertical leg is adjacent to 60°. So \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{12}{p}\), and \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{12}{p}=\frac{\sqrt{3}}{2}\), so \(p=\frac{24}{\sqrt{3}} = 8\sqrt{3}\) mm. Yes, that's correct. So the key was misidentifying which side is adjacent or opposite to the angle. So to correct: angle at top left is 60°, so the side opposite 60° is the horizontal leg (12 mm), the side adjacent to 60° is the vertical leg. So using sine: \(\sin(60^{\circ})=\frac{12}{p}\), so \(p=\frac{12}{\sin(60^{\circ})}=\frac{12}{\frac{\sqrt{3}}{2}}=\frac{24}{\sqrt{3}} = 8\sqrt{3}\) mm (rationalizing the denominator: \(\frac{24}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\)).

Step1: Identify the triangle type and sides

This is a 30 - 60 - 90 right - triangle. In a 30 - 60 - 90 triangle, the ratio of the sides (opposite \(30^{\circ}\), opposite \(60^{\circ}\), hypotenuse) is \(1:\sqrt{3}:2\). The side of length \(12\) mm is opposite the \(60^{\circ}\) angle (let's call this the longer leg), and \(p\) is the hypotenuse.

Step2: Relate the sides using the ratio

Let the length of the side opposite \(30^{\circ}\) (shorter leg) be \(x\). Then the longer leg (opposite \(60^{\circ}\)) is \(x\sqrt{3}\), and the hypotenuse \(p = 2x\). We know that the longer leg \(x\sqrt{3}=12\) mm.

Step3: Solve for \(x\)

From \(x\sqrt{3}=12\), we can solve for \(x\) by dividing both sides by \(\sqrt{3}\):
\(x=\frac{12}{\sqrt{3}}\)
Rationalize the denominator: \(x = \frac{12\sqrt{3}}{3}=4\sqrt{3}\)

Step4: Solve for \(p\) (hypotenuse)

Since the hypotenuse \(p = 2x\), substitute \(x = 4\sqrt{3}\) into the formula:
\(p = 2\times4\sqrt{3}=8\sqrt{3}\)

Answer:

\(8\sqrt{3}\)