Question

find p.

image of a right triangle with angles 30°, 60°, 90°, hypotenuse 38√3 ft, and side p opposite 30°

write your answer in simplest radical form.
blank feet

Explanation:

Step1: Identify triangle type

This is a 30-60-90 right triangle. In such a triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\) (opposite 30°, 60°, 90° respectively). The hypotenuse is opposite the 90° angle, the side opposite 30° is the shortest, and opposite 60° is \( \sqrt{3} \) times the shortest. Here, the side given \( 38\sqrt{3} \) ft is opposite 60°, and \( p \) is opposite 30°.

Step2: Relate sides using ratio

Let the side opposite 30° (which is \( p \)) be \( x \), then the side opposite 60° is \( x\sqrt{3} \). We know \( x\sqrt{3} = 38\sqrt{3} \). Divide both sides by \( \sqrt{3} \): \( x = 38 \). Wait, no—wait, actually, in 30-60-90, hypotenuse is \( 2x \), side opposite 30° is \( x \), side opposite 60° is \( x\sqrt{3} \). Wait, the given side: let's check angles. The right angle is at the top left, so the angles: 30° at the bottom, 60° at the top right. So the side opposite 30° (bottom angle) is the top side \( p \)? Wait, no—wait, the right angle is top left, so the sides: vertical side (left) is adjacent to 60°, horizontal side \( p \) is adjacent to 30°? Wait, maybe better to label: right angle at A, 60° at B, 30° at C. So side opposite 30° (angle C) is AB (which is \( p \)), side opposite 60° (angle B) is AC, and hypotenuse BC is \( 38\sqrt{3} \). Wait, no, hypotenuse is opposite right angle. So right angle at A, so hypotenuse is BC. Then angle at C is 30°, so side AB (opposite C, 30°) is \( p \), side AC (opposite B, 60°) is \( p\sqrt{3} \), hypotenuse BC is \( 2p \). Wait, but the given length is \( 38\sqrt{3} \). Wait, maybe I mixed up. Wait, the side labeled \( 38\sqrt{3} \) is the hypotenuse? No, hypotenuse is opposite right angle. Wait, the triangle: right angle at top left, so the two legs are vertical (left) and horizontal (top, \( p \)), and the hypotenuse is the slant side (length \( 38\sqrt{3} \)). Wait, no—wait, angles: 30° at the bottom vertex, 60° at the top right vertex, right angle at top left. So the sides: top side \( p \) (horizontal, between right angle and 60° angle) is adjacent to 60° angle, and opposite 30° angle. The slant side (hypotenuse) is between 60° and 30° angles, length \( 38\sqrt{3} \). Wait, in a right triangle, cos(60°) = adjacent/hypotenuse. Cos(60°) = \( \frac{1}{2} \), so \( \cos(60°) = \frac{p}{38\sqrt{3}} \)? No, that's wrong. Wait, no—wait, angle at top right is 60°, so the side adjacent to 60° is \( p \), and the hypotenuse is \( 38\sqrt{3} \). So \( \cos(60°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{p}{38\sqrt{3}} \). Since \( \cos(60°) = \frac{1}{2} \), then \( \frac{p}{38\sqrt{3}} = \frac{1}{2} \)? No, that can't be, because then \( p = \frac{38\sqrt{3}}{2} = 19\sqrt{3} \), but that contradicts earlier. Wait, I think I messed up the angle labeling. Let's use sine: \( \sin(30°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{p}{38\sqrt{3}} \). Since \( \sin(30°) = \frac{1}{2} \), so \( \frac{p}{38\sqrt{3}} = \frac{1}{2} \)? No, that would be \( p = \frac{38\sqrt{3}}{2} = 19\sqrt{3} \), but that's not right. Wait, no—wait, the side given is \( 38\sqrt{3} \), which is the hypotenuse? Wait, no, in a 30-60-90 triangle, the hypotenuse is twice the shortest side. Wait, maybe the given side is the side opposite 60°, so the side opposite 30° (which is \( p \)) is \( \frac{\text{side opposite 60°}}{\sqrt{3}} \). So side opposite 60° is \( 38\sqrt{3} \), so side opposite 30° ( \( p \)) is \( \frac{38\sqrt{3}}{\sqrt{3}} = 38 \). Wait, that makes sense. Because side opposite 60° is \( x\sqrt{3} \), so if \( x\sqrt{3} = 38\sqrt{3} \), then \( x = 38 \). So \( p = 38 \)? Wait, but let's check with hypotenuse. Hypotenuse would be \( 2x = 76 \), but the given side is \( 38\sqrt{3} \approx 65.8 \), and 76 is larger, so that can't be. Wait, I think I mixed up which side is which. Let's re-express: in 30-60-90, the sides are:
- Short leg (opposite 30°): \( x \)
- Long leg (opposite 60°): \( x\sqrt{3} \)
- Hypotenuse: \( 2x \)

So if the long leg (opposite 60°) is \( 38\sqrt{3} \), then \( x\sqrt{3} = 38\sqrt{3} \implies x = 38 \). So the short leg (opposite 30°) is \( x = 38 \), and hypotenuse is \( 2x = 76 \). But in the triangle, the side labeled \( p \) is the short leg (opposite 30°), so \( p = 38 \). Wait, but let's confirm with angles. The 30° angle is at the bottom, so the side opposite 30° is the top horizontal side \( p \), the side opposite 60° is the left vertical side, and hypotenuse is the slant side \( 38\sqrt{3} \). Wait, no—if the slant side is hypotenuse, then hypotenuse should be \( 2x \), so \( 2x = 38\sqrt{3} \implies x = 19\sqrt{3} \), but then long leg would be \( x\sqrt{3} = 19 \times 3 = 57 \), which is not \( 38\sqrt{3} \approx 65.8 \). So my initial angle labeling was wrong. Let's look at the triangle: right angle at top left, so the two legs are horizontal (top, \( p \)) and vertical (left), and hypotenuse is the right side (connecting top right to bottom). The angles: top right is 60°, bottom is 30°. So the vertical leg (left) is adjacent to 60° angle, horizontal leg \( p \) is adjacent to 30° angle. The hypotenuse is the right side, length \( 38\sqrt{3} \). So using trigonometry: \( \sin(60°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{p}{38\sqrt{3}} \)? No, \( \sin(60°) = \frac{\text{opposite}}{\text{hypotenuse}} \), where opposite to 60° is the horizontal leg \( p \)? Wait, no—angle at top right is 60°, so the side opposite to 60° is the vertical leg (left), and the side adjacent is \( p \) (horizontal). Wait, this is confusing. Let's use cosine: \( \cos(60°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{p}{38\sqrt{3}} \). Since \( \cos(60°) = 0.5 \), then \( p = 38\sqrt{3} \times 0.5 = 19\sqrt{3} \). Wait, now I'm confused. Wait, let's use the 30-60-90 ratios correctly. The angles: 30°, 60°, 90°. So the sides: the side opposite 30° is the shortest, opposite 60° is longer, hypotenuse longest. So if the angle at the bottom is 30°, then the side opposite to it (the top horizontal side \( p \)) is the shortest side (opposite 30°), the side opposite 60° (top right angle) is the vertical side (left), and hypotenuse is the slant side (right). So:
- Short side (opposite 30°): \( p \) (top horizontal)
- Long side (opposite 60°): vertical side (left) = \( p\sqrt{3} \)
- Hypotenuse (slant): \( 2p \)

But the hypotenuse is given as \( 38\sqrt{3} \), so \( 2p = 38\sqrt{3} \implies p = 19\sqrt{3} \). But then the long side would be \( p\sqrt{3} = 19 \times 3 = 57 \), and hypotenuse \( 38\sqrt{3} \approx 65.8 \), which is longer than 57, so that works (hypotenuse is longest). Wait, but earlier I thought the long side was \( 38\sqrt{3} \), but that was wrong. So which is it? Let's calculate the lengths:

If \( p = 19\sqrt{3} \approx 32.9 \), long side \( 57 \), hypotenuse \( 65.8 \). That makes sense (hypotenuse longest, long side longer than short side).

But wait, the problem says "Find p" with the side labeled \( 38\sqrt{3} \) as the hypotenuse? No, the side labeled \( 38\sqrt{3} \) is the hypotenuse? Wait, the triangle has a right angle, 60°, and 30°, so it's a 30-60-90 triangle. The side opposite 30° is \( p \), the side opposite 60° is, say, \( s \), and hypotenuse \( h = 38\sqrt{3} \). Then:

\( \sin(30°) = \frac{p}{h} \implies \frac{1}{2} = \frac{p}{38\sqrt{3}} \implies p = \frac{38\sqrt{3}}{2} = 19\sqrt{3} \)

Ah! That's the correct approach. Because \( \sin(\theta) = \text{opposite}/\text{hypotenuse} \). So angle 30°: opposite side is \( p \), hypotenuse is \( 38\sqrt{3} \). So \( \sin(30°) = p / (38\sqrt{3}) \). Since \( \sin(30°) = 1/2 \), then \( p = (38\sqrt{3}) \times (1/2) = 19\sqrt{3} \). Wait, now I see my earlier mistake: I misidentified which side is opposite which angle. The 30° angle is at the bottom, so the side opposite to it (the side not connected to it) is the top horizontal side \( p \), and the hypotenuse is the side connecting top right to bottom (length \( 38\sqrt{3} \)). So using sine of 30°: opposite over hypotenuse, so \( p = h \times \sin(30°) = 38\sqrt{3} \times 1/2 = 19\sqrt{3} \). Wait, but let's check with cosine of 60°: \( \cos(60°) = \text{adjacent}/\text{hypotenuse} = p / (38\sqrt{3}) \), and \( \cos(60°) = 1/2 \), so same result. So that's correct.

Answer:

\( 19\sqrt{3} \)