Question

find k.
right triangle with angles 30°, 60°, right angle; hypotenuse 8√3 yd, side k opposite 30°? wait, no, labels: 30° at one acute angle, 60° at the other, hypotenuse? wait, the side labeled 8√3 yd is... wait, the triangle has a right angle, 30°, 60°, so its a 30-60-90 triangle. the side labeled 8√3 yd: lets see, in a 30-60-90 triangle, sides are in ratio 1 : √3 : 2 (short leg : long leg : hypotenuse). lets identify the sides. the right angle, 30° at one vertex, 60° at another. the side opposite 30° is the short leg, opposite 60° is long leg, hypotenuse is opposite right angle. wait, the side labeled k: lets see the angles. the angle of 30°: the side adjacent to 30°? wait, maybe using trigonometry: cos(30°) = adjacent/hypotenuse, or sin(60°), etc. wait, the problem is to find k. lets parse the triangle: right angle, 30°, 60°, so angles are 30, 60, 90. the side labeled 8√3 yd: lets see, maybe its the long leg (opposite 60°), and k is the short leg (opposite 30°), or vice versa? wait, in a 30-60-90 triangle, short leg (opposite 30°) is x, long leg (opposite 60°) is x√3, hypotenuse is 2x. so if the long leg is 8√3, then x√3 = 8√3 ⇒ x = 8? wait, no, maybe k is adjacent to 60°? wait, lets use trigonometry. lets denote the right angle vertex, 30° vertex, 60° vertex. lets say the right angle is at (0,0), 30° at (0, b), 60° at (k, 0). then the hypotenuse is from (0, b) to (k, 0), length 8√3. the angle at (k, 0) is 60°, so the angle between the hypotenuse and the x-axis (side k) is 60°. so cos(60°) = adjacent/hypotenuse = k / (8√3). cos(60°) is 0.5, so k = 8√3 0.5 = 4√3? wait, no, maybe i got the angles wrong. alternatively, angle at the 30° vertex: the side adjacent to 30° is the long leg? wait, maybe better to recall 30-60-90 ratios. short leg (opposite 30°): x, long leg (opposite 60°): x√3, hypotenuse: 2x. so if the side labeled 8√3 is the hypotenuse, then hypotenuse is 2x = 8√3 ⇒ x = 4√3, which would be the short leg (opposite 30°). but wait, the angle of 30°: the side opposite 30° is the short leg. wait, maybe the side labeled 8√3 is the long leg (opposite 60°), so long leg = x√3 = 8√3 ⇒ x = 8, so short leg (opposite 30°) is 8, hypotenuse 16. but the problem is to find k. wait, the side labeled k: looking at the triangle, the side with length k is adjacent to the 60° angle? wait, the triangle has a right angle, 30°, 60°, so the sides: lets assign: angle 30°: opposite side is k, adjacent side is 8√3? no, that cant be. wait, maybe the side labeled 8√3 is the hypotenuse, and k is the side opposite 30°, so k = hypotenuse / 2 = (8√3)/2 = 4√3? wait, no, hypotenuse is 2x, so x = hypotenuse / 2. if hypotenuse is 8√3, then x = 4√3, which is the short leg (opposite 30°). so k would be 4√3? wait, but maybe i mixed up. alternatively, using cosine: cos(60°) = adjacent / hypotenuse. if the hypotenuse is 8√3, and adjacent to 60° is k, then cos(60°) = k / (8√3) ⇒ k = 8√3 0.5 = 4√3. alternatively, sin(30°) = opposite / hypotenuse = k / (8√3) ⇒ 0.5 = k / (8√3) ⇒ k = 4√3. so the problem is to find k in this 30-60-90 triangle, using trigonometry or the 30-60-90 side ratios. the ocr text is: \find k. triangle image: right triangle, angles 30°, 60°, right angle; side labeled 8√3 yd, side labeled k write your answer in simplest radical form. text box yards square root button\

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 \(8\sqrt{3}\) yd, and \(k\) is opposite the 30° angle.

Step2: Use 30-60-90 ratios

In a 30-60-90 triangle, the side opposite 30° (let's call it \(x\)) is half the hypotenuse (\(h\))? Wait, no—wait, the side opposite 30° is the shortest side, opposite 60° is \(x\sqrt{3}\), hypotenuse \(2x\). Wait, let's label: angle 30° opposite side \(k\), angle 60° opposite the other leg, hypotenuse \(8\sqrt{3}\). So hypotenuse \(= 2 \times\) (side opposite 30°)? Wait, no: side opposite 30° is \(x\), side opposite 60° is \(x\sqrt{3}\), hypotenuse \(2x\). Wait, here the hypotenuse is \(8\sqrt{3}\), so \(2x = 8\sqrt{3}\)? No, wait, maybe I mixed up. Wait, the side adjacent to 60° is \(k\) (opposite 30°). Wait, let's use trigonometry. \(\sin(30^\circ) = \frac{k}{\text{hypotenuse}}\). \(\sin(30^\circ) = \frac{1}{2}\), hypotenuse is \(8\sqrt{3}\). So \(\frac{1}{2} = \frac{k}{8\sqrt{3}}\), so \(k = \frac{8\sqrt{3}}{2} = 4\sqrt{3}\)? Wait, no, wait—wait, maybe the hypotenuse is not \(8\sqrt{3}\). Wait, the side labeled \(8\sqrt{3}\) is the side opposite 60°? Wait, no, let's look at the triangle: right angle, 30°, 60°. So the sides: the side opposite 30° is \(k\), the side opposite 60° is, say, \(m\), and hypotenuse \(h\). Then \(k : m : h = 1 : \sqrt{3} : 2\). So if \(m = 8\sqrt{3}\) (opposite 60°), then \(m = k\sqrt{3}\), so \(8\sqrt{3} = k\sqrt{3}\), so \(k = 8\)? Wait, that's conflicting. Wait, let's re-express. Let's denote: angle 30°: opposite side \(k\), angle 60°: opposite side \(a\), hypotenuse \(c = 8\sqrt{3}\). Then:

\(\sin(30^\circ) = \frac{k}{c}\) ⇒ \(k = c \times \sin(30^\circ) = 8\sqrt{3} \times \frac{1}{2} = 4\sqrt{3}\)

\(\sin(60^\circ) = \frac{a}{c}\) ⇒ \(a = 8\sqrt{3} \times \frac{\sqrt{3}}{2} = 8 \times \frac{3}{2} = 12\)

Wait, but maybe the side labeled \(8\sqrt{3}\) is the side opposite 60°, not the hypotenuse. Oh! That's the mistake. The triangle has a right angle, 30°, 60°, so the sides: the side opposite 30° is \(k\), the side opposite 60° is \(8\sqrt{3}\), and hypotenuse \(h\). Then, in 30-60-90, side opposite 60° is \(k\sqrt{3}\) (since side opposite 30° is \(k\), side opposite 60° is \(k\sqrt{3}\), hypotenuse \(2k\)). So \(k\sqrt{3} = 8\sqrt{3}\) ⇒ \(k = 8\). Ah, that's the correct approach. Because if the side opposite 60° is \(8\sqrt{3}\), then \(k\sqrt{3} = 8\sqrt{3}\), so \(k = 8\). Wait, why the confusion? Let's check the triangle: the right angle, 30° at the bottom, 60° at the top. So the side adjacent to 60° is \(k\) (opposite 30°), the side opposite 60° is the vertical leg (longer leg), and hypotenuse is the side labeled \(8\sqrt{3}\)? No, the side labeled \(8\sqrt{3}\) is the hypotenuse? Wait, the diagram: the right angle is at the top left, 60° at top right, 30° at bottom. So the sides: top side is \(k\) (horizontal), right side is hypotenuse \(8\sqrt{3}\), left side is vertical. So angle at bottom is 30°, so the side opposite 30° is \(k\) (top side), the side opposite 60° is the vertical leg, hypotenuse is \(8\sqrt{3}\). So in 30-60-90, side opposite 30° (\(k\)) is half the hypotenuse? Wait, no: hypotenuse is twice the side opposite 30°. So \(k = \frac{\text{hypotenuse}}{2}\)? But hypotenuse is \(8\sqrt{3}\), so \(k = 4\sqrt{3}\)? But that contradicts the other approach. Wait, no, let's use trigonometry. \(\cos(60^\circ) = \frac{k}{\text{hypotenuse}}\). \(\cos(60^\circ) = \frac{1}{2}\), so \(\frac{1}{2} = \frac{k}{8\sqrt{3}}\), so \(k = \frac{8\sqrt{3}}{2} = 4\sqrt{3}\). Wait, now I'm confused. Wait, let's draw the triangle: right angle at top left (so vertices: top left (right angle), top right (60°), bottom (30°)). So sides: top: \(k\) (between top left and top right), right: \(8\sqrt{3}\) (between top right and bottom), left: vertical (between top left and bottom). So angle at top right is 60°, so the side adjacent to 60° is \(k\), opposite is vertical leg, hypotenuse is \(8\sqrt{3}\). So \(\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{k}{8\sqrt{3}}\). \(\cos(60^\circ) = 0.5\), so \(k = 8\sqrt{3} \times 0.5 = 4\sqrt{3}\). Wait, but earlier when I thought the side opposite 60° is \(8\sqrt{3}\), that was wrong. The side labeled \(8\sqrt{3}\) is the hypotenuse. So the correct trigonometric approach: \(\cos(60^\circ) = \frac{k}{\text{hypotenuse}}\), so \(k = \text{hypotenuse} \times \cos(60^\circ) = 8\sqrt{3} \times \frac{1}{2} = 4\sqrt{3}\). Wait, but let's check with 30-60-90 ratios. In a 30-60-90 triangle, the sides are \(x\) (opposite 30°), \(x\sqrt{3}\) (opposite 60°), and \(2x\) (hypotenuse). So if hypotenuse is \(2x = 8\sqrt{3}\), then \(x = 4\sqrt{3}\), which is the side opposite 30° (which is \(k\)). So that matches. So the side opposite 30° is \(x = 4\sqrt{3}\), side opposite 60° is \(x\sqrt{3} = 4\sqrt{3} \times \sqrt{3} = 12\), hypotenuse \(2x = 8\sqrt{3}\). Yes, that works. So \(k = 4\sqrt{3}\)? Wait, no, wait: \(x\) is opposite 30°, so \(k = x = 4\sqrt{3}\). Wait, but earlier when I thought the side opposite 60° is \(8\sqrt{3}\), that was a mistake. The side labeled \(8\sqrt{3}\) is the hypotenuse. So the correct answer is \(4\sqrt{3}\)? Wait, no, wait the diagram: the side labeled \(8\sqrt{3}\) is the side between the 60° angle and the 30° angle, so that's the hypotenuse. So yes, hypotenuse is \(8\sqrt{3}\), so side opposite 30° (k) is half the hypotenuse? Wait, no, in 30-60-90, hypotenuse is twice the side opposite 30°, so side opposite 30° is hypotenuse / 2. So \(k = 8\sqrt{3} / 2 = 4\sqrt{3}\). Yes, that's correct.

Answer:

\(4\sqrt{3}\)