Question

find y.
right triangle with angles 30°, 60°, right angle; hypotenuse 6√3 yd, side y opposite 30°? wait, no: angles are 30°, 60°, right angle. so sides: in 30-60-90 triangle, ratios are 1 : √3 : 2 (opposite 30°, 60°, hypotenuse). wait, the hypotenuse is 6√3? wait, the side labeled 6√3 is the hypotenuse? wait, the triangle has angles 30°, 60°, right angle. so the side opposite 30° is the shortest side, opposite 60° is longer leg, hypotenuse is longest. wait, the side labeled y: lets see, angle 60°: the side adjacent to 60°? wait, no, lets label the triangle. right angle at top, 30° at bottom left, 60° at bottom right. so the sides: the side opposite 30° is y (since 30° is at bottom left, opposite side is the one with y? wait, no: right angle is top, so the two legs are vertical and horizontal? wait, the side labeled 6√3 is the hypotenuse? wait, the hypotenuse is opposite the right angle, so the side between 30° and 60° is hypotenuse, length 6√3. then, in 30-60-90 triangle, the sides are: opposite 30°: x, opposite 60°: x√3, hypotenuse: 2x. so if hypotenuse is 6√3, then 2x = 6√3 → x = 3√3? wait, no, wait: maybe y is the side opposite 30°, or adjacent? wait, the angle at bottom right is 60°, so the side adjacent to 60° is y, and the side opposite 60° is the other leg. wait, lets use trigonometry. lets denote: right angle at a, 30° at b, 60° at c. so side bc is hypotenuse, length 6√3. side ac is y (adjacent to 60°), side ab is opposite 60°. so cos(60°) = adjacent / hypotenuse = y / (6√3). cos(60°) is 0.5, so y = 0.5 6√3 = 3√3? wait, but in 30-60-90, cos(60°) is 0.5, so adjacent (y) = hypotenuse cos(60°) = 6√3 0.5 = 3√3. alternatively, sin(30°) = opposite / hypotenuse, but opposite to 30° is ab, so ab = 6√3 0.5 = 3√3? wait, no, maybe i mixed up. wait, 30-60-90 triangle: sides are in ratio 1 : √3 : 2, where 1 is opposite 30°, √3 opposite 60°, 2 opposite right angle (hypotenuse). so if hypotenuse is 2x, then opposite 30° is x, opposite 60° is x√3. so if hypotenuse is 6√3, then 2x = 6√3 → x = 3√3. so the side opposite 30° is x = 3√3, which would be y? wait, but the angle at b is 30°, so opposite side is ac (y). so yes, y = 3√3? wait, but lets check with trigonometry. sin(30°) = y / (6√3) → 0.5 = y / (6√3) → y = 3√3. yes. so the problem is to find y in a 30-60-90 triangle with hypotenuse 6√3, find the side opposite 30° (or adjacent to 60°). so the ocr text is: \find y. triangle image: right triangle, angles 30°, 60°, hypotenuse 6√3 yd, side y write your answer in simplest radical form. blank yards √ 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 \(6\sqrt{3}\) yd, and \(y\) is opposite the 30° angle.

Step2: Recall 30-60-90 ratios

In a 30-60-90 triangle, the side opposite 30° (let's call it \(x\)) is half the hypotenuse, and the side opposite 60° (let's call it \(x\sqrt{3}\)) is \(\sqrt{3}\) times the side opposite 30°. Here, the hypotenuse \(H = 6\sqrt{3}\), and \(y\) is opposite 30°, so \(y=\frac{H}{2}\)? Wait, no, wait. Wait, let's label the triangle: right angle, 30°, 60°. The side adjacent to 60° is \(y\), opposite to 60° is the other leg, hypotenuse is \(6\sqrt{3}\). Wait, maybe better to use trigonometry. Let's use cosine: \(\cos(60^\circ)=\frac{y}{6\sqrt{3}}\). Since \(\cos(60^\circ)=\frac{1}{2}\), then \(\frac{1}{2}=\frac{y}{6\sqrt{3}}\), so \(y = \frac{6\sqrt{3}}{2}=3\sqrt{3}\)? Wait, no, wait. Wait, 30-60-90 triangle: angles are 30, 60, 90. The sides: opposite 30 is \(a\), opposite 60 is \(a\sqrt{3}\), hypotenuse is \(2a\). So in this triangle, the hypotenuse is \(6\sqrt{3}\), so \(2a = 6\sqrt{3}\), so \(a = 3\sqrt{3}\). But wait, which side is \(y\)? The side with length \(y\) is adjacent to 60° and opposite to 30°, so that's the side opposite 30°, which is \(a\). Wait, but let's check with trigonometry. Let's take angle 30°: \(\sin(30^\circ)=\frac{y}{6\sqrt{3}}\), since \(\sin(30^\circ)=\frac{1}{2}\), then \(\frac{1}{2}=\frac{y}{6\sqrt{3}}\), so \(y = \frac{6\sqrt{3}}{2}=3\sqrt{3}\)? Wait, no, that can't be. Wait, maybe I mixed up the sides. Wait, the right angle, 30°, 60°. So the leg opposite 30° is the shorter leg, opposite 60° is the longer leg, hypotenuse is the longest side. So in the diagram, the hypotenuse is \(6\sqrt{3}\), the leg \(y\) is adjacent to 60° and opposite to 30°, so it's the shorter leg. So shorter leg \(= \frac{\text{hypotenuse}}{2}\)? Wait, hypotenuse is \(2\times\) shorter leg. So shorter leg \(= \frac{\text{hypotenuse}}{2}\). So hypotenuse is \(6\sqrt{3}\), so shorter leg \(y = \frac{6\sqrt{3}}{2}=3\sqrt{3}\)? Wait, but let's check with cosine of 60°: \(\cos(60^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{y}{6\sqrt{3}}\). \(\cos(60^\circ)=0.5\), so \(y = 6\sqrt{3}\times0.5 = 3\sqrt{3}\). Yes, that's correct.

Answer:

\(3\sqrt{3}\)