Question

#15 find the value of the indicated variable. if your answer is not an integer, express it in simplified radical form.
(right triangle with hypotenuse 10√3, angle 30°, right angle, and side x opposite the 30° angle? wait, no, looking at the triangle: right angle, one angle 30°, hypotenuse? wait, the side labeled 10√3 is the hypotenuse? wait, the triangle has a right angle, one angle 30°, and the side opposite the 30° angle? wait, no, the side with x: lets see, the right triangle, angle 30°, the side adjacent to 30°? wait, the triangle: right angle, angle 30°, so its a 30-60-90 triangle. the side opposite 30° is the shortest side, hypotenuse is twice that, and the other leg is √3 times the shortest side. wait, in the triangle, the side labeled 10√3: is that the hypotenuse or the longer leg? wait, the right angle, angle 30°, so the sides: lets denote the sides. let’s say the right angle is at the bottom, angle 30° at the top right, so the side opposite 30° is the left leg (x), the side adjacent to 30° is the bottom leg, and the hypotenuse is the top side (10√3)? wait, no, in a 30-60-90 triangle, the hypotenuse is twice the shorter leg (opposite 30°), and the longer leg (opposite 60°) is √3 times the shorter leg. wait, maybe the side labeled 10√3 is the longer leg (opposite 60°), so then the shorter leg (x) would be (10√3)/√3 = 10? wait, no, lets clarify. the triangle: right angle, angle 30°, so angles are 90°, 30°, 60°. so sides: opposite 30°: x (shorter leg), opposite 60°: lets say y (longer leg), hypotenuse: h. then h = 2x, y = x√3. now, in the diagram, the side labeled 10√3: is that y (longer leg) or h (hypotenuse)? looking at the triangle, the side with 10√3 is the one opposite the 60° angle? wait, no, the angle 30° is at the top, so the side opposite 30° is x (left leg), the side adjacent to 30° is the bottom leg (y), and the hypotenuse is the top side (10√3). wait, no, hypotenuse is opposite the right angle. so the right angle is at the bottom left? wait, the diagram: right angle at the bottom, angle 30° at the top right, so the sides: the side from right angle to 30° angle is adjacent to 30°, the side from right angle to x is opposite to 30°, and the hypotenuse is from x to 30° angle, labeled 10√3. so hypotenuse is 10√3, angle 30°, so the side opposite 30° (x) is half the hypotenuse? wait, no: in 30-60-90 triangle, hypotenuse = 2 * shorter leg (opposite 30°). so if hypotenuse is 10√3, then shorter leg (x) is (10√3)/2 = 5√3? wait, no, wait: maybe i got the sides wrong. wait, maybe the side labeled 10√3 is the longer leg (opposite 60°), so then longer leg = x√3, so x = (10√3)/√3 = 10. wait, this is confusing. anyway, the problem is to find x in a right triangle with angle 30° and one side 10√3, express in simplified radical form. the ocr text is: #15 find the value of the indicated variable. if your answer is not an integer, express it in simplified radical form.
(right triangle with right angle, angle 30°, side 10√3, and side x)

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 \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, the side opposite \(30^{\circ}\) is \(x\)? Wait, no, wait. Wait, the hypotenuse? Wait, no, the side labeled \(10\sqrt{3}\) is opposite the \(60^{\circ}\) angle? Wait, no, let's check the angles. The right angle, \(30^{\circ}\), so the third angle is \(60^{\circ}\). So the side opposite \(30^{\circ}\) is \(x\)? Wait, no, the side adjacent to \(30^{\circ}\) is the leg, and the hypotenuse? Wait, no, in a right triangle, \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\)? Wait, no, let's use sine or cosine. Wait, \(\sin(30^{\circ})=\frac{x}{10\sqrt{3}}\)? No, wait, maybe I got the sides wrong. Wait, in a 30 - 60 - 90 triangle, the sides are: opposite \(30^{\circ}\): \(a\), opposite \(60^{\circ}\): \(a\sqrt{3}\), hypotenuse: \(2a\). So if the side opposite \(60^{\circ}\) is \(10\sqrt{3}\), then \(a\sqrt{3}=10\sqrt{3}\), so \(a = 10\)? No, wait, no. Wait, the side labeled \(10\sqrt{3}\) is the hypotenuse? Wait, no, the right angle is at the bottom, so the hypotenuse is the side opposite the right angle, which is the side with \(10\sqrt{3}\)? Wait, no, the right angle is the square, so the hypotenuse is the side opposite the right angle, which is the side labeled \(10\sqrt{3}\). Then, the side opposite \(30^{\circ}\) is \(x\)? Wait, no, \(\sin(30^{\circ})=\frac{x}{10\sqrt{3}}\), because \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\). So \(\sin(30^{\circ})=\frac{1}{2}=\frac{x}{10\sqrt{3}}\), so \(x=\frac{10\sqrt{3}}{2}=5\sqrt{3}\)? Wait, no, that would be if \(x\) is opposite \(30^{\circ}\). But wait, maybe the side adjacent to \(30^{\circ}\) is \(x\). Wait, let's re - examine the triangle. The right angle is at the bottom left, the \(30^{\circ}\) angle is at the bottom right, so the sides: the leg opposite \(30^{\circ}\) is the vertical leg (left), and the leg adjacent to \(30^{\circ}\) is the horizontal leg (bottom). The hypotenuse is the top side, \(10\sqrt{3}\). So \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\text{horizontal leg}}{10\sqrt{3}}\), and \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{10\sqrt{3}}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), then \(x = 10\sqrt{3}\times\frac{1}{2}=5\sqrt{3}\). Wait, but the answer box is \(X=\square\sqrt{\square}\), so \(5\sqrt{3}\), so the first box is 5, the second box is 3.

Wait, let's do it properly. In a right triangle, \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The angle of \(30^{\circ}\) has the opposite side as \(x\) (the vertical leg), and the hypotenuse is \(10\sqrt{3}\). So \(\sin(30^{\circ})=\frac{x}{10\sqrt{3}}\). Since \(\sin(30^{\circ})=\frac{1}{2}\), we have \(\frac{1}{2}=\frac{x}{10\sqrt{3}}\). Solving for \(x\), we multiply both sides by \(10\sqrt{3}\): \(x=\frac{10\sqrt{3}}{2}=5\sqrt{3}\).

Step1: Identify the trigonometric ratio

We know that in a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here, \(\theta = 30^{\circ}\), the opposite side to \(\theta\) is \(x\), and the hypotenuse is \(10\sqrt{3}\).

Step2: Substitute the values

We know that \(\sin(30^{\circ})=\frac{1}{2}\). So we set up the equation \(\frac{1}{2}=\frac{x}{10\sqrt{3}}\).

Step3: Solve for \(x\)

To solve for \(x\), we multiply both sides of the equation by \(10\sqrt{3}\):
\(x = 10\sqrt{3}\times\frac{1}{2}\)
\(x = 5\sqrt{3}\)

Answer:

\(x = 5\sqrt{3}\), so the first box is \(5\) and the second box (under the square root) is \(3\).