Question

question 6
10 pts
g.gsr.6.1 (mc)
find the values of x and y.
image of a right triangle with angles 30°, 60°, hypotenuse 10, legs x and y
options:
x = 5, y = 5√3
x = 10, y = 10√3
x = 5√3, y = 5
x = 20, y = 20√3

Explanation:

Step1: Identify Triangle Type

This is a 30-60-90 right triangle. In such a triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite \(30^\circ\) is the shortest (let's call it \(a\)), opposite \(60^\circ\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\).

Step2: Determine Hypotenuse and Shortest Side

The hypotenuse here is given as \(10\). So, \(2a = 10\), which means \(a = 5\). The side opposite \(30^\circ\) (shortest side) is \(y\) (wait, no—wait, the angles: the right angle, \(60^\circ\), and \(30^\circ\). So the side opposite \(30^\circ\) is \(x\)? Wait, no, let's label the triangle. The right angle is at the top left. So the angles: bottom left is \(60^\circ\), top right is \(30^\circ\). So the side opposite \(30^\circ\) is the vertical side (x), and opposite \(60^\circ\) is the horizontal side (y), hypotenuse is 10. Wait, no—wait, in a right triangle, the side opposite \(30^\circ\) is the shortest. So hypotenuse is 10, so side opposite \(30^\circ\) (x) is \(10/2 = 5\)? Wait, no, wait: if the angle at the top right is \(30^\circ\), then the side opposite to it is the vertical side (x). So \(x = \text{hypotenuse} \times \sin(30^\circ)\). \(\sin(30^\circ) = 0.5\), so \(x = 10 \times 0.5 = 5\). Then the side opposite \(60^\circ\) (y) is \(\text{hypotenuse} \times \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\). Wait, but wait, alternatively, using the ratio: in 30-60-90, sides are \(a\) (opposite 30), \(a\sqrt{3}\) (opposite 60), \(2a\) (hypotenuse). So hypotenuse \(2a = 10\) ⇒ \(a = 5\). So side opposite 30 (x) is \(a = 5\), side opposite 60 (y) is \(a\sqrt{3} = 5\sqrt{3}\). Wait, but let's check the options. The first option is \(x = 5, y = 5\sqrt{3}\). Wait, but wait, maybe I mixed up x and y. Wait, the vertical side is x, horizontal is y. Wait, the angle at the bottom left is \(60^\circ\), so the side adjacent to \(60^\circ\) is x, opposite is y? Wait, no, let's use trigonometry. Let's denote the right angle at (0,0), top left at (0, x), top right at (y, x), bottom left at (0,0), bottom right at (y, 0). Wait, no, the triangle has vertices: top left (right angle), bottom left (60°), top right (30°). So the sides: vertical leg (x) from top left to bottom left, horizontal leg (y) from top left to top right, hypotenuse from bottom left to top right (length 10). So angle at bottom left is 60°, so in triangle, angle at bottom left: 60°, right angle, so angle at top right is 30°. So for angle at bottom left (60°), the adjacent side is x (vertical), opposite side is y (horizontal), hypotenuse 10. So \(\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{10}\), so \(x = 10 \times \cos(60^\circ) = 10 \times 0.5 = 5\). \(\sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{10}\), so \(y = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\). So that matches the first option: \(x = 5, y = 5\sqrt{3}\). Wait, but let's check the options again. The first option is \(x = 5, y = 5\sqrt{3}\), second is \(x=10, y=10\sqrt{3}\), third is \(x=5\sqrt{3}, y=5\), fourth is \(x=20, y=20\sqrt{3}\). So the correct one should be the first option. Wait, but wait, maybe I mixed up x and y. Wait, if the angle at the top right is 30°, then the side opposite to it is x (vertical), so \(x = 10 \times \sin(30°) = 5\), and the side adjacent to 30° is y (horizontal), so \(y = 10 \times \cos(30°) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\). Yes, that's correct. So x=5, y=5√3.

Answer:

A. \(x = 5, y = 5\sqrt{3}\)