Question

warm up #11 9/23
①
right triangle with horizontal leg 12, vertical leg y, hypotenuse x, and a 60° angle at the bottom right find x and y
②
isosceles right triangle with hypotenuse 9, legs x and y (marked equal)

Explanation:

Step1: Solve for \( x \) and \( y \) in the first triangle (30-60-90 triangle)

The first triangle is a right triangle with one angle \( 60^\circ \), so the other angle is \( 30^\circ \). 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, opposite \( 60^\circ \) is \( \sqrt{3} \) times the shortest, and the hypotenuse is twice the shortest. Here, the side adjacent to \( 60^\circ \) (opposite \( 30^\circ \)) is 12? Wait, no, wait. Wait, the right angle, one angle \( 60^\circ \), so the sides: let's see, the side labeled 12 is adjacent to \( 60^\circ \)? Wait, no, the right angle, so the two legs: one leg is 12, the other is \( y \), hypotenuse \( x \). The angle at the bottom is \( 60^\circ \), so the side opposite \( 60^\circ \) is 12? Wait, no, let's clarify. In a right triangle, the angles are \( 90^\circ \), \( 60^\circ \), so the third angle is \( 30^\circ \). So the side opposite \( 30^\circ \) is the shortest leg, let's say \( y \), the side opposite \( 60^\circ \) is \( 12 \), and hypotenuse \( x \). In 30-60-90 triangle, the ratio of sides is \( \text{short leg} : \text{long leg} : \text{hypotenuse} = 1 : \sqrt{3} : 2 \). So if the long leg (opposite \( 60^\circ \)) is 12, then the short leg (opposite \( 30^\circ \)) is \( \frac{12}{\sqrt{3}} = 4\sqrt{3} \)? Wait, no, wait, maybe I got the angles wrong. Wait, the angle at the bottom is \( 60^\circ \), so the side adjacent to \( 60^\circ \) is \( y \), and the side opposite is 12. Wait, no, let's use trigonometry. \( \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{y}{x} \), \( \sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{x} \), and \( \tan(60^\circ) = \frac{12}{y} \). We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), so \( \frac{12}{x} = \frac{\sqrt{3}}{2} \), solving for \( x \): \( x = \frac{12 \times 2}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} \)? Wait, no, that can't be. Wait, maybe the side labeled 12 is the adjacent to \( 30^\circ \). Wait, maybe I mixed up the angles. Let's re-express: the triangle has a right angle, one angle \( 60^\circ \), so the other is \( 30^\circ \). So the side opposite \( 30^\circ \) is the shortest leg. Let's assume that the side labeled 12 is the side opposite \( 60^\circ \), so the short leg (opposite \( 30^\circ \)) is \( y \), and hypotenuse \( x \). Then, \( \sin(60^\circ) = \frac{12}{x} \implies x = \frac{12}{\sin(60^\circ)} = \frac{12}{\frac{\sqrt{3}}{2}} = \frac{24}{\sqrt{3}} = 8\sqrt{3} \)? No, that's not right. Wait, maybe the side labeled 12 is the hypotenuse? No, the hypotenuse is \( x \). Wait, maybe I made a mistake. Wait, let's use the 30-60-90 ratios correctly. In a 30-60-90 triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \), where the side opposite 30° is \( s \), opposite 60° is \( s\sqrt{3} \), hypotenuse \( 2s \). So if the side opposite 60° is 12, then \( s\sqrt{3} = 12 \implies s = \frac{12}{\sqrt{3}} = 4\sqrt{3} \), so the short leg (opposite 30°) is \( 4\sqrt{3} \) (which is \( y \)), and hypotenuse \( x = 2s = 8\sqrt{3} \)? Wait, no, that doesn't seem right. Wait, maybe the side labeled 12 is the adjacent to 60°, so the adjacent side is 12, angle 60°, so \( \cos(60^\circ) = \frac{12}{x} \implies x = \frac{12}{\cos(60^\circ)} = \frac{12}{0.5} = 24 \). Then \( \tan(60^\circ) = \frac{y}{12} \implies y = 12 \tan(60^\circ) = 12\sqrt{3} \). Wait, that makes more sense. Because \( \cos(60^\circ) = 0.5 \), so hypotenuse \( x = 12 / 0.5 = 24 \). Then the other leg \( y = x \sin(60^\circ) = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3} \). Wait, no, \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), so the side opposite 60° is \( x \sin(60^\circ) = 24 \times \frac{\sqrt{3}}{2} = 12\sqrt{3} \), and the side adjacent is \( x \cos(60^\circ) = 24 \times 0.5 = 12 \), which matches the given 12. So yes, \( x = 24 \), \( y = 12\sqrt{3} \).

Step2: Solve for \( x \) and \( y \) in the second triangle (isosceles right triangle)

The second triangle is a right triangle with two legs equal (marked with \( x \) and \( y \), so \( x = y \)) and hypotenuse 9. In an isosceles right triangle (45-45-90), the legs are equal, and the hypotenuse is \( \text{leg} \times \sqrt{2} \). So if hypotenuse is 9, then \( x\sqrt{2} = 9 \implies x = y = \frac{9}{\sqrt{2}} = \frac{9\sqrt{2}}{2} \approx 6.364 \).

Wait, let's confirm: in a 45-45-90 triangle, the ratio of legs to hypotenuse is \( 1 : 1 : \sqrt{2} \), so each leg is \( \frac{\text{hypotenuse}}{\sqrt{2}} \). So \( x = y = \frac{9}{\sqrt{2}} = \frac{9\sqrt{2}}{2} \).

Answer:

For the first triangle: \( x = 24 \), \( y = 12\sqrt{3} \)
For the second triangle: \( x = y = \frac{9\sqrt{2}}{2} \)