Question

find the values of x and y. write your answers in simplest form. (there is a right triangle with one leg 9, another leg x, hypotenuse y, and an angle of 30° opposite the leg of length 9)

Explanation:

Step1: Identify triangle type

It's 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 \(a\)), the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\). Here, the side opposite \(30^{\circ}\) (the vertical leg) is 9? Wait, no, the right angle is at the top left, the angle at the bottom is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the vertical leg (length 9), the side adjacent to \(30^{\circ}\) is \(x\) (horizontal leg), and hypotenuse is \(y\).

Wait, in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is the shorter leg. So if the angle at the bottom is \(30^{\circ}\), then the side opposite to it (the vertical leg) is the shorter leg. So shorter leg \(= 9\), then:

Step2: Find hypotenuse \(y\)

In 30 - 60 - 90 triangle, hypotenuse \(= 2\times\) shorter leg. So \(y = 2\times9=18\)? Wait, no, wait. Wait, maybe I got the legs mixed up. Wait, the right angle is between the side of length 9 and \(x\). So the angle of \(30^{\circ}\) is at the end of \(x\) and \(y\). So the side opposite \(30^{\circ}\) is the side with length 9 (the vertical leg). Then the horizontal leg \(x\) is opposite \(60^{\circ}\), so \(x=\) shorter leg \(\times\sqrt{3}\), and hypotenuse \(y = 2\times\) shorter leg.

Wait, shorter leg (opposite \(30^{\circ}\)) is 9. Then:

Hypotenuse \(y = 2\times9 = 18\)? No, wait, no. Wait, maybe the side of length 9 is adjacent to \(30^{\circ}\). Wait, let's use trigonometry.

\(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{9}{x}\)

\(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(\frac{1}{\sqrt{3}}=\frac{9}{x}\), then \(x = 9\sqrt{3}\)

\(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{9}{y}\)

\(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{9}{y}\), then \(y = 18\)

Wait, that makes sense. Let's re - check:

In a right triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\theta = 30^{\circ}\), opposite side to \(30^{\circ}\) is 9, adjacent is \(x\). So \(\tan(30^{\circ})=\frac{9}{x}\), \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(x = 9\sqrt{3}\)

\(\sin(30^{\circ})=\frac{9}{y}\), \(\sin(30^{\circ})=\frac{1}{2}\), so \(y = 18\)

Or using the ratio: shorter leg (opposite \(30^{\circ}\)) is 9, then hypotenuse (y) is \(2\times9 = 18\), and the longer leg (x) is \(9\sqrt{3}\)

Step3: Verify

Check with Pythagoras: \(9^{2}+(9\sqrt{3})^{2}=81 + 243=324\), and \(18^{2}=324\). So it works.

Answer:

\(x = 9\sqrt{3}\), \(y = 18\)