Question

find the missing length(s) in each triangle. give the exact answer and the answer in simplest radical form. 1) triangle with hypotenuse 12√3, angle 30°, right angle, legs x and y x = type your answer... y = type your answer... √3

Explanation:

Step1: Identify the 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 hypotenuse is \(12\sqrt{3}\), and \(x\) is the side adjacent to \(30^{\circ}\) (opposite \(60^{\circ}\)), \(y\) is the side opposite \(30^{\circ}\).

Step2: Find the length of \(y\) (opposite \(30^{\circ}\))

In a 30 - 60 - 90 triangle, the hypotenuse \(h = 2y\) (since \(y\) is opposite \(30^{\circ}\)). Given \(h=12\sqrt{3}\), we have \(2y = 12\sqrt{3}\), so \(y=\frac{12\sqrt{3}}{2}=6\sqrt{3}\)? Wait, no, wait. Wait, the side opposite \(30^{\circ}\) is \(y\)? Wait, no, the right angle is between \(y\) and \(x\), so the hypotenuse is \(12\sqrt{3}\), angle \(30^{\circ}\) is at the end of \(x\) and hypotenuse. So the side opposite \(30^{\circ}\) is \(y\), side opposite \(60^{\circ}\) is \(x\), hypotenuse is \(12\sqrt{3}\). So hypotenuse \(= 2\times\) (side opposite \(30^{\circ}\)), so \(12\sqrt{3}=2y\), so \(y = 6\sqrt{3}\)? Wait, no, wait, maybe I mixed up. Wait, let's re - define: in a right triangle with angle \(30^{\circ}\), the side opposite \(30^{\circ}\) is the shortest side. Let's denote the side opposite \(30^{\circ}\) as \(a\), then hypotenuse is \(2a\), and the other leg (opposite \(60^{\circ}\)) is \(a\sqrt{3}\). Here, the hypotenuse is \(12\sqrt{3}\), so \(2a=12\sqrt{3}\), so \(a = 6\sqrt{3}\)? No, that can't be. Wait, no, maybe the side adjacent to \(30^{\circ}\) is \(x\), and the side opposite \(30^{\circ}\) is \(y\). So \(\cos(30^{\circ})=\frac{x}{\text{hypotenuse}}\), \(\sin(30^{\circ})=\frac{y}{\text{hypotenuse}}\).

\(\sin(30^{\circ})=\frac{1}{2}=\frac{y}{12\sqrt{3}}\), so \(y = 12\sqrt{3}\times\frac{1}{2}=6\sqrt{3}\)? Wait, no, that's not right. Wait, \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{y}{12\sqrt{3}}\), so \(y = 12\sqrt{3}\times\sin(30^{\circ})=12\sqrt{3}\times\frac{1}{2}=6\sqrt{3}\). And \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{12\sqrt{3}}\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(x = 12\sqrt{3}\times\frac{\sqrt{3}}{2}=\frac{12\times3}{2}=18\). Wait, that makes sense. Let's check: in a 30 - 60 - 90 triangle, if one leg is \(x\) (adjacent to \(30^{\circ}\)), one leg is \(y\) (opposite \(30^{\circ}\)), hypotenuse \(h\). Then:

\(\cos(30^{\circ})=\frac{x}{h}\), \(\sin(30^{\circ})=\frac{y}{h}\)

Given \(h = 12\sqrt{3}\), \(\sin(30^{\circ})=\frac{1}{2}\), so \(y=h\times\sin(30^{\circ})=12\sqrt{3}\times\frac{1}{2}=6\sqrt{3}\)? No, wait, no, \(\sin(30^{\circ})=\frac{1}{2}\), so \(y = 12\sqrt{3}\times\frac{1}{2}=6\sqrt{3}\)? But then the other leg \(x\) would be \(h\times\cos(30^{\circ})=12\sqrt{3}\times\frac{\sqrt{3}}{2}=\frac{12\times3}{2}=18\). Wait, let's verify with Pythagoras: \(x^{2}+y^{2}=h^{2}\). If \(x = 18\), \(y = 6\sqrt{3}\), then \(x^{2}+y^{2}=18^{2}+(6\sqrt{3})^{2}=324 + 108=432\), and \(h^{2}=(12\sqrt{3})^{2}=144\times3 = 432\). So that works.

Wait, but the problem says \(y=\text{[answer]}\sqrt{3}\). So \(y = 6\sqrt{3}\), so the number before \(\sqrt{3}\) is 6. And \(x = 18\). Let's re - do:

In 30 - 60 - 90 triangle:

- Side opposite \(30^{\circ}\): \(y\)
- Side opposite \(60^{\circ}\): \(x\)
- Hypotenuse: \(12\sqrt{3}\)

We know that hypotenuse \(= 2\times\) (side opposite \(30^{\circ}\)), so \(12\sqrt{3}=2y\), so \(y = 6\sqrt{3}\) (so the coefficient of \(\sqrt{3}\) is 6).

And side opposite \(60^{\circ}\) (\(x\)) \(=\sqrt{3}\times\) (side opposite \(30^{\circ}\)), so \(x=\sqrt{3}\times y=\sqrt{3}\times6\sqrt{3}=6\times3 = 18\).

Answer:

\(x = 18\)

\(y = 6\sqrt{3}\) (so the number in the \(y\) box is 6)