Question

find s.
10 in, 60°, s, 30°, right angle
write your answer in simplest radical form.
inches

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, opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shortest, and the hypotenuse is twice the shortest. Here, the side opposite \(30^{\circ}\) (let's say \(a\)) and the side opposite \(60^{\circ}\) is 10 in, and the hypotenuse is \(s\). Wait, no, let's check angles. The right angle, \(60^{\circ}\), and \(30^{\circ}\). So the side adjacent to \(60^{\circ}\) is 10 in (the leg), and we need to find the hypotenuse \(s\). In a right triangle, \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). \(\cos(60^{\circ})=\frac{1}{2}\), adjacent is 10, so \(\frac{1}{2}=\frac{10}{s}\), solving for \(s\), \(s = 20\)? Wait, no, maybe using sine. Wait, the side opposite \(30^{\circ}\) is the shorter leg. Wait, the angle of \(30^{\circ}\): the side opposite \(30^{\circ}\) is the leg with length (let's see), the other leg (opposite \(60^{\circ}\)) is 10. In 30 - 60 - 90 triangle, the sides are \(x\) (opposite \(30^{\circ}\)), \(x\sqrt{3}\) (opposite \(60^{\circ}\)), and \(2x\) (hypotenuse). So if the side opposite \(60^{\circ}\) is \(x\sqrt{3}=10\), then \(x = \frac{10}{\sqrt{3}}\), but that can't be. Wait, maybe I mixed up the angles. Wait, the right angle, one angle is \(60^{\circ}\), so the other is \(30^{\circ}\). So the side adjacent to \(30^{\circ}\) is the longer leg (opposite \(60^{\circ}\)), and the hypotenuse is \(s\). So \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\)? No, adjacent to \(30^{\circ}\) is the longer leg (opposite \(60^{\circ}\)), which is 10. Wait, \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{10}{s}\)? No, that would give \(s=\frac{20}{\sqrt{3}}\), which is not right. Wait, maybe using sine. \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), opposite to \(60^{\circ}\) is 10? No, the right angle is between the two legs. Wait, the triangle has a right angle, \(60^{\circ}\), and \(30^{\circ}\). So the legs are: one leg is 10 (let's say the leg adjacent to \(60^{\circ}\)), and the hypotenuse is \(s\). So \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\cos(60^{\circ}) = 0.5=\frac{10}{s}\), so \(s = 20\)? Wait, no, that seems too big. Wait, maybe I got the angle wrong. Wait, the angle of \(30^{\circ}\): the side opposite \(30^{\circ}\) is the shorter leg. So if the shorter leg is \(x\), then the longer leg is \(x\sqrt{3}\), hypotenuse \(2x\). So if the longer leg (opposite \(60^{\circ}\)) is 10, then \(x\sqrt{3}=10\), so \(x=\frac{10}{\sqrt{3}}\), and hypotenuse \(2x=\frac{20}{\sqrt{3}}=\frac{20\sqrt{3}}{3}\)? No, that contradicts. Wait, maybe the 10 is the shorter leg. Wait, the angle of \(60^{\circ}\): if the side adjacent to \(60^{\circ}\) is the shorter leg (opposite \(30^{\circ}\)), then \(x = 10\), so hypotenuse is \(2x = 20\), and the longer leg is \(10\sqrt{3}\). Ah, that makes sense. So the side with length 10 is the shorter leg (opposite \(30^{\circ}\)), so the hypotenuse \(s\) is twice that, so \(s = 20\)? Wait, no, the angle of \(30^{\circ}\) is at the bottom, so the side opposite \(30^{\circ}\) is the top leg (the one with length 10)? Wait, the diagram: right angle at the left, \(60^{\circ}\) at the top, \(30^{\circ}\) at the bottom. So the left leg is 10 (adjacent to \(60^{\circ}\) and \(30^{\circ}\)? No, the left leg is between the right angle and \(60^{\circ}\), so it's adjacent to \(60^{\circ}\), and opposite to \(30^{\circ}\). So the left leg (length 10) is opposite \(30^{\circ}\), so it's the shorter leg. Therefore, the hypotenuse \(s\) is twice the shorter leg, so \(s = 2\times10 = 20\). Wait, but let's check with trigonometry. \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{10}{s}\), \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{10}{s}\), so \(s = 20\). Yes, that's correct.

Step2: Verify with 30 - 60 - 90 ratios

In a 30 - 60 - 90 triangle, the hypotenuse is twice the length of the shorter leg (the leg opposite the \(30^{\circ}\) angle). Here, the leg opposite the \(30^{\circ}\) angle is 10 inches (since the \(30^{\circ}\) angle is at the bottom, the leg opposite to it is the left leg with length 10). Therefore, the hypotenuse \(s\) (opposite the right angle) is \(2\times10 = 20\) inches.

Answer:

\(20\)