Question

find t.
right triangle with 30°, 60°, right angle; side adjacent to 30° is ( 7sqrt{3} ) in, hypotenuse is t
write your answer in simplest radical form.
blank inches
radical symbol button

Explanation:

Step1: Identify triangle type

This is a 30-60-90 right triangle. In such a triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\) (opposite 30°, 60°, 90° respectively).

Step2: Relate sides to angles

The side \(7\sqrt{3}\) in is opposite the 30° angle? Wait, no. Wait, the angle of 30°: the side opposite 30° is the shorter leg, opposite 60° is the longer leg, and hypotenuse is opposite 90°. Wait, the right angle is between the two legs. The angle of 30°: the side adjacent to 30° is \(t\)? Wait, no. Let's use trigonometry. Let's see, the angle of 60°: the side adjacent to 60° is \(7\sqrt{3}\)? Wait, no. Wait, the right triangle has angles 30°, 60°, 90°. Let's label the sides: let the side opposite 30° be \(x\), opposite 60° be \(x\sqrt{3}\), hypotenuse \(2x\). Wait, in the triangle, the side with length \(7\sqrt{3}\) is adjacent to the 60° angle? Wait, no. Wait, the angle of 60°: the side opposite to 60° is \(t\)? Wait, no. Wait, let's use tangent. Wait, angle 30°: \(\tan(30°)=\frac{\text{opposite}}{\text{adjacent}}=\frac{7\sqrt{3}}{t}\)? Wait, no. Wait, \(\tan(30°)=\frac{1}{\sqrt{3}}\), \(\tan(60°)=\sqrt{3}\). Wait, the angle of 30°: the side opposite is \(7\sqrt{3}\)? No, wait the side with \(7\sqrt{3}\) is adjacent to 30°? Wait, no. Wait, the right angle is at the top. So the two legs: one leg is \(t\) (adjacent to 30°), the other leg is \(7\sqrt{3}\) (opposite to 30°). Wait, \(\tan(30°)=\frac{\text{opposite}}{\text{adjacent}}=\frac{7\sqrt{3}}{t}\). But \(\tan(30°)=\frac{1}{\sqrt{3}}\). So \(\frac{1}{\sqrt{3}}=\frac{7\sqrt{3}}{t}\). Cross-multiplying: \(t = 7\sqrt{3}\times\sqrt{3}=7\times3 = 21\)? Wait, no, that can't be. Wait, maybe I mixed up. Wait, angle 60°: \(\tan(60°)=\frac{\text{opposite}}{\text{adjacent}}=\frac{t}{7\sqrt{3}}\). Since \(\tan(60°)=\sqrt{3}\), so \(\sqrt{3}=\frac{t}{7\sqrt{3}}\). Then \(t = 7\sqrt{3}\times\sqrt{3}=7\times3 = 21\)? Wait, no, that's not right. Wait, no, in a 30-60-90 triangle, the sides are in ratio \(1 : \sqrt{3} : 2\). The side opposite 30° is the shortest side. Wait, the angle of 30°: the side opposite is \(7\sqrt{3}\)? No, that would make the hypotenuse \(14\sqrt{3}\), and the other leg \(21\). Wait, no, maybe the side \(7\sqrt{3}\) is the longer leg (opposite 60°), so the shorter leg (opposite 30°) is \(7\), and hypotenuse \(14\). Wait, no, the problem is to find \(t\). Wait, let's use sine. \(\sin(30°)=\frac{\text{opposite}}{\text{hypotenuse}}\), but \(t\) is a leg. Wait, \(\cos(30°)=\frac{\text{adjacent}}{\text{hypotenuse}}\), but \(t\) is adjacent to 30°. Wait, maybe the side \(7\sqrt{3}\) is opposite the 30° angle. Then \(\sin(30°)=\frac{7\sqrt{3}}{\text{hypotenuse}}\), but \(\sin(30°)=\frac{1}{2}\), so hypotenuse would be \(14\sqrt{3}\), and the other leg (adjacent to 30°) would be \(14\sqrt{3}\times\cos(30°)=14\sqrt{3}\times\frac{\sqrt{3}}{2}=14\times\frac{3}{2}=21\). But that's not matching. Wait, maybe the side \(7\sqrt{3}\) is adjacent to the 60° angle. So for angle 60°, the adjacent side is \(7\sqrt{3}\), and the opposite side is \(t\). Then \(\tan(60°)=\frac{t}{7\sqrt{3}}\). Since \(\tan(60°)=\sqrt{3}\), so \(t = 7\sqrt{3}\times\sqrt{3}=7\times3 = 21\)? No, that's 21, but that seems big. Wait, no, maybe I got the angles wrong. Wait, the triangle has angles 30°, 60°, 90°. The side opposite 30° is the shortest side. Let's say the side opposite 30° is \(x\), then opposite 60° is \(x\sqrt{3}\), hypotenuse \(2x\). In the diagram, the side with \(7\sqrt{3}\) is opposite the 30° angle? Then \(x = 7\sqrt{3}\), so the side opposite 60° (which is \(t\)) would be \(x\sqrt{3}=7\sqrt{3}\times\sqrt{3}=21\). Wait, but that would make the hypotenuse \(2x = 14\sqrt{3}\). But then the side \(t\) is 21. Alternatively, if the side \(7\sqrt{3}\) is the longer leg (opposite 60°), then the shorter leg (opposite 30°) is \(7\), and the hypotenuse is 14. But then \(t\) would be the hypotenuse? No, \(t\) is a leg. Wait, the diagram: the right angle is at the top, so the two legs are \(t\) (left leg) and \(7\sqrt{3}\) (right leg), and the base is the hypotenuse? No, no, in a right triangle, the hypotenuse is opposite the right angle. So the right angle is at the top, so the hypotenuse is the base (the side with 30° and 60° angles). So the two legs are \(t\) (vertical leg) and \(7\sqrt{3}\) (horizontal leg). Then angle at the bottom left is 30°, so \(\tan(30°)=\frac{\text{opposite}}{\text{adjacent}}=\frac{t}{7\sqrt{3}}\). \(\tan(30°)=\frac{1}{\sqrt{3}}\), so \(t = 7\sqrt{3}\times\frac{1}{\sqrt{3}}=7\). Wait, that makes sense! Oh, I see, I mixed up opposite and adjacent. The angle 30°: the opposite side is \(t\) (vertical leg), and the adjacent side is \(7\sqrt{3}\) (horizontal leg). So \(\tan(30°)=\frac{t}{7\sqrt{3}}\), so \(t = 7\sqrt{3}\times\tan(30°)=7\sqrt{3}\times\frac{1}{\sqrt{3}}=7\). Yes, that's correct. Because \(\tan(30°)=\frac{\text{opposite}}{\text{adjacent}}=\frac{t}{7\sqrt{3}}\), so solving for \(t\) gives \(t = 7\sqrt{3}\times\frac{1}{\sqrt{3}} = 7\).

Answer:

7