Question

find u.

write your answer in simplest radical form.

inches

Explanation:

Step1: Identify the triangle type

This is a right - angled triangle with angles \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\), so it is a 30 - 60 - 90 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 \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\).

Step2: Determine the sides relative to the angles

The side of length 7 inches is opposite the \(30^{\circ}\) angle? Wait, no. Wait, the right angle is at the bottom - right, the angle of \(60^{\circ}\) is at the bottom - left, and \(30^{\circ}\) is at the top. So the side adjacent to the \(60^{\circ}\) angle (the side with length 7 inches) is opposite the \(30^{\circ}\) angle? Wait, no. Let's label the triangle: let the right - angled vertex be \(C\), the \(60^{\circ}\) vertex be \(B\), and the \(30^{\circ}\) vertex be \(A\). So side \(BC = 7\) inches (adjacent to angle \(B = 60^{\circ}\)), side \(AC=u\) (adjacent to angle \(A = 30^{\circ}\)), and hypotenuse \(AB\).

In a right - angled triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). For angle \(B = 60^{\circ}\), \(\tan(60^{\circ})=\frac{AC}{BC}\). Since \(\tan(60^{\circ})=\sqrt{3}\) and \(BC = 7\) inches, we have \(\sqrt{3}=\frac{u}{7}\).

Step3: Solve for \(u\)

From \(\sqrt{3}=\frac{u}{7}\), we can solve for \(u\) by multiplying both sides by 7. So \(u = 7\sqrt{3}\)? Wait, no. Wait, maybe I mixed up the angles. Wait, in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is the shortest side. The side of length 7 is opposite the \(30^{\circ}\) angle? Wait, no. Let's use trigonometric ratios. \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), but we can also use \(\cot(60^{\circ})=\frac{\text{adjacent}}{\text{opposite}}\). Wait, \(\cot(60^{\circ})=\frac{1}{\sqrt{3}}\). The adjacent side to \(60^{\circ}\) is 7, and the opposite side is \(u\)? No, wait, the angle at the bottom is \(60^{\circ}\), the right angle is at the right. So the side with length 7 is adjacent to the \(60^{\circ}\) angle, and \(u\) is opposite the \(60^{\circ}\) angle? Wait, no. Let's use \(\tan(60^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\). The angle of \(60^{\circ}\): opposite side is \(u\), adjacent side is 7. Since \(\tan(60^{\circ})=\sqrt{3}\), then \(\tan(60^{\circ})=\frac{u}{7}\), so \(u = 7\tan(60^{\circ})\). Since \(\tan(60^{\circ})=\sqrt{3}\), then \(u = 7\sqrt{3}\)? Wait, no, wait. Wait, in a 30 - 60 - 90 triangle, if the side opposite \(30^{\circ}\) is \(x\), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\). If the side opposite \(30^{\circ}\) is 7, then the side opposite \(60^{\circ}\) is \(7\sqrt{3}\). Wait, maybe the side of length 7 is opposite \(30^{\circ}\). Let's check the angles: the angle at the top is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the side with length 7? No, the side opposite \(30^{\circ}\) should be the shortest side. Wait, the right angle is at the bottom - right, so the sides: the vertical side ( \(u\)) and the horizontal side (7) and the hypotenuse. The angle at the top is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the horizontal side (7 inches), the side opposite \(60^{\circ}\) is the vertical side (\(u\)), and the hypotenuse is \(2\times7 = 14\) inches. In a 30 - 60 - 90 triangle, the side opposite \(60^{\circ}\) is \(\sqrt{3}\) times the side opposite \(30^{\circ}\). So if the side opposite \(30^{\circ}\) is 7, then the side opposite \(60^{\circ}\) (which is \(u\)) is \(7\sqrt{3}\)? Wait, no, wait. Wait, \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), but we can also use \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, maybe a better approach: in a right - angled triangle, \(\tan(30^{\circ})=\frac{7}{u}\), because \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\) for angle \(30^{\circ}\). Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), we have \(\frac{1}{\sqrt{3}}=\frac{7}{u}\), then \(u = 7\sqrt{3}\). Wait, no, \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\), the opposite side to \(30^{\circ}\) is 7, and the adjacent side is \(u\). So \(\tan(30^{\circ})=\frac{7}{u}\), so \(u=\frac{7}{\tan(30^{\circ})}\). Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), then \(u = 7\sqrt{3}\). Wait, that makes sense.

Answer:

\(7\sqrt{3}\)