Question

find k.
write your answer in simplest radical form.
meters

Explanation:

Step1: Identify triangle type

This is a right - triangle with angles \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). In a \(30 - 60 - 90\) triangle, the ratios of the sides are \(1:\sqrt{3}:2\) (opposite to \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\) respectively). Let the side opposite \(30^{\circ}\) be \(x\), opposite \(60^{\circ}\) be \(x\sqrt{3}\), and hypotenuse be \(2x\).

Step2: Relate given side to ratio

The side adjacent to \(30^{\circ}\) and opposite to \(60^{\circ}\) is \(19\sqrt{3}\) m. In the \(30 - 60 - 90\) triangle ratio, the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the side opposite \(30^{\circ}\) is \(x\), and the hypotenuse (the side we need to find \(k\) which is opposite \(90^{\circ}\) or adjacent to \(30^{\circ}\)? Wait, no. Wait, the side with length \(19\sqrt{3}\) is adjacent to the \(30^{\circ}\) angle? Wait, no. Let's use trigonometry. Let's consider the angle of \(30^{\circ}\). The side adjacent to \(30^{\circ}\) is \(19\sqrt{3}\), and the side \(k\) is the hypotenuse? Wait, no. Wait, in the right - triangle, the angles are \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). Let's use cosine of \(30^{\circ}\). \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, the adjacent side to \(30^{\circ}\) is \(19\sqrt{3}\), and the hypotenuse is \(k\)? No, wait, \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), and \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, maybe better to use tangent or sine. Wait, the side with length \(19\sqrt{3}\) is opposite the \(60^{\circ}\) angle. In a \(30 - 60 - 90\) triangle, the side opposite \(60^{\circ}\) is \(\sqrt{3}\) times the side opposite \(30^{\circ}\), and the hypotenuse is twice the side opposite \(30^{\circ}\). Let the side opposite \(30^{\circ}\) be \(a\), then the side opposite \(60^{\circ}\) is \(a\sqrt{3}\), and hypotenuse is \(2a\). We know that \(a\sqrt{3}=19\sqrt{3}\), so \(a = 19\). Then the hypotenuse (which is \(k\)) is \(2a=38\)? Wait, no, wait. Wait, maybe I mixed up. Wait, the side \(k\) is the hypotenuse? Wait, no, the side \(k\) is one of the legs? Wait, no, the right angle is at the bottom - right. So the sides: the side with length \(19\sqrt{3}\) is one leg (adjacent to \(30^{\circ}\)), the side \(k\) is the other leg (opposite to \(60^{\circ}\))? No, wait, no. Let's use trigonometry. Let's take the angle of \(30^{\circ}\). \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\). The opposite side to \(30^{\circ}\) is \(x\), adjacent is \(19\sqrt{3}\). \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}=\frac{x}{19\sqrt{3}}\), so \(x = 19\). Then the hypotenuse (which is \(k\)) is \(2x = 38\)? Wait, no, the hypotenuse is the side opposite the right angle. Wait, the right angle is at the bottom - right, so the hypotenuse is the side opposite the right angle, which is the side between the \(30^{\circ}\) and \(60^{\circ}\) angles. Wait, maybe I made a mistake. Let's use sine of \(60^{\circ}\). \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(60^{\circ}\) is \(19\sqrt{3}\), and the hypotenuse is \(k\). \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{19\sqrt{3}}{k}\). Cross - multiply: \(k\times\sqrt{3}=2\times19\sqrt{3}\). Divide both sides by \(\sqrt{3}\): \(k = 38\). Wait, that's simpler. Using \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), where opposite side to \(60^{\circ}\) is \(19\sqrt{3}\), hypotenuse is \(k\). So \(\sin(60^{\circ})=\frac{19\sqrt{3}}{k}\), and \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\). So \(\frac{\sqrt{3}}{2}=\frac{19\sqrt{3}}{k}\). Multiply both sides by \(k\): \(\frac{\sqrt{3}}{2}k = 19\sqrt{3}\). Divide both sides by \(\sqrt{3}\): \(\frac{1}{2}k=19\). Multiply both sides by \(2\): \(k = 38\).

Answer:

\(38\)