Question

find k.
(image of a right triangle with angles 30°, 60°, 90°, one leg 7 ft, the other leg k)
write your answer in simplest radical form.
feet
√
submit

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 (let's call it \(x\)), the side opposite \(60^\circ\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\).

Step2: Determine which side is given

The side of length 7 ft is opposite the \(30^\circ\) angle? Wait, no. Wait, the right angle, the \(30^\circ\) angle, and the \(60^\circ\) angle. Let's label the triangle: let the right angle be \(C\), \(30^\circ\) be \(A\), \(60^\circ\) be \(B\). Then side opposite \(A\) (30°) is \(BC\), opposite \(B\) (60°) is \(AC\), and hypotenuse \(AB\) is \(k\). Wait, the given side is 7 ft, which is adjacent to the \(60^\circ\) angle? Wait, no, looking at the diagram: the right angle is at the top right, \(30^\circ\) at the top left, \(60^\circ\) at the bottom, and the side adjacent to \(60^\circ\) is 7 ft? Wait, no, let's use trigonometry. In a right triangle, \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}\). Wait, the side opposite \(60^\circ\) is \(k\)? No, wait, the right angle, so the two legs: one is 7 ft (adjacent to \(60^\circ\)), and the other leg (opposite \(60^\circ\)) is \(k\)? Wait, no, let's use sine or cosine. Wait, angle \(60^\circ\): \(\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos(60^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, the side of 7 ft: let's see, the angle at the bottom is \(60^\circ\), the right angle is at the top right, so the side of 7 ft is adjacent to the \(60^\circ\) angle, and the hypotenuse is \(k\). Wait, no, \(\cos(60^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\), so \(\cos(60^\circ)=\frac{7}{k}\)? But \(\cos(60^\circ)=\frac{1}{2}\), so \(\frac{1}{2}=\frac{7}{k}\), then \(k = 14\)? No, that can't be. Wait, maybe I got the angles wrong. Wait, the angle at the top left is \(30^\circ\), bottom is \(60^\circ\), right angle at top right. So the side opposite \(30^\circ\) is 7 ft? Because in a 30-60-90 triangle, the side opposite 30° is half the hypotenuse. So if the side opposite 30° is 7 ft, then hypotenuse \(k = 2\times7 = 14\)? No, wait, no. Wait, the side opposite 30° is the shortest side. Wait, let's re-examine: the angle of \(30^\circ\) has its opposite side as 7 ft? Then hypotenuse is \(2\times7 = 14\), and the other leg is \(7\sqrt{3}\). But in the diagram, the side labeled 7 ft is adjacent to the \(60^\circ\) angle. Wait, maybe the side of 7 ft is opposite the \(30^\circ\) angle. Let's confirm: in a 30-60-90 triangle, sides are in ratio \(x : x\sqrt{3} : 2x\), where \(x\) is opposite 30°, \(x\sqrt{3}\) opposite 60°, and \(2x\) hypotenuse. If the side opposite 30° is 7 ft, then hypotenuse is \(2x = 14\), and the side opposite 60° is \(7\sqrt{3}\). But wait, the problem says "Find k", which is the hypotenuse? Wait, no, looking at the diagram: the side labeled \(k\) is one of the legs? Wait, no, the right angle is at the top right, so the two legs are the vertical and horizontal, and the hypotenuse is \(k\) (the slant side from top left to bottom). The side of 7 ft is the vertical leg (adjacent to the \(30^\circ\) angle). Wait, angle at top left is \(30^\circ\), so \(\sin(30^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{7}{k}\). Since \(\sin(30^\circ)=\frac{1}{2}\), then \(\frac{1}{2}=\frac{7}{k}\), so \(k = 14\)? No, that's not right. Wait, maybe the side of 7 ft is opposite the \(60^\circ\) angle. Wait, let's use \(\tan(60^\circ)=\frac{k}{7}\), since \(\tan(60^\circ)=\sqrt{3}\), so \(k = 7\sqrt{3}\). Ah, that makes sense. Because in the right triangle, angle \(60^\circ\) has opposite side \(k\) and adjacent side 7 ft. So \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{k}{7}\). Since \(\tan(60^\circ)=\sqrt{3}\), then \(k = 7\sqrt{3}\). Wait, let's check: \(\tan(60^\circ)=\sqrt{3}\), so \(\sqrt{3}=\frac{k}{7}\), so \(k = 7\sqrt{3}\). Yes, that's correct. Because the side adjacent to \(60^\circ\) is 7 ft, and the side opposite is \(k\), so tangent of 60° is opposite over adjacent, so \(k = 7\tan(60^\circ)=7\sqrt{3}\).

Answer:

\(7\sqrt{3}\)