Question

find z.
right triangle with angles 30°, 60°, 90°, one leg 7 km, hypotenuse z
write your answer in simplest radical form.
blank kilometers
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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). The side of length 7 km is opposite the 30° angle? Wait, no. Wait, the right angle is at the top. Let's label the triangle: let the right angle be \(C\), 60° at \(B\), 30° at \(A\). Then side \(BC = 7\) km (adjacent to 60°, opposite 30°), side \(z\) is the hypotenuse? Wait, no, wait. Wait, in a 30-60-90 triangle, the hypotenuse is twice the shorter leg (opposite 30°). Wait, the angle at the bottom is 30°, so the side opposite 30° is the side with length 7 km? Wait, no, let's check angles. The triangle has angles 30°, 60°, 90°. So the side opposite 30° is the shortest side. Let's see: the side with length 7 km is adjacent to the 60° angle and opposite the 30° angle? Wait, no, the right angle is at the top, so the sides: the leg adjacent to 60° is 7 km, the leg opposite 60° is the other leg, and the hypotenuse is \(z\). Wait, in a 30-60-90 triangle, the hypotenuse \(c\) is related to the shorter leg (opposite 30°) \(a\) by \(c = 2a\), and the longer leg (opposite 60°) \(b = a\sqrt{3}\). Wait, maybe I got the angles wrong. Wait, the angle at the bottom is 30°, so the side opposite 30° is the side with length 7 km? Wait, no, the right angle is at the top, so the two legs are: one leg is 7 km (between 60° and right angle), the other leg is between 30° and right angle, and the hypotenuse is \(z\) (between 30° and 60°). Wait, the angle at 60°: in a triangle, angles sum to 180°, so 90 + 60 + 30 = 180, correct. So the side opposite 30° is the leg of length 7 km? Wait, no, the side opposite 30° would be the side between 60° and right angle? Wait, no, let's use trigonometry. Let's use sine or cosine. Let's take angle 60°: \(\cos(60°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{z}\). Since \(\cos(60°) = \frac{1}{2}\), so \(\frac{1}{2} = \frac{7}{z}\), so \(z = 14\)? Wait, no, that can't be. Wait, maybe angle 30°: \(\cos(30°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{z \cos(30°)}{z}\)? Wait, no, let's re-express. Wait, the side with length 7 km is adjacent to the 60° angle, so adjacent side is 7, hypotenuse is \(z\). So \(\cos(60°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{7}{z}\). Since \(\cos(60°) = \frac{1}{2}\), then \(\frac{1}{2} = \frac{7}{z}\) → \(z = 14\)? But that's not a radical. Wait, maybe I mixed up the angle. Wait, maybe the side of length 7 km is opposite the 60° angle. Let's try that. Then \(\sin(60°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{z}\). \(\sin(60°) = \frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2} = \frac{7}{z}\) → \(z = \frac{14}{\sqrt{3}} = \frac{14\sqrt{3}}{3}\)? No, that doesn't match. Wait, maybe the side of length 7 km is the longer leg (opposite 60°). Wait, in a 30-60-90 triangle, longer leg (opposite 60°) is \(a\sqrt{3}\), shorter leg (opposite 30°) is \(a\), hypotenuse is \(2a\). So if the longer leg is 7 km, then \(a\sqrt{3} = 7\) → \(a = \frac{7}{\sqrt{3}}\), then hypotenuse \(z = 2a = \frac{14}{\sqrt{3}} = \frac{14\sqrt{3}}{3}\)? But that doesn't seem right. Wait, maybe I got the angles reversed. Wait, the angle at the bottom is 30°, so the side opposite 30° is the shorter leg. Let's look at the triangle again: the right angle is at the top, so the two legs are: one leg is from top right (60°) to top (right angle) to bottom left (30°), and the other leg is from top (right angle) to top left (60°) to bottom (30°). Wait, no, the labels: let's call the vertices: \(A\) (bottom, 30°), \(B\) (top right, 60°), \(C\) (top, right angle). So sides: \(AC\) is from \(A\) to \(C\) (leg), \(BC\) is from \(B\) to \(C\) (leg, length 7 km), \(AB\) is the hypotenuse \(z\). Angle at \(B\) is 60°, so in triangle \(ABC\), angle at \(B\) is 60°, angle at \(A\) is 30°, angle at \(C\) is 90°. So side \(BC = 7\) km (adjacent to angle \(B\), opposite angle \(A\)). So side \(BC\) is opposite angle \(A\) (30°), so \(BC = 7 = \text{opposite 30°}\), so it's the shorter leg. Then the hypotenuse \(AB = z = 2 \times \text{shorter leg} = 2 \times 7 = 14\) km? Wait, that makes sense. Because in a 30-60-90 triangle, hypotenuse is twice the shorter leg (opposite 30°). So if the shorter leg (opposite 30°) is 7 km, then hypotenuse \(z = 2 \times 7 = 14\) km. Wait, but that's not a radical. But the problem says "simplest radical form". Wait, maybe I made a mistake. Wait, maybe the side of length 7 km is the longer leg (opposite 60°). Let's check: if the longer leg (opposite 60°) is 7 km, then shorter leg \(a = \frac{7}{\sqrt{3}}\), hypotenuse \(z = 2a = \frac{14}{\sqrt{3}} = \frac{14\sqrt{3}}{3}\). But that's a radical. Wait, the problem says "simplest radical form", so maybe that's the case. Wait, let's use trigonometry. Let's use cosine of 60°: \(\cos(60°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{7}{z}\). \(\cos(60°) = 0.5 = \frac{1}{2}\), so \(\frac{1}{2} = \frac{7}{z}\) → \(z = 14\). But that's not a radical. Alternatively, use sine of 30°: \(\sin(30°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{7}{z}\). \(\sin(30°) = \frac{1}{2}\), so same result. Wait, maybe the side of length 7 km is adjacent to 30°? Wait, no, angle at \(A\) is 30°, so adjacent side to \(A\) is \(AC\), opposite is \(BC\). So \(BC = 7\) is opposite 30°, so hypotenuse is 14. But the problem says "simplest radical form", which makes me think maybe I messed up. Wait, maybe the triangle is labeled differently. Wait, the angle at the bottom is 30°, so the side opposite 30° is the leg with length 7 km, and the hypotenuse is \(z\), so \(z = 2 \times 7 = 14\). But 14 is an integer, no radical. Maybe the problem has a typo, or I misread the angle. Wait, maybe the angle at the top right is 30° and bottom is 60°? Wait, no, the diagram shows 30° at the bottom and 60° at the top right. Wait, let's check the triangle again. The right angle is at the top, so the two legs are: one leg (horizontal) is 7 km (between 60° and right angle), the other leg (vertical) is between 30° and right angle, and the hypotenuse is \(z\) (between 30° and 60°). So angle at 60°: the side adjacent to 60° is 7 km (horizontal leg), the side opposite 60° is the vertical leg, and hypotenuse is \(z\). In a 30-60-90 triangle, the horizontal leg (adjacent to 60°) is the shorter leg? Wait, no, 60° is larger than 30°, so the side opposite 60° is longer. Wait, no, in a right triangle, the larger angle has the longer opposite side. So 60° is larger than 30°, so the side opposite 60° (vertical leg) is longer than the side opposite 30° (horizontal leg, 7 km). So horizontal leg (opposite 30°) is 7 km, so hypotenuse is 14 km, vertical leg is \(7\sqrt{3}\) km. So \(z\) is the hypotenuse, so \(z = 14\) km. But that's not a radical. Wait, maybe the problem meant the other leg? No, the problem says "Find z", which is the side labeled \(z\) in the diagram, which is the hypotenuse. So maybe the answer is 14, even though it's not a radical. Wait, but the problem says "simplest radical form", which is confusing. Wait, maybe I made a mistake in angle identification. Let's re-express:

In triangle \(ABC\), right-angled at \(C\), \(\angle A = 30^\circ\), \(\angle B = 60^\circ\), \(BC = 7\) km (side opposite \(\angle A\)). Then by definition of sine: \(\sin(\angle A) = \frac{BC}{AB}\) → \(\sin(30^\circ) = \frac{7}{z}\) → \(\frac{1}{2} = \frac{7}{z}\) → \(z = 14\) km. So that's the answer. Maybe the "simplest radical form" is just 14, since it's an integer.

Step2: Apply 30-60-90 triangle ratios

In a 30-60-90 right triangle, the hypotenuse is twice the length of the shorter leg (the leg opposite the 30° angle). Here, the leg opposite the 30° angle is 7 km (since \(\angle A = 30^\circ\) and \(BC\) is opposite \(\angle A\)). Thus, the hypotenuse \(z = 2 \times 7 = 14\) km.

Answer:

\(14\)