Question

find c.
right triangle with angles 60°, 30°, right angle; side adjacent to 60° is √2 m, hypotenuse is c
write your answer in simplest radical form.
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Explanation:

Step1: Identify triangle type

This is a right - triangle with angles \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). In a \(30 - 60 - 90\) right - triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (opposite to \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\) respectively) or we can also use trigonometric ratios. Let's use the cosine function. The side adjacent to the \(60^{\circ}\) angle is \(\sqrt{2}\) m, and the hypotenuse is \(c\). We know that \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\theta = 60^{\circ}\), \(\cos60^{\circ}=\frac{1}{2}\), and the adjacent side to \(60^{\circ}\) is \(\sqrt{2}\) m, and hypotenuse is \(c\). So \(\cos60^{\circ}=\frac{\sqrt{2}}{c}\)

Step2: Solve for \(c\)

Since \(\cos60^{\circ}=\frac{1}{2}\), we have the equation \(\frac{1}{2}=\frac{\sqrt{2}}{c}\). Cross - multiplying gives us \(c\times1 = 2\times\sqrt{2}\), so \(c = 2\sqrt{2}\)? Wait, no, wait. Wait, maybe I mixed up the angles. Let's re - examine the triangle. The right angle is between the side of length \(\sqrt{2}\) and the other leg. The angle of \(60^{\circ}\) is at the vertex with the side \(\sqrt{2}\). So the side \(\sqrt{2}\) is adjacent to the \(60^{\circ}\) angle, and the hypotenuse is \(c\). Wait, or maybe we can use the sine function. The side \(\sqrt{2}\) is opposite to the \(30^{\circ}\) angle? Wait, no. Let's find the angles: the right angle is \(90^{\circ}\), one angle is \(60^{\circ}\), so the third angle is \(30^{\circ}\). So the side opposite \(30^{\circ}\) is the shortest side. Wait, the side with length \(\sqrt{2}\) is adjacent to \(60^{\circ}\), so let's use \(\sin\) for the \(30^{\circ}\) angle. \(\sin30^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}\). The side opposite \(30^{\circ}\) is the side with length \(\sqrt{2}\)? No, wait, no. Wait, in a right - triangle, the side opposite \(30^{\circ}\) is half the hypotenuse. Wait, maybe I made a mistake in identifying the sides. Let's label the triangle: let the right - angle be at \(B\), angle at \(A\) be \(60^{\circ}\), angle at \(C\) be \(30^{\circ}\), and \(AB=\sqrt{2}\) m, \(BC\) is the other leg, and \(AC = c\) (hypotenuse). Then, \(\cos A=\frac{AB}{AC}\), where \(A = 60^{\circ}\), \(AB=\sqrt{2}\), \(AC = c\). \(\cos60^{\circ}=\frac{\sqrt{2}}{c}\), and \(\cos60^{\circ}=\frac{1}{2}\), so \(\frac{1}{2}=\frac{\sqrt{2}}{c}\), so \(c = 2\sqrt{2}\)? Wait, no, that can't be. Wait, maybe I got the adjacent and opposite wrong. Let's use the sine of \(60^{\circ}\). \(\sin60^{\circ}=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(60^{\circ}\) is \(BC\), and hypotenuse is \(c\). \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\). But we know the adjacent side is \(\sqrt{2}\). Wait, another approach: in a \(30 - 60 - 90\) triangle, the sides are in the ratio \(x:x\sqrt{3}:2x\), where \(x\) is the side opposite \(30^{\circ}\), \(x\sqrt{3}\) opposite \(60^{\circ}\), and \(2x\) hypotenuse. Let's see which side is which. The side of length \(\sqrt{2}\): if the angle of \(30^{\circ}\) is opposite to a side, then the side opposite \(30^{\circ}\) is \(x\), opposite \(60^{\circ}\) is \(x\sqrt{3}\), hypotenuse \(2x\). Wait, the side \(\sqrt{2}\) is adjacent to \(60^{\circ}\), so it should be equal to \(x\) (since adjacent to \(60^{\circ}\) is the side opposite \(30^{\circ}\)). Wait, no, in a right - triangle, the side adjacent to \(60^{\circ}\) is the side opposite \(30^{\circ}\). So if the side adjacent to \(60^{\circ}\) is \(\sqrt{2}\), then the side opposite \(30^{\circ}\) is \(\sqrt{2}\), so \(x=\sqrt{2}\), and the hypotenuse \(c = 2x=2\sqrt{2}\)? Wait, no, that's not right. Wait, let's use trigonometry correctly. Let's take angle \(60^{\circ}\): \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). Adjacent side to \(60^{\circ}\) is \(\sqrt{2}\), hypotenuse is \(c\). \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{\sqrt{2}}{c}\), so \(c = 2\sqrt{2}\)? Wait, but let's check with sine. \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(60^{\circ}\) is the other leg. \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{\text{opposite}}{c}\), so opposite \(=\frac{\sqrt{3}}{2}c\). And by Pythagoras, \((\sqrt{2})^2+(\frac{\sqrt{3}}{2}c)^2=c^2\). \(2+\frac{3}{4}c^2=c^2\), \(2=c^2-\frac{3}{4}c^2=\frac{1}{4}c^2\), so \(c^2 = 8\), \(c = 2\sqrt{2}\). Yes, that's correct.

Wait, another way: in a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). The angle of \(30^{\circ}\): the side opposite \(30^{\circ}\) is \(\sqrt{2}\) (wait, no, if the angle is \(30^{\circ}\), the side opposite is the shortest side. Wait, maybe I had the angle wrong. Wait, the triangle has angles \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\). Let's see the side given: \(\sqrt{2}\) is adjacent to \(60^{\circ}\), so the angle adjacent to \(\sqrt{2}\) is \(60^{\circ}\), so the angle opposite to \(\sqrt{2}\) is \(30^{\circ}\). So \(\sin(30^{\circ})=\frac{\sqrt{2}}{c}\), since \(\sin(30^{\circ})=\frac{1}{2}\), then \(\frac{1}{2}=\frac{\sqrt{2}}{c}\), so \(c = 2\sqrt{2}\). Yes, that's correct.

Answer:

\(2\sqrt{2}\)