Question

find d.
right triangle with right angle, one angle 30°, one angle 60°, side adjacent to 60° is ( 3sqrt{3} ) mi, hypotenuse is d
write your answer in simplest radical form.
blank miles
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Explanation:

Step1: Identify the triangle type

This is a right - triangle with angles \(30^{\circ}\), \(60^{\circ}\), and \(90^{\circ}\). 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 one, the side opposite \(60^{\circ}\) is \(\sqrt{3}\) times the shortest side, and the hypotenuse (opposite \(90^{\circ}\)) is twice the shortest side.

We are given the side opposite \(60^{\circ}\) which is \(3\sqrt{3}\) miles. Let the side opposite \(30^{\circ}\) be \(x\). We know that the side opposite \(60^{\circ}=\sqrt{3}\times\) (side opposite \(30^{\circ}\)). So, \(3\sqrt{3}=\sqrt{3}x\). Solving for \(x\), we divide both sides by \(\sqrt{3}\), getting \(x = 3\) miles.

Step2: Find the hypotenuse \(d\)

The hypotenuse \(d\) (opposite \(90^{\circ}\)) is twice the side opposite \(30^{\circ}\). Since the side opposite \(30^{\circ}\) is \(3\) miles, then \(d = 2\times3=6\)? Wait, no, wait. Wait, maybe I mixed up the sides. Wait, let's use trigonometry. Let's consider the angle of \(60^{\circ}\). The side adjacent to \(60^{\circ}\) is \(3\sqrt{3}\)? No, wait, the right - angle is between the side of length \(3\sqrt{3}\) and the other leg, and \(d\) is the hypotenuse. Wait, let's use cosine. \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(60^{\circ}\) is \(3\sqrt{3}\)? No, wait, \(\cos(60^{\circ})=\frac{1}{2}\), and if we take the angle of \(30^{\circ}\), \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). Wait, the side of length \(3\sqrt{3}\) is opposite to \(30^{\circ}\)? No, wait, in a right - triangle, the side opposite \(30^{\circ}\) is the shortest side. Let's label the triangle: let the right angle be at \(C\), angle at \(A = 30^{\circ}\), angle at \(B=60^{\circ}\), so side \(BC\) (opposite \(A\)) is the shortest side, side \(AC\) (opposite \(B\)) is \(3\sqrt{3}\), and side \(AB = d\) (hypotenuse).

Using the sine function: \(\sin(60^{\circ})=\frac{AC}{AB}\). We know that \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), \(AC = 3\sqrt{3}\), and \(AB = d\). So, \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{d}\). Cross - multiply: \(\sqrt{3}d=2\times3\sqrt{3}\). Divide both sides by \(\sqrt{3}\): \(d = 6\)? Wait, no, that can't be. Wait, or use cosine of \(30^{\circ}\). \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(30^{\circ}\) is \(3\sqrt{3}\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\). So, \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{d}\). Cross - multiplying gives \(\sqrt{3}d = 2\times3\sqrt{3}\), so \(d = 6\)? Wait, but let's check with the \(30 - 60 - 90\) triangle ratios. In a \(30 - 60 - 90\) triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite \(30^{\circ}\) is \(x\), opposite \(60^{\circ}\) is \(x\sqrt{3}\), and hypotenuse is \(2x\). If the side opposite \(60^{\circ}\) is \(3\sqrt{3}\), then \(x\sqrt{3}=3\sqrt{3}\), so \(x = 3\) (side opposite \(30^{\circ}\)), and hypotenuse \(d = 2x=6\)? Wait, no, that seems wrong. Wait, no, maybe the side of length \(3\sqrt{3}\) is opposite \(30^{\circ}\). Then the side opposite \(60^{\circ}\) would be \(3\sqrt{3}\times\sqrt{3}=9\), and hypotenuse would be \(2\times3\sqrt{3}=6\sqrt{3}\). Ah! Here is the mistake. I mixed up which side is which.

Let's start over. Let the right - triangle have angles: \(30^{\circ}\), \(60^{\circ}\), \(90^{\circ}\). Let the side opposite \(30^{\circ}\) be \(a\), opposite \(60^{\circ}\) be \(b\), and hypotenuse \(c\). Then \(a:b:c = 1:\sqrt{3}:2\), so \(b=a\sqrt{3}\), \(c = 2a\).

Now, in the given triangle, the side with length \(3\sqrt{3}\) is opposite the \(30^{\circ}\) angle? No, if the side opposite \(30^{\circ}\) is \(a\), then \(b=a\sqrt{3}\). If \(a = 3\sqrt{3}\), then \(b=3\sqrt{3}\times\sqrt{3}=9\), and \(c = 2\times3\sqrt{3}=6\sqrt{3}\). Wait, now I am confused. Let's use trigonometry with angle \(30^{\circ}\). \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). Let the side opposite \(30^{\circ}\) be \(3\sqrt{3}\), then \(\sin(30^{\circ})=\frac{3\sqrt{3}}{d}\), but \(\sin(30^{\circ})=\frac{1}{2}\), so \(\frac{1}{2}=\frac{3\sqrt{3}}{d}\), then \(d = 6\sqrt{3}\). Wait, that makes more sense.

Wait, let's look at the triangle again. The right angle is between the side of length \(3\sqrt{3}\) and the other leg, and \(d\) is the hypotenuse. The angle at the bottom is \(60^{\circ}\), so the side of length \(3\sqrt{3}\) is adjacent to the \(60^{\circ}\) angle. So, \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). \(\cos(60^{\circ})=\frac{1}{2}\), adjacent side is \(3\sqrt{3}\), hypotenuse is \(d\). So, \(\frac{1}{2}=\frac{3\sqrt{3}}{d}\), then \(d = 6\sqrt{3}\)? No, wait, \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), if the adjacent side to \(60^{\circ}\) is \(3\sqrt{3}\), then \(\frac{1}{2}=\frac{3\sqrt{3}}{d}\), so \(d = 6\sqrt{3}\)? Wait, no, I think I had the adjacent and opposite wrong. Let's label the triangle properly:

- Let the right angle be at vertex \(C\).
- Angle at vertex \(A = 30^{\circ}\), angle at vertex \(B = 60^{\circ}\).
- Side \(BC\): opposite angle \(A\) ( \(30^{\circ}\) ), length \(x\).
- Side \(AC\): opposite angle \(B\) ( \(60^{\circ}\) ), length \(3\sqrt{3}\).
- Side \(AB\): hypotenuse, length \(d\).

Using the sine rule: \(\frac{AC}{\sin(30^{\circ})}=\frac{AB}{\sin(90^{\circ})}\). Wait, no, sine rule is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). So, \(\frac{AC}{\sin A}=\frac{AB}{\sin C}\). \(AC = 3\sqrt{3}\), \(\sin A=\sin(30^{\circ})=\frac{1}{2}\), \(\sin C=\sin(90^{\circ}) = 1\), \(AB = d\). So, \(\frac{3\sqrt{3}}{\frac{1}{2}}=\frac{d}{1}\), so \(d=3\sqrt{3}\times2 = 6\sqrt{3}\).

Yes, that's correct. Because in the sine rule, \(\frac{AC}{\sin(30^{\circ})}=\frac{AB}{\sin(90^{\circ})}\), so \(AB=\frac{AC\times\sin(90^{\circ})}{\sin(30^{\circ})}=\frac{3\sqrt{3}\times1}{\frac{1}{2}} = 6\sqrt{3}\).

Answer:

\(6\sqrt{3}\)