Question

find k.
right triangle with angles 30°, 60°, right angle; side adjacent to 30° is 3√3 in, hypotenuse is k
write your answer in simplest radical form.
blank inches

Explanation:

Step1: Identify 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 side (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 given is \(3\sqrt{3}\) inches, and it is opposite the \(30^{\circ}\) angle? Wait, no. Wait, the right angle is between the side of length \(3\sqrt{3}\) and the other leg. Wait, let's label the triangle. Let the right angle be at \(C\), the \(30^{\circ}\) angle at \(A\), and \(60^{\circ}\) angle at \(B\). Then side \(BC = 3\sqrt{3}\) (adjacent to \(60^{\circ}\), opposite to \(30^{\circ}\)), side \(AC\) is the other leg, and hypotenuse \(AB=k\).

In a 30 - 60 - 90 triangle, \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(60^{\circ}\) is \(3\sqrt{3}\), and hypotenuse is \(k\). We know that \(\cos(60^{\circ})=\frac{1}{2}\). So \(\cos(60^{\circ})=\frac{3\sqrt{3}}{k}\)

Step3: Solve for \(k\)

We have \(\frac{1}{2}=\frac{3\sqrt{3}}{k}\). Cross - multiplying gives \(k = 2\times3\sqrt{3}\)? Wait, no, that's wrong. Wait, maybe using sine. \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(30^{\circ}\) is \(3\sqrt{3}\)? Wait, no, let's re - examine.

Wait, in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is the shortest side. Let's say the side opposite \(30^{\circ}\) is \(a\), opposite \(60^{\circ}\) is \(a\sqrt{3}\), hypotenuse is \(2a\).

Looking at the triangle, the side with length \(3\sqrt{3}\) is opposite the \(30^{\circ}\) angle? No, wait the angle of \(30^{\circ}\) is at the vertex where the hypotenuse \(k\) and the other leg meet. Wait, the right angle is between the leg of length \(3\sqrt{3}\) and the leg opposite to \(30^{\circ}\). Wait, maybe using tangent. \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\). The opposite side to \(30^{\circ}\) is \(3\sqrt{3}\)? No, I think I made a mistake.

Wait, another approach: In a 30 - 60 - 90 triangle, the hypotenuse is twice the length of the shorter leg (the leg opposite \(30^{\circ}\)). Let the shorter leg (opposite \(30^{\circ}\)) be \(x\), then hypotenuse \(= 2x\), and the longer leg (opposite \(60^{\circ}\)) is \(x\sqrt{3}\).

Looking at the triangle, the side of length \(3\sqrt{3}\) is the longer leg (opposite \(60^{\circ}\)), so \(x\sqrt{3}=3\sqrt{3}\), which means \(x = 3\). Then the hypotenuse \(k = 2x=6\)? Wait, no, wait. Wait, if the longer leg (opposite \(60^{\circ}\)) is \(x\sqrt{3}=3\sqrt{3}\), then \(x = 3\). Then the hypotenuse (opposite the right angle) is \(2x = 6\)? Wait, no, the hypotenuse is opposite the right angle. Wait, the right angle is between the two legs. So the legs are \(x\) (opposite \(30^{\circ}\)) and \(x\sqrt{3}\) (opposite \(60^{\circ}\)), hypotenuse is \(2x\).

Given that the leg opposite \(60^{\circ}\) is \(3\sqrt{3}\), so \(x\sqrt{3}=3\sqrt{3}\implies x = 3\). Then the hypotenuse \(k = 2x=6\)? Wait, no, let's use trigonometry. Let's take angle \(60^{\circ}\). \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(60^{\circ}\) is the leg opposite, which is the leg adjacent to \(30^{\circ}\). Wait, maybe I confused the sides.

Wait, let's label the triangle properly. Let \(\angle A = 30^{\circ}\), \(\angle B=60^{\circ}\), \(\angle C = 90^{\circ}\). Then side \(BC\) is opposite \(\angle A\) (30°), so \(BC=x\), side \(AC\) is opposite \(\angle B\) (60°), so \(AC = x\sqrt{3}\), and hypotenuse \(AB = k=2x\).

We are given that \(AC = 3\sqrt{3}\) (since \(AC\) is the side with length \(3\sqrt{3}\)). So \(x\sqrt{3}=3\sqrt{3}\implies x = 3\). Then hypotenuse \(k = 2x=6\)? Wait, no, that can't be. Wait, if \(AC = 3\sqrt{3}\) (opposite 60°), then \(x\sqrt{3}=3\sqrt{3}\implies x = 3\). Then hypotenuse \(AB=2x = 6\). But let's check with cosine. \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to \(30^{\circ}\) is \(AC = 3\sqrt{3}\), hypotenuse is \(k\). \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{k}\). Cross - multiplying: \(k\times\sqrt{3}=2\times3\sqrt{3}\implies k = 6\). Yes, that's correct.

Wait, another way: \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to \(60^{\circ}\) is \(AC = 3\sqrt{3}\)? No, opposite side to \(60^{\circ}\) is \(BC\)? Wait, I think I messed up the labeling. Let's start over.

In a right - angled triangle, the hypotenuse is the side opposite the right angle. Let the right angle be at \(C\), so \(AB\) is the hypotenuse (\(k\)). Angle at \(A\) is \(30^{\circ}\), angle at \(B\) is \(60^{\circ}\). Then side \(BC\) is opposite angle \(A\) (30°), so \(BC\) is the shorter leg. Side \(AC\) is opposite angle \(B\) (60°), so \(AC\) is the longer leg. We are given that \(AC = 3\sqrt{3}\) (the side with length \(3\sqrt{3}\)).

In a 30 - 60 - 90 triangle, \(\frac{\text{longer leg}}{\text{shorter leg}}=\sqrt{3}\), and \(\frac{\text{hypotenuse}}{\text{shorter leg}} = 2\). Also, \(\cos(30^{\circ})=\frac{\text{longer leg}}{\text{hypotenuse}}\). \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{k}\). Solving for \(k\):

Multiply both sides by \(k\): \(\frac{\sqrt{3}}{2}k=3\sqrt{3}\)

Divide both sides by \(\sqrt{3}\): \(\frac{1}{2}k = 3\)

Multiply both sides by 2: \(k = 6\)

Answer:

\(6\)