Question

find w.
12 mm
write your answer in simplest radical form.
□ millimeters

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, then the side opposite \(60^{\circ}\), then the hypotenuse. Here, the side of length \(12\) mm is opposite the \(60^{\circ}\) angle, and \(w\) is opposite the \(30^{\circ}\) angle? Wait, no, wait. Wait, the right angle is at the corner, so let's label the triangle. Let the right angle be \(C\), the \(30^{\circ}\) angle be \(A\), and the \(60^{\circ}\) angle be \(B\). Then side \(BC\) (opposite \(A\)) is \(w\), side \(AC\) (opposite \(B\)) is \(12\) mm, and side \(AB\) is the hypotenuse. In a 30 - 60 - 90 triangle, \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{w}{12}\)? Wait, no, \(\tan(60^{\circ})=\frac{12}{w}\) because \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\). \(\tan(60^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{12}{w}\), so \(w = \frac{12}{\sqrt{3}}\). Rationalize the denominator: \(w=\frac{12\sqrt{3}}{3}=4\sqrt{3}\)? Wait, no, wait, maybe I mixed up the angles. Wait, the angle at \(A\) is \(30^{\circ}\), so the side opposite \(30^{\circ}\) is the shorter leg. Wait, the side of length \(12\) is adjacent to the \(30^{\circ}\) angle? Wait, let's use sine or cosine. \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\), but maybe better to recall the ratios. In a 30 - 60 - 90 triangle, if the shorter leg (opposite \(30^{\circ}\)) is \(x\), the longer leg (opposite \(60^{\circ}\)) is \(x\sqrt{3}\), and hypotenuse is \(2x\). Here, the longer leg is \(12\) mm (opposite \(60^{\circ}\)), so \(x\sqrt{3}=12\), so \(x = \frac{12}{\sqrt{3}} = 4\sqrt{3}\)? Wait, no, the shorter leg is opposite \(30^{\circ}\), so \(w\) is the shorter leg? Wait, no, wait the angle of \(30^{\circ}\) is at the top, so the side opposite \(30^{\circ}\) is the side opposite the \(30^{\circ}\) angle, which is the side adjacent to the \(60^{\circ}\) angle. Wait, maybe I should use trigonometric ratios. Let's define: angle \(A = 30^{\circ}\), angle \(B = 60^{\circ}\), right angle at \(C\). Then side \(BC\) (opposite \(A\)) is \(w\), side \(AC\) (opposite \(B\)) is \(12\) mm, side \(AB\) is hypotenuse. Then \(\sin(30^{\circ})=\frac{w}{AB}\), \(\sin(60^{\circ})=\frac{12}{AB}\). Since \(\sin(30^{\circ})=\frac{1}{2}\) and \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{w}{AB}=\frac{1}{2}\) and \(\frac{12}{AB}=\frac{\sqrt{3}}{2}\). From the second equation, \(AB=\frac{24}{\sqrt{3}} = 8\sqrt{3}\). Then from the first equation, \(w=\frac{AB}{2}=\frac{8\sqrt{3}}{2}=4\sqrt{3}\)? Wait, no, that can't be. Wait, maybe I had the longer leg wrong. Wait, no, if the angle at \(A\) is \(30^{\circ}\), then the side opposite \(30^{\circ}\) (BC) is the shorter leg, and side AC (opposite \(60^{\circ}\)) is the longer leg. So longer leg = shorter leg \(\times\sqrt{3}\). So if longer leg is \(12\), then shorter leg \(w=\frac{12}{\sqrt{3}} = 4\sqrt{3}\). Wait, but let's check with tangent. \(\tan(30^{\circ})=\frac{w}{12}\)? No, \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}=\frac{w}{12}\)? Then \(w=\frac{12}{\sqrt{3}} = 4\sqrt{3}\). Yes, that's correct. So the steps:

Step1: Recall 30 - 60 - 90 triangle ratios

In a 30 - 60 - 90 right triangle, the ratio of the shorter leg (opposite \(30^{\circ}\)) to the longer leg (opposite \(60^{\circ}\)) is \(1:\sqrt{3}\). Let the shorter leg (opposite \(30^{\circ}\)) be \(w\), and the longer leg (opposite \(60^{\circ}\)) be \(12\) mm. Then \(\frac{w}{12}=\frac{1}{\sqrt{3}}\).

Step2: Solve for \(w\)

Cross - multiply: \(w\sqrt{3}=12\). Then \(w = \frac{12}{\sqrt{3}}\). Rationalize the denominator by multiplying numerator and denominator by \(\sqrt{3}\): \(w=\frac{12\sqrt{3}}{3}=4\sqrt{3}\).

Answer:

\(4\sqrt{3}\)