Question

find j.
right triangle with angles 30°, 60°, right angle; side opposite 30° is 12√3 mm, hypotenuse is j
write your answer in simplest radical form.
blank millimeters

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°, hypotenuse respectively).

Step2: Relate given side to ratio

The side opposite 60° is \(12\sqrt{3}\) mm. In the ratio, the side opposite 60° is \(\sqrt{3}\) times the side opposite 30°, and the hypotenuse \(j\) is twice the side opposite 30°. Let the side opposite 30° be \(x\). Then \(\sqrt{3}x = 12\sqrt{3}\), so \(x = 12\).

Step3: Find hypotenuse \(j\)

Since hypotenuse \(j = 2x\), substitute \(x = 12\) to get \(j = 2\times12 = 24\). Wait, alternatively, using cosine: \(\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\), but adjacent to 30° is \(12\sqrt{3}\)? Wait no, wait: in 30-60-90, the side opposite 30° is shortest, opposite 60° is longer leg, hypotenuse is longest. Wait, the side given is \(12\sqrt{3}\), which is opposite 60° (since angle 30° is at the bottom left, so the side adjacent to 30° is the side with \(12\sqrt{3}\)? Wait no, let's label the triangle: right angle at top, 30° at bottom left, 60° at bottom right. So the side between 30° and right angle is adjacent to 30°, opposite to 60°. The side between 60° and right angle is adjacent to 60°, opposite to 30°. The hypotenuse is \(j\). So \(\sin(30^\circ)=\frac{\text{opposite to 30°}}{j}\), and \(\cos(30^\circ)=\frac{\text{opposite to 60°}}{j}\). Wait, \(\cos(30^\circ)=\frac{\sqrt{3}}{2}=\frac{12\sqrt{3}}{j}\). Solve for \(j\): \(j=\frac{12\sqrt{3}}{\cos(30^\circ)}=\frac{12\sqrt{3}}{\frac{\sqrt{3}}{2}} = 12\sqrt{3}\times\frac{2}{\sqrt{3}} = 24\).

Answer:

24