Question

find j.
right triangle with angles 60°, 30°, right angle; one leg is 2√6 mi, hypotenuse is j
write your answer in simplest radical form.
blank miles

Explanation:

Step1: Identify triangle type

This is a 30-60-90 right triangle. In such a triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite \(30^\circ\) is the shortest leg (\(x\)), opposite \(60^\circ\) is \(x\sqrt{3}\), and hypotenuse is \(2x\).

Step2: Determine the leg opposite \(60^\circ\)

The given leg (\(2\sqrt{6}\)) is adjacent to \(30^\circ\), so it's the leg opposite \(60^\circ\)? Wait, no: the right angle, \(30^\circ\), \(60^\circ\). The side labeled \(2\sqrt{6}\) is adjacent to \(30^\circ\) and opposite \(60^\circ\)? Wait, no: in a right triangle, the side opposite \(30^\circ\) is the shorter leg. Wait, the angle at the bottom is \(30^\circ\), so the side opposite \(30^\circ\) is the top leg (horizontal), and the side opposite \(60^\circ\) is the vertical leg (\(2\sqrt{6}\))? Wait, no, let's clarify angles. The right angle is at the top left. So the angles are: top left: \(90^\circ\), top right: \(60^\circ\), bottom: \(30^\circ\). So the sides: the side opposite \(30^\circ\) (top right angle? No, wait, angles: the triangle has angles \(90^\circ\), \(60^\circ\), \(30^\circ\). So the side opposite \(30^\circ\) is the side opposite the \(30^\circ\) angle (bottom angle), which is the horizontal leg (top side). The side opposite \(60^\circ\) (top right angle) is the vertical leg (left side, length \(2\sqrt{6}\)). The hypotenuse is \(j\), opposite the right angle.

Wait, in 30-60-90 triangle, the hypotenuse is twice the shorter leg (opposite \(30^\circ\)). Wait, maybe I mixed up. Let's use trigonometry. Let's take the angle at the bottom: \(30^\circ\). The adjacent side to \(30^\circ\) is the vertical leg (\(2\sqrt{6}\)), and the hypotenuse is \(j\). So \(\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2\sqrt{6}}{j}\). We know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\). So:

$$ \frac{\sqrt{3}}{2} = \frac{2\sqrt{6}}{j} $$

Step3: Solve for \(j\)

Cross-multiplying:

$$ j \cdot \sqrt{3} = 2 \cdot 2\sqrt{6} $$

$$ j \cdot \sqrt{3} = 4\sqrt{6} $$

Divide both sides by \(\sqrt{3}\):

$$ j = \frac{4\sqrt{6}}{\sqrt{3}} $$

Simplify \(\frac{\sqrt{6}}{\sqrt{3}} = \sqrt{\frac{6}{3}} = \sqrt{2}\), so:

$$ j = 4\sqrt{2} $$

Wait, alternatively, using the 30-60-90 ratios. Wait, maybe I made a mistake in identifying the sides. Let's re-express: in a 30-60-90 triangle, the sides are \(x\) (opposite \(30^\circ\)), \(x\sqrt{3}\) (opposite \(60^\circ\)), and \(2x\) (hypotenuse). Let's see: the vertical leg (length \(2\sqrt{6}\)) is opposite the \(60^\circ\) angle (top right), so that should be \(x\sqrt{3}\). So \(x\sqrt{3} = 2\sqrt{6}\), so \(x = \frac{2\sqrt{6}}{\sqrt{3}} = 2\sqrt{2}\). Then the hypotenuse \(j = 2x = 2 \cdot 2\sqrt{2} = 4\sqrt{2}\). Yes, that matches. So \(j = 4\sqrt{2}\).

Answer:

\(4\sqrt{2}\)