Question

find d.
image of a right triangle with angles 30°, 60°, right angle, one side labeled 3√6 yd, side d to find
write your answer in simplest radical form.
blank yards
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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°, 90° respectively). The side given (\(3\sqrt{6}\) yd) is opposite the 60° angle, and \(d\) is opposite the 90° angle (hypotenuse).

Step2: Use sine ratio

\(\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{3\sqrt{6}}{d}\). We know \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{6}}{d}\).

Step3: Solve for \(d\)

Cross - multiply: \(d\times\sqrt{3}=2\times3\sqrt{6}\). Then \(d = \frac{6\sqrt{6}}{\sqrt{3}}\). Rationalize the denominator: \(d=\frac{6\sqrt{6}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{6\sqrt{18}}{3}=\frac{6\times3\sqrt{2}}{3}=6\sqrt{2}\)? Wait, no, wait. Wait, in 30 - 60 - 90 triangle, the hypotenuse is twice the shorter leg, and the longer leg (opposite 60°) is \(\sqrt{3}\) times the shorter leg. Wait, maybe I mixed up. Let's re - identify: the angle of 30°: the side opposite 30° is the shorter leg, opposite 60° is longer leg, opposite 90° is hypotenuse. The given side \(3\sqrt{6}\) is adjacent to 30°? Wait, no, the right angle is at the vertex with the right angle symbol. So the angle of 30°: the side adjacent to 30° is \(3\sqrt{6}\)? Wait, no, let's look at the triangle: the right angle is between the side of length \(3\sqrt{6}\) and the other leg, and the angles are 30°, 60°, 90°. So the side of length \(3\sqrt{6}\) is opposite the 60° angle? Wait, no, the angle of 30°: the side opposite 30° is the shorter leg, the side opposite 60° is longer leg (length \(=\) shorter leg \(\times\sqrt{3}\)), hypotenuse \(=\) 2 \(\times\) shorter leg. Wait, maybe using cosine: \(\cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to 30° is \(3\sqrt{6}\), hypotenuse is \(d\). \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{6}}{d}\), then \(d=\frac{3\sqrt{6}\times2}{\sqrt{3}}=\frac{6\sqrt{6}}{\sqrt{3}} = 6\sqrt{2}\)? No, wait, \(\sqrt{6}/\sqrt{3}=\sqrt{2}\), so \(6\sqrt{2}\)? Wait, no, wait, let's do it again. \(d=\frac{3\sqrt{6}}{\cos(30^\circ)}\), \(\cos(30^\circ)=\frac{\sqrt{3}}{2}\), so \(d = 3\sqrt{6}\times\frac{2}{\sqrt{3}}=3\times2\times\frac{\sqrt{6}}{\sqrt{3}}=6\times\sqrt{2}=6\sqrt{2}\)? Wait, no, that's wrong. Wait, maybe the given side is the shorter leg? Wait, no, the angle of 30°: if the side \(d\) is the hypotenuse, and the side \(3\sqrt{6}\) is opposite 60°, then in 30 - 60 - 90 triangle, longer leg (opposite 60°) \(= \) shorter leg \(\times\sqrt{3}\), hypotenuse \(= 2\times\) shorter leg. Let shorter leg \(= x\), then longer leg \(=x\sqrt{3}=3\sqrt{6}\), so \(x=\frac{3\sqrt{6}}{\sqrt{3}} = 3\sqrt{2}\). Then hypotenuse \(d = 2x=6\sqrt{2}\)? Wait, no, that's the same as before. Wait, but let's check with sine: \(\sin(60^\circ)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{3\sqrt{6}}{d}\), \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{6}}{d}\), cross - multiply: \(d\sqrt{3}=6\sqrt{6}\), \(d=\frac{6\sqrt{6}}{\sqrt{3}}=6\sqrt{2}\) (since \(\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}\)). Wait, but another way: in 30 - 60 - 90 triangle, the hypotenuse is \(\frac{\text{longer leg}}{\sin(60^\circ)}\). Longer leg is \(3\sqrt{6}\), \(\sin(60^\circ)=\frac{\sqrt{3}}{2}\), so \(d=\frac{3\sqrt{6}}{\frac{\sqrt{3}}{2}}=3\sqrt{6}\times\frac{2}{\sqrt{3}} = 6\sqrt{2}\). Wait, but I think I made a mistake earlier. Wait, no, let's calculate \(\frac{3\sqrt{6}}{\sqrt{3}}\): \(\sqrt{6}=\sqrt{2}\times\sqrt{3}\), so \(\frac{3\sqrt{2}\times\sqrt{3}}{\sqrt{3}} = 3\sqrt{2}\), then hypotenuse is \(2\times3\sqrt{2}=6\sqrt{2}\). Yes, that's correct. So \(d = 6\sqrt{2}\)? Wait, no, wait, no! Wait, I think I mixed up the sides. Let's re - draw the triangle mentally: right angle, one angle 30°, one 60°. The side labeled \(3\sqrt{6}\) is adjacent to the 30° angle? No, the angle of 30°: the side opposite 30° is the shortest side. Let's use the cosine of 30°: \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to 30° is \(3\sqrt{6}\), hypotenuse is \(d\). \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{\sqrt{3}}{2}=\frac{3\sqrt{6}}{d}\), solving for \(d\): \(d=\frac{3\sqrt{6}\times2}{\sqrt{3}}=\frac{6\sqrt{6}}{\sqrt{3}} = 6\sqrt{2}\) (because \(\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}\)). Wait, but let's check with the 30 - 60 - 90 ratios. If the hypotenuse is \(d\), the side opposite 30° is \(\frac{d}{2}\), the side opposite 60° is \(\frac{d\sqrt{3}}{2}\). We know that the side opposite 60° is \(3\sqrt{6}\), so \(\frac{d\sqrt{3}}{2}=3\sqrt{6}\). Solve for \(d\): \(d\sqrt{3}=6\sqrt{6}\), \(d=\frac{6\sqrt{6}}{\sqrt{3}} = 6\sqrt{2}\) (rationalizing: \(\frac{6\sqrt{6}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{6\sqrt{18}}{3}=\frac{6\times3\sqrt{2}}{3}=6\sqrt{2}\)). Yes, that's correct.

Answer:

\(6\sqrt{2}\)