Question

find d.
image of a right triangle with angles 30°, 60°, right angle, hypotenuse 2√3 cm, side d
write your answer in simplest radical form.
blank centimeters
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Explanation:

Step1: Identify triangle type

It's 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 hypotenuse is \(2\sqrt{3}\) cm, and \(d\) is opposite the 30° angle? Wait, no: angle 30° has opposite side? Wait, let's label the triangle. The right angle is between the side \(d\) and the other leg. The hypotenuse is \(2\sqrt{3}\) cm. The angle of 30°: in a 30 - 60 - 90 triangle, the side opposite 30° is the shortest leg, let's call it \(x\), opposite 60° is \(x\sqrt{3}\), hypotenuse is \(2x\). Wait, here, the hypotenuse is \(2\sqrt{3}\), so \(2x = 2\sqrt{3}\)? No, wait, maybe I mixed up. Wait, the angle at the bottom is 60°, so the side \(d\): let's use trigonometry. Let's denote the right angle vertex, the 30° vertex, and 60° vertex. So, for the 30° angle, the adjacent side to 30° is \(d\)? Wait, no. Let's use cosine or sine. Let's take the 30° angle: \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to 30° is \(d\)? Wait, no, the hypotenuse is \(2\sqrt{3}\). Wait, the right angle is between the two legs: one leg is \(d\), the other leg (let's say \(a\)), and hypotenuse \(2\sqrt{3}\). The angle at the left is 30°, so the side opposite 30° is \(a\) (the leg opposite 30°), and the side adjacent to 30° is \(d\). So \(\cos(30^{\circ})=\frac{d}{\text{hypotenuse}}\), so \(d=\text{hypotenuse}\times\cos(30^{\circ})\). Wait, \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), hypotenuse is \(2\sqrt{3}\). So \(d = 2\sqrt{3}\times\frac{\sqrt{3}}{2}\). Wait, that would be \(\frac{2\sqrt{3}\times\sqrt{3}}{2}=\frac{2\times3}{2}=3\)? No, that can't be. Wait, maybe I used the wrong angle. Let's use the 60° angle. For the 60° angle, \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). The adjacent side to 60° is \(d\), hypotenuse \(2\sqrt{3}\). \(\cos(60^{\circ})=\frac{1}{2}\), so \(d = 2\sqrt{3}\times\frac{1}{2}=\sqrt{3}\)? No, that's not right. Wait, maybe I should use sine. For the 30° angle: \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to 30° is the leg that's not \(d\), let's say \(a\), so \(\sin(30^{\circ})=\frac{a}{2\sqrt{3}}\), so \(a = 2\sqrt{3}\times\frac{1}{2}=\sqrt{3}\). Then, for the 60° angle, \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{d}{2\sqrt{3}}\), so \(d = 2\sqrt{3}\times\frac{\sqrt{3}}{2}=\frac{2\times3}{2}=3\)? Wait, no, that's conflicting. Wait, maybe the triangle is labeled as: right angle at the top, 30° at the left, 60° at the bottom. So the legs: left leg (vertical) is \(d\), bottom leg (horizontal) is \(a\), hypotenuse is \(2\sqrt{3}\) (the side connecting left and bottom vertices). So angle at left is 30°, so the side opposite 30° is the bottom leg \(a\), side adjacent to 30° is \(d\) (the vertical leg). So \(\tan(30^{\circ})=\frac{a}{d}\), \(\sin(30^{\circ})=\frac{a}{2\sqrt{3}}\), so \(a = 2\sqrt{3}\times\frac{1}{2}=\sqrt{3}\). Then \(\tan(30^{\circ})=\frac{\sqrt{3}}{d}=\frac{1}{\sqrt{3}}\), so \(d=\sqrt{3}\times\sqrt{3}=3\)? No, that's not. Wait, I think I messed up the trigonometry. Wait, let's recall the 30 - 60 - 90 triangle ratios. In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\), where the side opposite 30° is \(x\) (shortest leg), opposite 60° is \(x\sqrt{3}\) (longer leg), hypotenuse \(2x\). So here, the hypotenuse is \(2\sqrt{3}\), so \(2x = 2\sqrt{3}\)? No, that would make \(x=\sqrt{3}\), but then the longer leg would be \(x\sqrt{3}=\sqrt{3}\times\sqrt{3}=3\), and shortest leg \(x=\sqrt{3}\). Wait, but which leg is \(d\)? If \(d\) is the longer leg (opposite 60°), then \(d = 3\)? No, wait, the angle at the bottom is 60°, so the side opposite 60° is the longer leg, which would be \(d\) if \(d\) is opposite 60°. Wait, the angle at the bottom is 60°, so the side opposite 60° is the leg from the right angle to the left vertex, which is \(d\)? No, maybe I need to re - draw mentally. The triangle has a right angle (red square), left angle 30°, bottom angle 60°. So the sides: the side opposite 30° is the leg between right angle and bottom angle (let's call it \(a\)), the side opposite 60° is the leg between right angle and left angle (which is \(d\)), and hypotenuse between left and bottom angles (length \(2\sqrt{3}\)). So in 30 - 60 - 90 triangle, side opposite 30° (\(a\)) : side opposite 60° (\(d\)) : hypotenuse = \(1:\sqrt{3}:2\). So \(a:x:2x\), where \(x\) is hypotenuse? No, hypotenuse is \(2x\) when the shortest leg is \(x\). Wait, no, the ratio is shortest leg (\(x\)) : longer leg (\(x\sqrt{3}\)) : hypotenuse (\(2x\)). So here, hypotenuse is \(2\sqrt{3}\), so \(2x = 2\sqrt{3}\) → \(x=\sqrt{3}\). Then the longer leg (opposite 60°) is \(x\sqrt{3}=\sqrt{3}\times\sqrt{3}=3\)? No, that's not. Wait, no, if hypotenuse is \(2x\), then \(x\) is the shortest leg (opposite 30°). So shortest leg \(x\), longer leg \(x\sqrt{3}\), hypotenuse \(2x\). So if hypotenuse is \(2\sqrt{3}\), then \(2x = 2\sqrt{3}\) → \(x=\sqrt{3}\) (shortest leg, opposite 30°). Then longer leg (opposite 60°) is \(x\sqrt{3}=\sqrt{3}\times\sqrt{3}=3\)? But that would mean \(d = 3\)? Wait, but let's use sine. For angle 30°: \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\text{shortest leg}}{\text{hypotenuse}}=\frac{x}{2\sqrt{3}}=\frac{1}{2}\) → \(x = \sqrt{3}\) (shortest leg). Then the longer leg (opposite 60°) is \(x\sqrt{3}=\sqrt{3}\times\sqrt{3}=3\)? But that seems off. Wait, maybe I made a mistake in identifying which leg is which. Let's use trigonometry with angle 30°: \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). The opposite side to 30° is the leg that's not \(d\), let's say \(a\), so \(a = 2\sqrt{3}\times\sin(30^{\circ})=2\sqrt{3}\times\frac{1}{2}=\sqrt{3}\). Then, using Pythagoras: \(a^{2}+d^{2}=(2\sqrt{3})^{2}\). So \((\sqrt{3})^{2}+d^{2}=12\) → \(3 + d^{2}=12\) → \(d^{2}=9\) → \(d = 3\)? Wait, but that's a whole number. But the problem says "simplest radical form". Wait, maybe I messed up. Wait, no, 3 is an integer, which is in simplest form. Wait, but let's check with cosine for angle 60°: \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{d}{2\sqrt{3}}\). \(\cos(60^{\circ})=\frac{1}{2}\), so \(d = 2\sqrt{3}\times\frac{1}{2}=\sqrt{3}\)? Wait, now I'm confused. Wait, no, \(\cos(60^{\circ})\) adjacent side to 60° is \(d\)? Wait, angle 60°: the adjacent side is \(d\), hypotenuse \(2\sqrt{3}\). So \(\cos(60^{\circ})=\frac{d}{2\sqrt{3}}\) → \(d = 2\sqrt{3}\times\frac{1}{2}=\sqrt{3}\). But then Pythagoras: \(d^{2}+a^{2}=(2\sqrt{3})^{2}\). If \(d=\sqrt{3}\), then \(a^{2}=12 - 3 = 9\), \(a = 3\). Then, for angle 30°, \(\sin(30^{\circ})=\frac{a}{2\sqrt{3}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\), which is \(\sin(60^{\circ})\), not 30°. Oh! Here's the mistake. I mixed up the opposite and adjacent sides. For angle 30°: the opposite side is \(a\) (the leg opposite 30°), and adjacent side is \(d\) (the leg adjacent to 30°). So \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{2\sqrt{3}}=\frac{1}{2}\) → \(a = \sqrt{3}\). Then \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{d}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\) → \(d = 2\sqrt{3}\times\frac{\sqrt{3}}{2}=\frac{2\times3}{2}=3\). Then, for angle 60°: \(\sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{d}{2\sqrt{3}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\), which is correct, and \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{a}{2\sqrt{3}}=\frac{\sqrt{3}}{2\sqrt{3}}=\frac{1}{2}\), which is correct. So the mistake was in identifying which side is opposite which angle. So \(d\) is the adjacent side to 30° and opposite side to 60°, so using \(\cos(30^{\circ})\) or \(\sin(60^{\circ})\) gives \(d = 3\). Wait, but 3 is an integer, which is in simplest radical form (since it has no radical). So the value of \(d\) is 3.

Step2: Verify with Pythagoras

We have \(d = 3\), \(a=\sqrt{3}\), hypotenuse \(2\sqrt{3}\). Check \(3^{2}+(\sqrt{3})^{2}=9 + 3=12\), and \((2\sqrt{3})^{2}=12\). So it satisfies Pythagoras' theorem. Also, the angles: \(\tan(30^{\circ})=\frac{a}{d}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\), which is correct for 30° angle. \(\tan(60^{\circ})=\frac{d}{a}=\frac{3}{\sqrt{3}}=\sqrt{3}\), which is correct for 60° angle. So the calculation is correct.

Answer:

\(3\)