Question

find the length of side x in simplest radical form with a rational denominator. triangle image with right angle, 5 on one leg, 60° and 30° angles, x on another leg answer attempt 1 out of 2 x = input box submit answer

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 of length 5 is opposite the 60° angle, and \(x\) is opposite the 30° angle.

Step2: Use tangent ratio or side ratios

Using the tangent of 60°: \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{5}{x}\). Since \(\tan(60^\circ)=\sqrt{3}\), we have \(\sqrt{3}=\frac{5}{x}\).

Step3: Solve for \(x\)

Rearrange the equation: \(x = \frac{5}{\sqrt{3}}\). Rationalize the denominator by multiplying numerator and denominator by \(\sqrt{3}\): \(x=\frac{5\sqrt{3}}{3}\)? Wait, no—wait, in 30-60-90, the side opposite 30° is the shortest, opposite 60° is longer leg. Wait, maybe I mixed up. Wait, the right angle, one angle 60°, one 30°. So the side adjacent to 60° is \(x\) (opposite 30°), and the side opposite 60° is 5. So in 30-60-90, longer leg (opposite 60°) is \(\sqrt{3}\) times shorter leg (opposite 30°). So \(5 = \sqrt{3} \times x\)? Wait, no—wait, shorter leg is opposite 30°, longer leg opposite 60°. So if \(x\) is opposite 30°, then longer leg (5) is \(x\sqrt{3}\). So \(5 = x\sqrt{3}\), so \(x=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}\)? Wait, no, that can't be. Wait, maybe I had the angles wrong. Wait, the right angle, angle at the bottom left is 60°, bottom right is 30°. So the side with length 5 is adjacent to 60°? Wait, no, the right angle is at the top. So the sides: the leg with length 5 is adjacent to the 60° angle, and \(x\) is adjacent to the 30° angle. Wait, using tangent: \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{5}\)? Wait, no, opposite of 60° is \(x\)? Wait, I think I messed up the labels. Let's re-express: right triangle, right angle at top. So the two legs: one is 5 (left leg), one is \(x\) (right leg). The angles: bottom left is 60°, bottom right is 30°. So the left leg (5) is opposite the 30° angle? Wait, no: in a triangle, the side opposite an angle is across from it. So bottom left angle (60°) is opposite the right leg (\(x\)). Bottom right angle (30°) is opposite the left leg (5). Ah! There we go. So opposite 30° is 5, opposite 60° is \(x\). In 30-60-90 triangle, the sides are: opposite 30°: \(a\), opposite 60°: \(a\sqrt{3}\), hypotenuse: \(2a\). So here, opposite 30° is 5, so \(a = 5\). Then opposite 60° (which is \(x\)) is \(a\sqrt{3}=5\sqrt{3}\)? Wait, no, that contradicts. Wait, no: if opposite 30° is 5, then opposite 60° is \(5\sqrt{3}\), hypotenuse \(10\). Wait, but let's check with tangent. \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{5}\). \(\tan(60^\circ)=\sqrt{3}\), so \(x = 5\sqrt{3}\). Wait, now I'm confused. Wait, let's draw the triangle: right angle at top, left vertex 60°, right vertex 30°. So the left leg (from top to left vertex) is 5, right leg (top to right vertex) is \(x\), base is hypotenuse. So angle at left vertex (60°): between left leg (5) and hypotenuse. So the side opposite 60° is the right leg (\(x\)), and the side adjacent to 60° is the left leg (5). So \(\tan(60^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{x}{5}\), so \(x = 5\tan(60^\circ)=5\sqrt{3}\). Ah! That's correct. Because in the triangle, angle at left is 60°, so the side opposite to it is the right leg (\(x\)), and adjacent is left leg (5). So \(\tan(60^\circ)=\sqrt{3}=\frac{x}{5}\), so \(x = 5\sqrt{3}\). Wait, that makes sense. Because 30-60-90: if the side opposite 30° is 5 (left leg? No, wait, angle at right is 30°, so opposite to 30° is left leg (5). So left leg (5) is opposite 30°, so it's the shorter leg. Then the right leg (x) is opposite 60°, so it's the longer leg, which is \(\sqrt{3}\) times shorter leg. So \(x = 5\sqrt{3}\). Yes, that's correct. So my initial mistake was misidentifying which angle is opposite which side. So angle at right is 30°, so opposite side is left leg (5) (shorter leg). Angle at left is 60°, opposite side is right leg (x) (longer leg), which is \(\sqrt{3}\) times shorter leg. So \(x = 5\sqrt{3}\).

Answer:

\(5\sqrt{3}\) (Wait, no—wait, let's confirm with sine. \(\sin(60^\circ)=\frac{x}{\text{hypotenuse}}\), but we don't know hypotenuse. Alternatively, \(\cos(60^\circ)=\frac{5}{\text{hypotenuse}}\), \(\cos(60^\circ)=0.5\), so hypotenuse is 10. Then by Pythagoras: \(5^2 + x^2 = 10^2\) → \(25 + x^2 = 100\) → \(x^2 = 75\) → \(x = \sqrt{75} = 5\sqrt{3}\). Yes! That's correct. So earlier mistake was in the 30-60-90 ratio: shorter leg (opposite 30°) is 5, hypotenuse is 10, longer leg (opposite 60°) is \(5\sqrt{3}\). So \(x = 5\sqrt{3}\).