Question

find c.
right triangle with angles 30°, 60°, 90°, hypotenuse? wait, no, the side adjacent to 30° is 14√3 in, and side c is adjacent to 60°? wait, the triangle has a right angle, 30°, 60°, one leg is 14√3 in (opposite 60°?), and c is the other leg (opposite 30°?). wait, the ocr text: the triangle has 30°, 60°, right angle, side labeled 14√3 in, and c. then: write your answer in simplest radical form. blank inches, and a square root button.

Explanation:

Step1: Identify triangle type

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

Looking at the triangle, the side with length \(14\sqrt{3}\) is opposite the \(60^{\circ}\) angle (since the right angle is between the side \(c\) and the hypotenuse, and the \(30^{\circ}\) angle is at the top, so the side \(14\sqrt{3}\) is opposite \(60^{\circ}\)). Let the side opposite \(30^{\circ}\) be \(c\) (wait, no: let's correct. The right angle is at the bottom right, so the angle at the top left is \(30^{\circ}\), the angle at the bottom left is \(60^{\circ}\). So the side adjacent to \(30^{\circ}\) is \(14\sqrt{3}\)? Wait, no, in a right triangle, the sides: the side opposite \(30^{\circ}\) is the shortest leg. Let's denote:

Let \(\angle A = 30^{\circ}\), \(\angle B=60^{\circ}\), right angle at \(C\). Then side opposite \(30^{\circ}\) (side \(BC\)) is \(c\), side opposite \(60^{\circ}\) (side \(AC\)) is \(14\sqrt{3}\), and hypotenuse \(AB\).

In 30 - 60 - 90 triangle, \(\text{side opposite }60^{\circ}=\text{side opposite }30^{\circ}\times\sqrt{3}\). So if side opposite \(60^{\circ}\) is \(14\sqrt{3}\), then side opposite \(30^{\circ}\) (which is \(c\)) is \(\frac{\text{side opposite }60^{\circ}}{\sqrt{3}}\).

Step2: Calculate \(c\)

We know that in a 30 - 60 - 90 triangle, \(\text{side opposite }60^{\circ}=\text{short leg}\times\sqrt{3}\). Let the short leg (opposite \(30^{\circ}\)) be \(c\). Then:

\(14\sqrt{3}=c\times\sqrt{3}\)

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

\(c = \frac{14\sqrt{3}}{\sqrt{3}}\)

The \(\sqrt{3}\) in numerator and denominator cancels out, so \(c = 14\)? Wait, no, wait. Wait, maybe I mixed up the sides. Wait, the hypotenuse is opposite the right angle. Wait, the side with length \(14\sqrt{3}\): is it the hypotenuse? No, because the hypotenuse is the longest side. Wait, no, in the triangle, the right angle is at the bottom right, so the hypotenuse is the side from the top left to the bottom right? Wait, no, the right angle is between the side \(c\) and the side \(14\sqrt{3}\). So the hypotenuse is the side opposite the right angle, which is the side from the top left to the bottom left. Wait, maybe I got the angles wrong. Let's re - label:

Let the right angle be at \(C\), so \(\angle C = 90^{\circ}\), \(\angle A=30^{\circ}\), \(\angle B = 60^{\circ}\). Then side \(AC = 14\sqrt{3}\) (adjacent to \(30^{\circ}\)), side \(BC=c\) (opposite to \(30^{\circ}\)), and hypotenuse \(AB\).

In a right triangle, \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{c}{14\sqrt{3}}\)

We know that \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\)

So \(\frac{1}{\sqrt{3}}=\frac{c}{14\sqrt{3}}\)

Multiply both sides by \(14\sqrt{3}\):

\(c=\frac{14\sqrt{3}}{\sqrt{3}} = 14\)? Wait, no, that can't be. Wait, maybe the side \(14\sqrt{3}\) is the hypotenuse? No, because in a 30 - 60 - 90 triangle, the hypotenuse is twice the short leg. Wait, maybe I made a mistake in identifying the sides.

Wait, another approach: in a 30 - 60 - 90 triangle, the ratios are: short leg (opposite 30°): \(x\), long leg (opposite 60°): \(x\sqrt{3}\), hypotenuse: \(2x\).

Looking at the triangle, the side with length \(14\sqrt{3}\) is the long leg (opposite 60°), so \(x\sqrt{3}=14\sqrt{3}\), so \(x = 14\). The short leg (opposite 30°) is \(x = 14\)? Wait, no, the side \(c\) is the short leg? Wait, the right angle is at the bottom right, so the side \(c\) is adjacent to the 60° angle and opposite to the 30° angle. So yes, \(c\) is the short leg (opposite 30°), and the long leg (opposite 60°) is \(14\sqrt{3}\). So since long leg \(=x\sqrt{3}\), and long leg is \(14\sqrt{3}\), then \(x\sqrt{3}=14\sqrt{3}\implies x = 14\). So \(c=x = 14\)? Wait, but let's check with trigonometry.

Using sine: \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\). Wait, what's the hypotenuse? Wait, maybe the side \(14\sqrt{3}\) is the hypotenuse? No, because \(\sin(30^{\circ})=\frac{1}{2}\), so if hypotenuse is \(14\sqrt{3}\), then opposite side (c) would be \(\frac{1}{2}\times14\sqrt{3}=7\sqrt{3}\). Wait, now I'm confused. Let's re - examine the triangle.

The triangle has a right angle, a 30° angle, and a 60° angle. The side labeled \(14\sqrt{3}\) is one of the legs, and \(c\) is the other leg. Let's use trigonometric ratios. Let's take the 30° angle:

\(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{c}{14\sqrt{3}}\)

\(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so:

\(\frac{1}{\sqrt{3}}=\frac{c}{14\sqrt{3}}\)

Multiply both sides by \(14\sqrt{3}\):

\(c=\frac{14\sqrt{3}}{\sqrt{3}} = 14\)? No, that would be if we use tan, but maybe we should use sine or cosine.

Wait, another way: \(\cos(30^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}\). Wait, what's the hypotenuse? Let's call the hypotenuse \(h\). Then \(\cos(30^{\circ})=\frac{14\sqrt{3}}{h}\), and \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so:

\(\frac{\sqrt{3}}{2}=\frac{14\sqrt{3}}{h}\)

Cross - multiply:

\(h\times\sqrt{3}=2\times14\sqrt{3}\)

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

\(h = 28\)

Then, \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{c}{28}\)

Since \(\sin(30^{\circ})=\frac{1}{2}\), then:

\(\frac{1}{2}=\frac{c}{28}\)

Multiply both sides by 28:

\(c = 14\)

Wait, that's the same result. So \(c = 14\).

Wait, but let's verify with the ratio. If hypotenuse is \(28\) (which is \(2x\), so \(x = 14\)), short leg is \(x = 14\), long leg is \(x\sqrt{3}=14\sqrt{3}\), which matches the given side. So yes, \(c = 14\).

Answer:

\(14\)