Question

fill in the blank 1 point
if a = 12\sqrt{3}, find b and c.
b= choose your answer... c= choose your answer...

Explanation:

Step1: Identify triangle type

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

Here, side \(a\) is opposite \(30^\circ\)? Wait, no. Wait, the right angle is between \(a\) and \(b\). So angle \(30^\circ\) is at the bottom right, so side opposite \(30^\circ\) is \(a\)? Wait no, let's check angles. The right angle is between \(a\) (vertical leg) and \(b\) (horizontal leg). The angle at the bottom is \(30^\circ\), so the side opposite \(30^\circ\) is \(a\) (vertical leg), side opposite \(60^\circ\) is \(b\) (horizontal leg), and hypotenuse is \(c\). Wait, no: in a right triangle, the side opposite \(30^\circ\) is the shorter leg. Let's confirm: angle at bottom is \(30^\circ\), so the side opposite to it is \(a\) (vertical leg). Then side opposite \(60^\circ\) (top angle) is \(b\) (horizontal leg), and hypotenuse is \(c\).

In 30-60-90 triangle, the ratios are:
- Shorter leg (opposite \(30^\circ\)): \(x\)
- Longer leg (opposite \(60^\circ\)): \(x\sqrt{3}\)
- Hypotenuse: \(2x\)

Wait, but here \(a = 12\sqrt{3}\) is the side opposite \(30^\circ\)? Wait no, if angle at bottom is \(30^\circ\), then the side opposite to it is \(a\) (vertical leg). Then the longer leg (opposite \(60^\circ\)) is \(b\), and hypotenuse \(c\). Wait, but if \(a\) is opposite \(30^\circ\), then \(a = x\), longer leg \(b = x\sqrt{3}\), hypotenuse \(c = 2x\). But \(a = 12\sqrt{3}\), so \(x = 12\sqrt{3}\)? Wait that can't be, because then longer leg would be \(x\sqrt{3} = 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36\), and hypotenuse \(2x = 24\sqrt{3}\). Wait, but that seems reversed. Wait maybe I mixed up the angles. Let's re-examine the triangle.

The right angle is between \(a\) (vertical) and \(b\) (horizontal). The top angle is \(60^\circ\), bottom angle is \(30^\circ\). So the side opposite \(30^\circ\) is \(a\) (vertical leg), side opposite \(60^\circ\) is \(b\) (horizontal leg), hypotenuse \(c\). Wait, but in a 30-60-90 triangle, the side opposite \(30^\circ\) is the shorter leg. So if \(a\) is opposite \(30^\circ\), then \(a\) is the shorter leg. But \(a = 12\sqrt{3}\), which is longer than, say, if \(b\) were the shorter leg. Wait maybe I got the opposite sides wrong. Let's use trigonometry.

Using trigonometric ratios:

For angle \(30^\circ\) (bottom angle):

\(\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b}\)

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

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

\(\sin(30^\circ) = \frac{1}{2}\), so \(\frac{1}{2} = \frac{a}{c}\) => \(c = 2a\)

Wait, but if \(a = 12\sqrt{3}\), then:

\(b = a\sqrt{3} = 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36\)

\(c = 2a = 2 \times 12\sqrt{3} = 24\sqrt{3}\)? Wait no, that can't be, because then the hypotenuse would be longer than the leg, but \(24\sqrt{3}\) is longer than \(36\)? Wait \(24\sqrt{3} \approx 24 \times 1.732 \approx 41.57\), and \(36\) is less than that, so hypotenuse is longer, which is correct. Wait, but let's check with the other angle, \(60^\circ\) (top angle):

\(\tan(60^\circ) = \frac{b}{a}\) (since for angle \(60^\circ\), opposite is \(b\), adjacent is \(a\))

\(\tan(60^\circ) = \sqrt{3}\), so \(\sqrt{3} = \frac{b}{a}\) => \(b = a\sqrt{3}\), which matches the earlier result.

\(\sin(60^\circ) = \frac{b}{c}\) => \(\frac{\sqrt{3}}{2} = \frac{b}{c}\) => \(c = \frac{2b}{\sqrt{3}}\)

Alternatively, \(\cos(30^\circ) = \frac{b}{c}\) => \(\frac{\sqrt{3}}{2} = \frac{b}{c}\) => \(c = \frac{2b}{\sqrt{3}}\)

But let's confirm with the ratio. Wait maybe I had the ratio reversed. Let's recall: 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 (length \(x\)), side opposite \(60^\circ\) is \(x\sqrt{3}\) (longer leg), and hypotenuse is \(2x\). So if the side opposite \(30^\circ\) is \(a\), then \(a = x\), longer leg \(b = x\sqrt{3}\), hypotenuse \(c = 2x\). But \(a = 12\sqrt{3}\), so \(x = 12\sqrt{3}\), then \(b = x\sqrt{3} = 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36\), and \(c = 2x = 24\sqrt{3}\). Wait, but let's check with \(\tan(30^\circ) = \frac{a}{b} = \frac{12\sqrt{3}}{36} = \frac{\sqrt{3}}{3} = \tan(30^\circ)\), which is correct. And \(\sin(30^\circ) = \frac{a}{c} = \frac{12\sqrt{3}}{24\sqrt{3}} = \frac{1}{2}\), which is correct. So that works.

Wait, but maybe I had the legs reversed. Let's check the angles again. The top angle is \(60^\circ\), so the side opposite to it is \(b\) (horizontal leg), so \(b\) is opposite \(60^\circ\), so \(b = x\sqrt{3}\), and \(a\) is opposite \(30^\circ\), so \(a = x\). Then with \(a = 12\sqrt{3}\), \(x = 12\sqrt{3}\), so \(b = x\sqrt{3} = 36\), \(c = 2x = 24\sqrt{3}\). That makes sense.

Step2: Calculate \(b\)

Using the ratio: longer leg (opposite \(60^\circ\)) is \(x\sqrt{3}\), where \(x = a = 12\sqrt{3}\) (shorter leg, opposite \(30^\circ\)). So \(b = x\sqrt{3} = 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36\).

Step3: Calculate \(c\)

Hypotenuse \(c = 2x = 2 \times 12\sqrt{3} = 24\sqrt{3}\). Wait, no: wait, if \(x\) is the shorter leg (opposite \(30^\circ\)), then hypotenuse is \(2x\). So \(c = 2 \times 12\sqrt{3} = 24\sqrt{3}\)? Wait but let's check with Pythagoras: \(a^2 + b^2 = c^2\). \(a = 12\sqrt{3}\), \(b = 36\), so \(a^2 = (12\sqrt{3})^2 = 144 \times 3 = 432\), \(b^2 = 36^2 = 1296\), sum is \(432 + 1296 = 1728\). \(c^2 = (24\sqrt{3})^2 = 576 \times 3 = 1728\). So that works.

Wait, but alternatively, if we consider \(a\) as the longer leg (opposite \(60^\circ\)), then shorter leg (opposite \(30^\circ\)) would be \(x\), longer leg \(x\sqrt{3} = a = 12\sqrt{3}\), so \(x = 12\), then \(b = x = 12\), and \(c = 2x = 24\). Wait, that's a different result. Wait, now I'm confused. Which is it?

Wait let's re-express the triangle. Let's label the angles: right angle at the bottom left, so vertices: bottom left (right angle), top left (angle \(60^\circ\)), bottom right (angle \(30^\circ\)). So sides:

- Bottom left to top left: \(a\) (vertical leg)
- Bottom left to bottom right: \(b\) (horizontal leg)
- Top left to bottom right: \(c\) (hypotenuse)

So angle at bottom right is \(30^\circ\), so the side opposite to it is \(a\) (vertical leg). So side opposite \(30^\circ\) is \(a\), so \(a\) is the shorter leg. Then side opposite \(60^\circ\) (top left angle) is \(b\) (horizontal leg), which is the longer leg. Then hypotenuse \(c\).

In 30-60-90 triangle, shorter leg (opposite \(30^\circ\)): \(x\)

Longer leg (opposite \(60^\circ\)): \(x\sqrt{3}\)

Hypotenuse: \(2x\)

So if \(a = x = 12\sqrt{3}\) (shorter leg), then:

Longer leg \(b = x\sqrt{3} = 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36\)

Hypotenuse \(c = 2x = 2 \times 12\sqrt{3} = 24\sqrt{3}\)

But let's check with trigonometry:

\(\tan(30^\circ) = \frac{a}{b}\) => \(\frac{1}{\sqrt{3}} = \frac{12\sqrt{3}}{b}\) => \(b = 12\sqrt{3} \times \sqrt{3} = 36\) (correct)

\(\sin(30^\circ) = \frac{a}{c}\) => \(\frac{1}{2} = \frac{12\sqrt{3}}{c}\) => \(c = 24\sqrt{3}\) (correct)

Alternatively, if we thought \(a\) was the longer leg (opposite \(60^\circ\)), then:

Longer leg \(a = x\sqrt{3} = 12\sqrt{3}\) => \(x = 12\) (shorter leg, opposite \(30^\circ\))

Then shorter leg \(b = x = 12\), hypotenuse \(c = 2x = 24\)

But then check Pythagoras: \(a^2 + b^2 = (12\sqrt{3})^2 + 12^2 = 432 + 144 = 576\), and \(c^2 = 24^2 = 576\). Wait, that also works! Wait, so now I see the confusion: which leg is which?

Ah! The key is to determine which angle is opposite which leg. Let's use the angle at the bottom right: \(30^\circ\). The side opposite to it is the vertical leg \(a\). The side adjacent to it is the horizontal leg \(b\). So:

\(\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c}\)

\(\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c}\)

\(\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b}\)

Similarly, for the top angle \(60^\circ\):

\(\sin(60^\circ) = \frac{b}{c}\)

\(\cos(60^\circ) = \frac{a}{c}\)

\(\tan(60^\circ) = \frac{b}{a}\)

Let's use \(\tan(60^\circ) = \sqrt{3} = \frac{b}{a}\). Given \(a = 12\sqrt{3}\), so \(b = a \times \sqrt{3} = 12\sqrt{3} \times \sqrt{3} = 12 \times 3 = 36\). Then use \(\sin(30^\circ) = \frac{1}{2} = \frac{a}{c}\) => \(c = 2a = 2 \times 12\sqrt{3} = 24\sqrt{3}\). Wait, but earlier when I assumed \(a\) was the longer leg, I got \(b = 12\) and \(c = 24\), but that contradicts \(\tan(60^\circ) = \frac{b}{a} = \frac{12}{12\sqrt{3}} = \frac{1}{\sqrt{3}} = \tan(30^\circ)\), which is wrong. So that means my initial assumption was wrong. So the correct approach is:

Since angle at bottom right is \(30^\circ\), the side opposite to it is \(a\) (vertical leg), so \(a\) is the shorter leg (opposite \(30^\circ\)), \(b\) is the longer leg (opposite \(60^\circ\)), and \(c\) is hypotenuse.

Thus, using \(\tan(60^\circ) = \frac{b}{a}\) (since \(60^\circ\) is at the top, opposite \(b\)), so \(b = a \times \tan(60^\circ) = 12\sqrt{3} \times \sqrt{3} = 36\).

Using \(\sin(30^\circ) = \frac{a}{c}\) => \(c = \frac{a}{\sin(30^\circ)} = \frac{12\sqrt{3}}{\frac{1}{2}} = 24\sqrt{3}\).

Alternatively, using the 30-60-90 ratio: shorter leg (opposite \(30^\circ\)) is \(a = 12\sqrt{3}\), so longer leg (opposite \(60^\circ\)) is \(a \times \sqrt{3} = 36\), hypotenuse is \(a \times 2 = 24\sqrt{3}\).

Yes, that makes sense. So \(b = 36\) and \(c = 24\sqrt{3}\). Wait, but let's check with Pythagoras:

\(a^2 + b^2 = (12\sqrt{3})^2 + 36^2 = 144 \times 3 + 1296 = 432 + 1296 = 1728\)

\(c^2 = (24\sqrt{3})^2 = 576 \times 3 = 1728\)

So it checks out.

Answer:

\(b = 36\), \(c = 24\sqrt{3}\)