Question

fill in the blank 1 point if b = 24, find a and c. a= choose your answer... c= choose your answer...

Explanation:

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 one (let's call it \(x\)), the side opposite \(60^{\circ}\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\).

Step 1: Identify the sides

In the given triangle, angle \(30^{\circ}\) is opposite side \(a\), angle \(60^{\circ}\) is opposite side \(b = 24\), and the hypotenuse is \(c\).

From the ratio of 30 - 60 - 90 triangle:
- Side opposite \(60^{\circ}\) (which is \(b\)) \(=x\sqrt{3}\)
- Side opposite \(30^{\circ}\) (which is \(a\)) \(=x\)
- Hypotenuse \(c = 2x\)

We know that \(b=24=x\sqrt{3}\), so we can solve for \(x\):

\(x=\frac{24}{\sqrt{3}}\)

Rationalize the denominator:

\(x=\frac{24\sqrt{3}}{3}=8\sqrt{3}\times\sqrt{3}?\) Wait, no. Wait, \(\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3} = 8\sqrt{3}\)? Wait, no, let's correct that.

Wait, if \(b\) (opposite \(60^{\circ}\)) is \(x\sqrt{3}=24\), then \(x=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3} = 8\sqrt{3}\)? Wait, no, that's wrong. Wait, let's re - identify the sides.

Wait, in the right - triangle, the right angle is between \(a\) and \(b\). So angle \(30^{\circ}\) is at the end of side \(b\), so side \(a\) is opposite \(30^{\circ}\), side \(b\) is opposite \(60^{\circ}\), and hypotenuse \(c\) is opposite the right angle.

So:
- \(\tan(30^{\circ})=\frac{a}{b}\)
- \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), and \(b = 24\)

So \(\frac{a}{24}=\frac{1}{\sqrt{3}}\), then \(a=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\times\sqrt{3}?\) No, \(\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\)? Wait, no, \(24\div\sqrt{3}=8\sqrt{3}\times\sqrt{3}\) is wrong. Wait, \(\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\)? Wait, \(24\div\sqrt{3}\): multiply numerator and denominator by \(\sqrt{3}\), we get \(\frac{24\sqrt{3}}{3}=8\sqrt{3}\)? No, \(24\div3 = 8\), so \(\frac{24\sqrt{3}}{3}=8\sqrt{3}\)? Wait, no, \(24\sqrt{3}\div3 = 8\sqrt{3}\). But also, we can use the ratio of 30 - 60 - 90 triangle.

In a 30 - 60 - 90 triangle, the sides are in the ratio \(a:b:c=1:\sqrt{3}:2\) (where \(a\) is opposite \(30^{\circ}\), \(b\) is opposite \(60^{\circ}\), \(c\) is hypotenuse).

So if \(b\) (opposite \(60^{\circ}\)) is \(\sqrt{3}\) part, and \(a\) (opposite \(30^{\circ}\)) is \(1\) part, \(c\) (hypotenuse) is \(2\) part.

Let the length of the side opposite \(30^{\circ}\) (i.e., \(a\)) be \(x\). Then the side opposite \(60^{\circ}\) (i.e., \(b\)) is \(x\sqrt{3}\), and hypotenuse \(c = 2x\).

We know that \(b = 24=x\sqrt{3}\), so \(x=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\)? Wait, no, that's not correct. Wait, \(\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\)? Wait, no, \(24\div\sqrt{3}=8\sqrt{3}\times\sqrt{3}\) is wrong. Wait, let's calculate \(\frac{24}{\sqrt{3}}\):

Multiply numerator and denominator by \(\sqrt{3}\): \(\frac{24\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\). Wait, but if we use \(\tan(30^{\circ})=\frac{a}{b}\), \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(a = b\times\tan(30^{\circ})=24\times\frac{1}{\sqrt{3}}=\frac{24}{\sqrt{3}} = 8\sqrt{3}\times\sqrt{3}\)? No, \(8\sqrt{3}\) is approximately \(13.856\), but let's check with another trigonometric function.

Alternatively, \(\cos(30^{\circ})=\frac{b}{c}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(\frac{24}{c}=\frac{\sqrt{3}}{2}\), then \(c=\frac{24\times2}{\sqrt{3}}=\frac{48}{\sqrt{3}}=\frac{48\sqrt{3}}{3}=16\sqrt{3}\)? Wait, no, I think I mixed up the sides.

Wait, let's start over.

In the right - triangle, angle \(30^{\circ}\), angle \(60^{\circ}\), right angle.

So:
- \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}\)
- \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), \(b = 24\)
- So \(a=b\times\tan(30^{\circ})=24\times\frac{1}{\sqrt{3}}=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\times\sqrt{3}\)? No, \(\frac{24}{\sqrt{3}} = 8\sqrt{3}\approx13.856\)? Wait, no, \(24\div\sqrt{3}\approx24\div1.732\approx13.856\), and \(8\sqrt{3}\approx13.856\).

- For the hypotenuse \(c\), we can use \(\cos(30^{\circ})=\frac{b}{c}\), \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\), so \(c=\frac{b}{\cos(30^{\circ})}=\frac{24}{\frac{\sqrt{3}}{2}}=\frac{48}{\sqrt{3}}=\frac{48\sqrt{3}}{3}=16\sqrt{3}\approx27.712\)? Wait, no, that's not right. Wait, or we can use the Pythagorean theorem. If \(a = 8\sqrt{3}\) and \(b = 24\), then \(c=\sqrt{a^{2}+b^{2}}=\sqrt{(8\sqrt{3})^{2}+24^{2}}=\sqrt{192 + 576}=\sqrt{768}=16\sqrt{3}\approx27.712\). But also, in a 30 - 60 - 90 triangle, the hypotenuse is twice the shorter leg. Wait, the shorter leg is \(a\) (opposite \(30^{\circ}\)), so \(c = 2a\). Let's check: if \(a = 8\sqrt{3}\), then \(c=16\sqrt{3}\), and \(2a = 16\sqrt{3}\), which matches. And \(b=\sqrt{c^{2}-a^{2}}=\sqrt{(16\sqrt{3})^{2}-(8\sqrt{3})^{2}}=\sqrt{768 - 192}=\sqrt{576}=24\), which is correct.

Wait, but there is a simpler way. In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (shorter leg : longer leg : hypotenuse). The longer leg (opposite \(60^{\circ}\)) is \(b = 24\), which is \(\sqrt{3}\) times the shorter leg \(a\). So \(a=\frac{b}{\sqrt{3}}=\frac{24}{\sqrt{3}} = 8\sqrt{3}\times\sqrt{3}\)? No, \(a=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\), and the hypotenuse \(c = \frac{2b}{\sqrt{3}}=\frac{48}{\sqrt{3}} = 16\sqrt{3}\), or \(c = 2a\) (since \(c = 2\times8\sqrt{3}=16\sqrt{3}\)).

Wait, but we can also use the fact that \(\tan(60^{\circ})=\frac{b}{a}\), \(\tan(60^{\circ})=\sqrt{3}\), so \(\sqrt{3}=\frac{24}{a}\), then \(a=\frac{24}{\sqrt{3}}=8\sqrt{3}\approx13.856\), and \(c=\frac{a}{\sin(30^{\circ})}\), since \(\sin(30^{\circ})=\frac{1}{2}\), so \(c = 2a=16\sqrt{3}\approx27.712\).

Step 1: Find \(a\)

We know that in a 30 - 60 - 90 triangle, \(\tan(30^{\circ})=\frac{a}{b}\). Since \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\) and \(b = 24\), we have:

\(a=b\times\tan(30^{\circ})=24\times\frac{1}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}\)

Step 2: Find \(c\)

We know that \(\cos(30^{\circ})=\frac{b}{c}\). Since \(\cos(30^{\circ})=\frac{\sqrt{3}}{2}\) and \(b = 24\), we have:

\(\frac{\sqrt{3}}{2}=\frac{24}{c}\)

Cross - multiply: \(c\times\sqrt{3}=24\times2 = 48\)

Then \(c=\frac{48}{\sqrt{3}}=\frac{48\sqrt{3}}{3}=16\sqrt{3}\)

Or we can use the Pythagorean theorem: \(c=\sqrt{a^{2}+b^{2}}=\sqrt{(8\sqrt{3})^{2}+24^{2}}=\sqrt{192 + 576}=\sqrt{768}=16\sqrt{3}\)

Answer:

\(a = 8\sqrt{3}\) (or approximately \(13.86\)) and \(c = 16\sqrt{3}\) (or approximately \(27.71\))

If we rationalize or simplify further:

\(a = 8\sqrt{3}\approx13.86\), \(c = 16\sqrt{3}\approx27.71\)

But if we want to write them in exact form:

\(a = 8\sqrt{3}\), \(c = 16\sqrt{3}\)

Alternatively, we can write \(a = 8\sqrt{3}\approx13.86\) and \(c = 16\sqrt{3}\approx27.71\), or we can note that in a 30 - 60 - 90 triangle, if the longer leg (opposite \(60^{\circ}\)) is \(24\), the shorter leg (opposite \(30^{\circ}\)) is \(\frac{24}{\sqrt{3}} = 8\sqrt{3}\) and the hypotenuse is \(\frac{48}{\sqrt{3}}=16\sqrt{3}\)

Wait, another way: In a 30 - 60 - 90 triangle, the sides are in the ratio \(1:\sqrt{3}:2\) (shorter leg : longer leg : hypotenuse). Let the shorter leg (a) be \(x\), longer leg (b) be \(x\sqrt{3}\), hypotenuse (c) be \(2x\).

Given \(b = 24=x\sqrt{3}\), so \(x=\frac{24}{\sqrt{3}} = 8\sqrt{3}\) (this is \(a\)), and \(c = 2x=16\sqrt{3}\)

So, \(a = 8\sqrt{3}\) (or approximately \(13.86\)) and \(c = 16\sqrt{3}\) (or approximately \(27.71\))