Question

find the sine, cosine, and tangent of $\angle e$.
triangle image with right angle at f, side ef = $3\sqrt{5}$, side eg = $2\sqrt{17}$
write your answer in simplified, rationalized form. do not round.
$\sin(e) = \square$
$\cos(e) = \square$
$\tan(e) = \square$

Explanation:

Step1: Find the length of FG

In right triangle \( EFG \) with right angle at \( F \), we use the Pythagorean theorem \( EG^{2}=EF^{2}+FG^{2} \). Let \( EF = 3\sqrt{5} \), \( EG=2\sqrt{17} \). Then \( FG=\sqrt{EG^{2}-EF^{2}} \).

First, calculate \( EG^{2}=(2\sqrt{17})^{2}=4\times17 = 68 \) and \( EF^{2}=(3\sqrt{5})^{2}=9\times5=45 \).

So \( FG=\sqrt{68 - 45}=\sqrt{23} \)? Wait, no, \( 68-45 = 23 \)? Wait, no, \( 2\sqrt{17}\) squared is \( 4\times17 = 68 \), \( 3\sqrt{5}\) squared is \( 9\times5 = 45 \), \( 68 - 45=23 \)? Wait, that can't be. Wait, maybe I made a mistake. Wait, \( EG \) is the hypotenuse? Wait, no, in the right triangle, the hypotenuse is the side opposite the right angle. The right angle is at \( F \), so the hypotenuse is \( EG \). So \( EG^{2}=EF^{2}+FG^{2} \), so \( FG^{2}=EG^{2}-EF^{2} \). Let's recalculate:

\( EG = 2\sqrt{17} \), so \( EG^{2}=(2)^{2}\times(\sqrt{17})^{2}=4\times17 = 68 \)

\( EF = 3\sqrt{5} \), so \( EF^{2}=(3)^{2}\times(\sqrt{5})^{2}=9\times5 = 45 \)

Then \( FG^{2}=68 - 45 = 23 \)? Wait, that seems odd. Wait, maybe the right angle is at \( F \), so the sides: \( EF \) and \( FG \) are legs, \( EG \) is hypotenuse. So \( FG=\sqrt{EG^{2}-EF^{2}}=\sqrt{68 - 45}=\sqrt{23} \). Wait, but maybe I mixed up the sides. Wait, angle \( E \): we need to find sine, cosine, tangent of \( \angle E \). So for \( \angle E \), the opposite side is \( FG \), adjacent side is \( EF \), hypotenuse is \( EG \).

Step2: Calculate \( \sin(E) \)

\( \sin(E)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{FG}{EG} \)

We have \( FG=\sqrt{23} \)? Wait, no, wait, \( 68 - 45 = 23 \), so \( FG=\sqrt{23} \). Wait, but let's check again. Wait, maybe I made a mistake in identifying the sides. Wait, the right angle is at \( F \), so:

- Opposite to \( \angle E \): \( FG \)
- Adjacent to \( \angle E \): \( EF \)
- Hypotenuse: \( EG \)

So \( \sin(E)=\frac{FG}{EG} \), \( \cos(E)=\frac{EF}{EG} \), \( \tan(E)=\frac{FG}{EF} \)

Wait, but let's recalculate \( FG \):

\( EG^{2}=EF^{2}+FG^{2} \implies FG^{2}=EG^{2}-EF^{2}=(2\sqrt{17})^{2}-(3\sqrt{5})^{2}=4\times17 - 9\times5=68 - 45 = 23 \implies FG=\sqrt{23} \)

Wait, but \( \sqrt{23} \) is correct? Wait, maybe the problem has a typo, but let's proceed.

Step3: Calculate \( \sin(E) \)

\( \sin(E)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{FG}{EG}=\frac{\sqrt{23}}{2\sqrt{17}} \). Rationalize the denominator: multiply numerator and denominator by \( \sqrt{17} \), we get \( \frac{\sqrt{23}\times\sqrt{17}}{2\times17}=\frac{\sqrt{391}}{34} \)

Wait, that can't be right. Wait, maybe I mixed up the sides. Wait, maybe \( EF \) is not the adjacent side. Wait, let's look at the triangle again. The vertices are \( E \), \( F \), \( G \), right angle at \( F \). So \( \angle E \) is at vertex \( E \), so the sides:

- \( EF \): one leg (from \( E \) to \( F \))
- \( FG \): another leg (from \( F \) to \( G \))
- \( EG \): hypotenuse (from \( E \) to \( G \))

So for \( \angle E \):

- Opposite side: \( FG \)
- Adjacent side: \( EF \)
- Hypotenuse: \( EG \)

Wait, but maybe I made a mistake in the Pythagorean theorem. Let's check again:

\( EG = 2\sqrt{17} \), so \( EG^2 = 4 \times 17 = 68 \)

\( EF = 3\sqrt{5} \), so \( EF^2 = 9 \times 5 = 45 \)

Then \( FG^2 = EG^2 - EF^2 = 68 - 45 = 23 \), so \( FG = \sqrt{23} \)

Then:

\( \sin(E) = \frac{FG}{EG} = \frac{\sqrt{23}}{2\sqrt{17}} = \frac{\sqrt{23 \times 17}}{2 \times 17} = \frac{\sqrt{391}}{34} \)

\( \cos(E) = \frac{EF}{EG} = \frac{3\sqrt{5}}{2\sqrt{17}} = \frac{3\sqrt{5 \times 17}}{2 \times 17} = \frac{3\sqrt{85}}{34} \)

\( \tan(E) = \frac{FG}{EF} = \frac{\sqrt{23}}{3\sqrt{5}} = \frac{\sqrt{23 \times 5}}{3 \times 5} = \frac{\sqrt{115}}{15} \)

Wait, but this seems complicated. Maybe I made a mistake in identifying the hypotenuse. Wait, maybe \( EG \) is not the hypotenuse. Wait, the right angle is at \( F \), so the hypotenuse is \( EG \), so that part is correct. Alternatively, maybe the length of \( FG \) is calculated wrong. Wait, let's check the problem again. The triangle has \( EF = 3\sqrt{5} \), \( EG = 2\sqrt{17} \), right angle at \( F \). So \( EG \) is the hypotenuse. Then \( FG = \sqrt{EG^2 - EF^2} = \sqrt{(2\sqrt{17})^2 - (3\sqrt{5})^2} = \sqrt{68 - 45} = \sqrt{23} \). So that's correct.

So:

\( \sin(E) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{FG}{EG} = \frac{\sqrt{23}}{2\sqrt{17}} = \frac{\sqrt{23 \times 17}}{2 \times 17} = \frac{\sqrt{391}}{34} \)

\( \cos(E) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{EF}{EG} = \frac{3\sqrt{5}}{2\sqrt{17}} = \frac{3\sqrt{85}}{34} \)

\( \tan(E) = \frac{\text{opposite}}{\text{adjacent}} = \frac{FG}{EF} = \frac{\sqrt{23}}{3\sqrt{5}} = \frac{\sqrt{115}}{15} \)

Wait, but maybe I made a mistake in the Pythagorean theorem. Let's check the values again. \( 2\sqrt{17} \) is approximately \( 2\times4.123 = 8.246 \), \( 3\sqrt{5} \) is approximately \( 3\times2.236 = 6.708 \). Then \( \sqrt{8.246^2 - 6.708^2} = \sqrt{68 - 45} = \sqrt{23} \approx 4.796 \). Let's check if \( 6.708^2 + 4.796^2 \approx 45 + 23 = 68 \), and \( 8.246^2 \approx 68 \), so that's correct.

So the calculations are:

\( \sin(E) = \frac{\sqrt{23}}{2\sqrt{17}} = \frac{\sqrt{391}}{34} \)

\( \cos(E) = \frac{3\sqrt{5}}{2\sqrt{17}} = \frac{3\sqrt{85}}{34} \)

\( \tan(E) = \frac{\sqrt{23}}{3\sqrt{5}} = \frac{\sqrt{115}}{15} \)

Wait, but maybe the problem has a different configuration. Wait, maybe \( EF \) is not the adjacent side. Wait, no, in angle \( E \), the adjacent side is the leg that is part of angle \( E \) and not the hypotenuse. So \( EF \) is adjacent, \( FG \) is opposite, \( EG \) is hypotenuse.

Answer:

\( \sin(E) = \boldsymbol{\frac{\sqrt{391}}{34}} \)

\( \cos(E) = \boldsymbol{\frac{3\sqrt{85}}{34}} \)

\( \tan(E) = \boldsymbol{\frac{\sqrt{115}}{15}} \)