Question

find the sine, cosine, and tangent of $\angle e$.
simplify your answers and write them as proper fractions, improper fractions, or whole numbers.
$\sin (e) = \square$
$\cos (e) = \square$
$\tan (e) = \square$

Explanation:

Step1: Find the length of EF

In right triangle \( EFG \) with right angle at \( F \), hypotenuse \( EG = 73 \), and \( FG = 55 \). Using the Pythagorean theorem \( EF^{2}+FG^{2}=EG^{2} \), we can find \( EF \).

$$ EF=\sqrt{EG^{2}-FG^{2}}=\sqrt{73^{2}-55^{2}}=\sqrt{(73 + 55)(73 - 55)}=\sqrt{128\times18}=\sqrt{2304}=48 $$

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

In a right triangle, \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \). For \( \angle E \), the opposite side is \( FG = 55 \) and the hypotenuse is \( EG = 73 \). So,

$$ \sin(E)=\frac{FG}{EG}=\frac{55}{73} $$

Step3: Calculate \( \cos(E) \)

In a right triangle, \( \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}} \). For \( \angle E \), the adjacent side is \( EF = 48 \) and the hypotenuse is \( EG = 73 \). So,

$$ \cos(E)=\frac{EF}{EG}=\frac{48}{73} $$

Step4: Calculate \( \tan(E) \)

In a right triangle, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). For \( \angle E \), the opposite side is \( FG = 55 \) and the adjacent side is \( EF = 48 \). So,

$$ \tan(E)=\frac{FG}{EF}=\frac{55}{48} $$

Answer:

\( \sin(E)=\frac{55}{73} \)
\( \cos(E)=\frac{48}{73} \)
\( \tan(E)=\frac{55}{48} \)