Question

find the sine, cosine, and tangent of ∠f.
write your answer in simplified, rationalized form. do not round.
sin(f) =
cos(f) =
tan(f) =

Explanation:

Step1: Calculate side $HG$

Use Pythagorean theorem:
$$HG = \sqrt{FH^2 - FG^2} = \sqrt{(4\sqrt{6})^2 - (\sqrt{61})^2} = \sqrt{96 - 61} = \sqrt{35}$$

Step2: Find $\sin(F)$

Opposite over hypotenuse:
$$\sin(F) = \frac{HG}{FH} = \frac{\sqrt{35}}{4\sqrt{6}} = \frac{\sqrt{35} \cdot \sqrt{6}}{4 \cdot 6} = \frac{\sqrt{210}}{24}$$

Step3: Find $\cos(F)$

Adjacent over hypotenuse:
$$\cos(F) = \frac{FG}{FH} = \frac{\sqrt{61}}{4\sqrt{6}} = \frac{\sqrt{61} \cdot \sqrt{6}}{4 \cdot 6} = \frac{\sqrt{366}}{24}$$

Step4: Find $\tan(F)$

Opposite over adjacent:
$$\tan(F) = \frac{HG}{FG} = \frac{\sqrt{35}}{\sqrt{61}} = \frac{\sqrt{35} \cdot \sqrt{61}}{61} = \frac{\sqrt{2135}}{61}$$

Answer:

$\sin(F) = \frac{\sqrt{210}}{24}$
$\cos(F) = \frac{\sqrt{366}}{24}$
$\tan(F) = \frac{\sqrt{2135}}{61}$