find the sine, cosine, and tangent of $angle h$.\nsimplify your answers and write them as proper fractions…

Explanation:

Step1: Find hypotenuse $GH$

Use Pythagorean theorem: $$GH = \sqrt{GF^2 + HF^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$

Step2: Calculate $\sin(H)$

Opposite over hypotenuse: $$\sin(H) = \frac{\text{Opposite to } \angle H}{\text{Hypotenuse}} = \frac{12}{13}$$

Step3: Calculate $\cos(H)$

Adjacent over hypotenuse: $$\cos(H) = \frac{\text{Adjacent to } \angle H}{\text{Hypotenuse}} = \frac{5}{13}$$

Step4: Calculate $\tan(H)$

Opposite over adjacent: $$\tan(H) = \frac{\text{Opposite to } \angle H}{\text{Adjacent to } \angle H} = \frac{12}{5}$$

Answer:

$\sin(H) = \frac{12}{13}$ $\cos(H) = \frac{5}{13}$ $\tan(H) = \frac{12}{5}$