Right triangle trigonometry

The trig ratios of a right triangle each use two of its three sides.
$\sin\theta$ is opposite over hypotenuse, $\cos\theta$ is adjacent over hypotenuse, and $\tan\theta$ is opposite over adjacent.
SOH-CAH-TOA captures all three.

The figure labels the sides relative to angle $\theta$.
The side across from $\theta$ is the opposite; the side beside $\theta$ that is not the hypotenuse is the adjacent; and the hypotenuse faces the right angle.
Choose the ratio that uses the sides you care about.

To find a missing side, pick the ratio linking the side you want to the one you know.
If the hypotenuse is $18$ and you want the side opposite a $30^\circ$ angle, use $\sin 30^\circ = \frac{\text{opposite}}{18}$.
Since $\sin 30^\circ = \frac{1}{2}$, the opposite side is $9$.

Often you combine a trig ratio with the Pythagorean theorem.
If $\tan\theta = \frac{8}{15}$, take the opposite as $8$ and the adjacent as $15$, so the hypotenuse is $\sqrt{8^2 + 15^2} = 17$.
Then any other ratio, like $\cos\theta = \frac{15}{17}$, follows.

In a right triangle, the side opposite angle $Z$ is $5$ and the hypotenuse is $13$. What is $\sin Z$?
$\sin Z = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}$.
So $\sin Z = \frac{5}{13}$.

A right triangle has hypotenuse $20$, and angle $A = 30^\circ$. What is the side opposite $A$?
Use $\sin 30^\circ = \frac{\text{opposite}}{20}$, and $\sin 30^\circ = \frac{1}{2}$.
So the opposite side is $20 \times \frac{1}{2} = 10$.

In a right triangle, $\tan\theta = \frac{5}{12}$. What is $\cos\theta$?
Take opposite $= 5$ and adjacent $= 12$, so the hypotenuse is $\sqrt{5^2 + 12^2} = 13$.
Then $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}$.