There are two basic approaches to trigonometry:
Both approaches were introduced in Introduction to Trigonometry.Be sure to read this prior section, since it covers important notation and conventions.

using the familiar $\,30^\circ$$60^\circ$$90^\circ\,$ triangle:
$$
\begin{alignat}{2}
\sin 30^\circ&\ \ =\ \ && \frac{s}{2s} &\ \ =\ \ && \frac 12\cr\cr
\cos 30^\circ &\ \ =\ \ && \frac{\sqrt 3s}{2s} &\ \ =\ \ && \frac{\sqrt 3}2\cr\cr
\sin 60^\circ &\ \ =\ \ && \frac{\sqrt 3s}{2s} &\ \ =\ \ && \frac{\sqrt 3}2\cr\cr
\cos 60^\circ &\ \ =\ \ && \frac{s}{2s} &\ \ =\ \ && \frac 12
\end{alignat}
$$

Consider a right triangle with legs of lengths $\,5\,$ and $\,7\,.$
Let $\,\theta\,$ be the largest acute angle.
Find $\,\sin\theta\,,$ $\,\cos\theta\,,$ and $\,\tan\theta\,.$
(It is not necessary to simplify radicals or to rationalize denominators.)
SOLUTION: By the Pythagorean Theorem: ${\text{HYP}}^2 = 5^2 + 7^2$ Therefore: $\text{HYP} = \sqrt{5^2 + 7^2} = \sqrt{74}$ The largest acute angle is opposite the longest leg. Thus: $$ \begin{gather} \cssId{s63}{\sin\theta = \frac{7}{\sqrt{74}}}\cr\cr \cssId{s64}{\cos\theta = \frac{5}{\sqrt{74}}}\cr\cr \cssId{s65}{\tan\theta = \frac{7}{5}} \end{gather} $$ NOTE: Suppose that the right triangle had legs of lengths $\,500\,$ and $\,700\,$ (instead of $\,5\,$ and $\,7\,$). Then, you would first ‘shrink’ the triangle to a more manageable size, by dividing the lengths by $\,100\,.$ Similar triangles have the same angles! Don't work with big numbers when you don't have to! Or, suppose the right triangle had legs of lengths $\,0.00005\,$ and $\,0.00007\,$ (instead of $\,5\,$ and $\,7\,$). You would first ‘stretch’ the triangle by multiplying the lengths by $\,100{,}000\,.$ Similar triangles have the same angles! Don't work with very small numbers when you don't have to! 

On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
