When you work with right triangles, it's convenient to use sine and
cosine to ‘scale’ the hypotenuse
and give you the lengths of the
shorter sides.
This is the simple thought process:

Want the side adjacent to a known angle $\,\theta\,$?
Scale the hypotenuse by $\,\cos\theta\,$, giving $\,h\cos\theta\,$.

Want the side opposite a known angle $\,\theta\,$?
Scale the hypotenuse by $\,\sin\theta\,$, giving $\,h\sin\theta\,$.
The sketches below illustrate the idea.
These scaling factors are an immediate consequence of the
right triangle definitions of sine and cosine:
Want the side adjacent to $\,\theta\,$?
Scale the hypotenuse by $\,\cos\theta\,$:
$$
\begin{gather}
\cos\theta = \frac{\text{ADJ}}{h}\cr\cr
\text{ADJ} = h\cos\theta
\end{gather}
$$

Want the side opposite $\,\theta\,$?
Scale the hypotenuse by $\,\sin\theta\,$:
$$
\begin{gather}
\sin\theta = \frac{\text{OPP}}{h}\cr\cr
\text{OPP} = h\sin\theta
\end{gather}
$$

Given a right triangle with known hypotenuse and one known acute angle,
you should immediately be able to write down exact lengths for
the remaining two sides.
You shouldn't have to pass through the extra
step of thinking:
$\,\displaystyle\text{sine} = \frac{\text{OPP}}{\text{HYP}}\,$ or
$\,\displaystyle\text{cosine} = \frac{\text{ADJ}}{\text{HYP}}\,$
The scaling factors are also an immediate consequence of the
unit circle definitions of sine and cosine.
To get the desired right triangle with hypotenuse $\,h\,$ and acute angle $\,\theta\,$,
start with a
similar right triangle inside the unit circle.
Then, scale it to the correct size, as follows:
 Lay off the acute angle $\,\theta\,$ in the unit circle.
The hypotenuse of the blue triangle has length $\,1\,$.
 By definition:
 the $x$value of the terminal point is $\,\cos\theta\,$.
 the $y$value of the terminal point is $\,\sin\theta\,$.
 Consequently, in the blue triangle:
 the side adjacent to $\,\theta\,$ has length $\,\cos\theta\,$
 the side opposite from $\,\theta\,$ has length $\,\sin\theta\,$
 Scale the blue triangle by a factor of
$\,h\,$, giving the yellow triangle.
That is, all three sides of the blue triangle get multiplied by $\,h\,$.
 Consequently, in the yellow triangle:
 the hypotenuse has length $\,h\,$
 the side adjacent to $\,\theta\,$ has length $\,h\cos\theta\,$
(so the hypotenuse has been scaled by $\,\cos\theta\,$)
 the side opposite from $\,\theta\,$ has length $\,h\sin\theta\,$
(so the hypotenuse has been scaled by $\,\sin\theta\,$)


SUMMARY:
Suppose you're given a right triangle with known hypotenuse $\,h\,$
and one known acute angle $\,\theta\,$.
The remaining two sides are shorter than the hypotenuse.
How much shorter?
Well, $\,0 < \cos\theta < 1\,$.
The side adjacent to $\,\theta\,$ has length $\,h\cos\theta\,$.
So, $\,\cos\theta\,$ gives the appropriate ‘shrinking factor’ for the adjacent side!
Also, $\,0 < \sin\theta < 1\,$.
The side opposite from $\,\theta\,$ has length $\,h\sin\theta\,$.
So, $\,\sin\theta\,$ gives the appropriate ‘shrinking factor’ for the opposite side!
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Area of a Triangle