Aus Online Mathematik Brückenkurs 1

In the triangle, we seek the hypotenuse x, knowing the angle 35° and that the adjacent has length 11.

The definition of sine gives

\displaystyle \sin 35^{\circ} = \frac{11}{x}

and thus

\displaystyle x = \frac{11}{\sin 35^{\circ}}\quad ({} \approx 19\textrm{.}2)\,\textrm{.}