Aus Online Mathematik Brückenkurs 1

In the triangle, we seek the hypotenuse \displaystyle x, knowing the angle 35o 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 \left( \approx 19.2 \right)