Solution 4.2:1e
From Förberedande kurs i matematik 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{.} | 

