Chapter 12 - Vectors and the Geometry of Space - 12.2 Vectors - 12.2 Exercises - Page 845: 30

$F_{horizontal}= (50N)\cos(38 ^\circ)\approx34.9N$ $F_{vertical}= (50N)\sin(38 ^\circ)\approx30.8N$

For a force $F=50N$ at $\theta= 38 ^\circ$ from horizontal we can break down the horizontal and vertical components using the formulas: $F_{horizontal}= F\cos(\theta)$ and $F_{vertical}= F\sin(\theta)$ $F_{horizontal}= (50N)cos(38 ^\circ)\approx34.9N$ $F_{vertical}= (50N)sin(38 ^\circ)\approx30.8N$