Trigonometry (11th Edition) Clone

The incline makes an angle of $18.7^{\circ}$ with the horizontal.
The wheelbarrow's weight of $50.0~lb$ is directed straight down. The force of $16.0~lb$ is directed up the incline at an angle of $\theta$ above the horizontal. We can draw a triangle with the $50.0~lb$ as the hypotenuse of the triangle and the force of $16.0~lb$ opposite the angle $\theta$. We can find $\theta$: $sin~\theta = \frac{16.0}{50.0}$ $\theta = arcsin(\frac{16.0}{50.0})$ $\theta = 18.7^{\circ}$ The incline makes an angle of $18.7^{\circ}$ with the horizontal.