#### Answer

The incline makes an angle of $18.7^{\circ}$ with the horizontal.

#### Work Step by Step

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.