## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Since the $40^{\circ}$ angle is below the positive x-axis, the x-component of the vector is positive. We can find the x-component. Let $y$ be the magnitude of the y-component; $tan(\theta) = \frac{y}{x}$ $x = \frac{y}{tan(\theta)}$ $x = \frac{10~m/s}{tan(40^{\circ})}$ $x = 12~m/s$ The x-component of the vector is thus 12 m/s.