## Trigonometry (10th Edition)

The magnitude of the horizontal component of $v$ is $25\sqrt{2}$ The magnitude of the vertical component of $v$ is $25\sqrt{2}$
We can find the magnitude of the horizontal component of $v$: $v_x = v~cos~\theta$ $v_x = (50)~cos~45^{\circ}$ $v_x = (50)~(\frac{\sqrt{2}}{2})$ $v_x = 25\sqrt{2}$ The magnitude of the horizontal component of $v$ is $25\sqrt{2}$ We can find the magnitude of the vertical component of $v$: $v_y = v~sin~\theta$ $v_y = (50)~sin~45^{\circ}$ $v_y = (50)~(\frac{\sqrt{2}}{2})$ $v_y = 25\sqrt{2}$ The magnitude of the vertical component of $v$ is $25\sqrt{2}$