Find the magnitudes of the horizontal and vertical components for the vector $\mathbf{v}$, if $\alpha$ is the direction angle of $\mathbf{v}$ from the horizontal. \[ \alpha=43^{\circ},|\mathbf{v}|=75 \] The magnitude of the horizontal component of $\mathbf{v}$ is (Round to the nearest integer as needed.) The magnitude of the vertical component of $v$ is (Round to the nearest integer as needed.)
Solution
Step 1 ：Given that the direction angle of the vector $\mathbf{v}$ from the horizontal is $\alpha=43^\circ$ and the magnitude of the vector is $|\mathbf{v}|=75$.
Step 2 ：The magnitude of the horizontal component of a vector can be found using the formula $|\mathbf{v}|\cos(\alpha)$, where $|\mathbf{v}|$ is the magnitude of the vector and $\alpha$ is the angle the vector makes with the horizontal.
Step 3 ：Similarly, the magnitude of the vertical component can be found using the formula $|\mathbf{v}|\sin(\alpha)$.
Step 4 ：Substituting the given values into the formulas, we find that the magnitude of the horizontal component of $\mathbf{v}$ is approximately 55 and the magnitude of the vertical component of $\mathbf{v}$ is approximately 51.
Step 5 ：Final Answer: The magnitude of the horizontal component of $\mathbf{v}$ is \(\boxed{55}\). The magnitude of the vertical component of $\mathbf{v}$ is \(\boxed{51}\).
