Question

find the magnitudes of the horizontal and vertical components for the vector v, if α is the direction angle of v from the horizontal. α = 192.7°, |v| = 168 the magnitude of the horizontal component of v is 164 (round to the nearest integer as needed.) the magnitude of the vertical component of v is (round to the nearest integer as needed.)

Explanation:

Step1: Recall the formula for horizontal and vertical components

The horizontal component \(v_x = |\mathbf{v}|\cos\alpha\) and the vertical component \(v_y=|\mathbf{v}|\sin\alpha\), where \(|\mathbf{v}|\) is the magnitude of the vector and \(\alpha\) is the direction angle.
Given \(|\mathbf{v}| = 168\) and \(\alpha=192.7^{\circ}\)

Step2: Calculate the vertical component

We know that \(v_y = |\mathbf{v}|\sin\alpha\)
Substitute \(|\mathbf{v}| = 168\) and \(\alpha = 192.7^{\circ}\) into the formula:
\(v_y=168\times\sin(192.7^{\circ})\)
Since \(\sin(192.7^{\circ})=\sin(180^{\circ} + 12.7^{\circ})=-\sin(12.7^{\circ})\)
\(\sin(12.7^{\circ})\approx0.22\)
\(v_y=168\times(- 0.22)\approx - 37\)
The magnitude of the vertical component is \(|v_y|\)

Answer:

The magnitude of the vertical component of \(\mathbf{v}\) is \(37\)