Find the component form of v given its magnitude and the angle it makes with the positive x-axis. Round your answer to four decimals.‖v‖=9,θ=150°

Find the component form of v given its magnitude and the angle it makes with the positive x-axis.‖v‖=13,v in the direction of 4i+4j

If v lies in the first quadrant and makes an angle 𝜋/3 with the positive x-axis and |v| = 10, find v in component form.

Let θ be the angle that the vector A makes with the +x axis, measured counterclockwise from that axis. Find θ for a vector that has the following components: Ax=75.4,Ay=42.5. Note the answer should be in radian. the answer has no unit

A vector has a magnitude of 10 units and its y-component is 5 units. What is its angle with positive x-axis?Group of answer choices60o45o30o0.52o90o

To find the component form of the vector \(\mathbf{v}\) with a magnitude of 4 and a bearing of 125°, we need to convert the polar coordinates (magnitude and direction) into Cartesian coordinates (components along the \(i\) and \(j\) directions). 1. **Convert the bearing to a standard angle:** Bearings are measured clockwise from the north direction. A bearing of 125° means the angle is 125° clockwise from the north. To convert this to a standard angle (measured counterclockwise from the positive x-axis), we use: \[ \text{Standard angle} = 360° - 125° = 235° \] 2. **Calculate the components:** The components of the vector can be found using trigonometry: \[ v_x = v \cos(\theta) \] \[ v_y = v \sin(\theta) \] where \(v\) is the magnitude and \(\theta\) is the standard angle. Given: \[ v = 4, \quad \theta = 235° \] Converting 235° to radians: \[ \theta = 235° \times \frac{\pi}{180°} = \frac{235\pi}{180} \approx 4.1015 \text{ radians} \] Now, calculate the components: \[ v_x = 4 \cos(235°) = 4 \cos(4.1015) \approx 4 \times (-0.5736) \approx -2.29 \] \[ v_y = 4 \sin(235°) = 4 \sin(4.1015) \approx 4 \times (-0.8192) \approx -3.28 \] 3. **Write the vector in component form:** \[ \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} = -2.29\mathbf{i} - 3.28\mathbf{j} \] Therefore, the correct answer is: \[ \boxed{E} \]