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 is 21.8 m long and makes an angle of 58.9° counterclockwise from the y-axis (on the side of the -x-axis). What is the x-component of this vector?

A vector A has components x-component of 12 m and y-component 5 m. What is the angle that vector A makes with the x axis?Group of answer choices2.6 degrees12.6 degrees22.6 degrees32.6 degrees42.6 degrees

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

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°

To find the resultant vector using the numerical method, we need to break each vector into its components and then sum the components. ### Step 1: Break each vector into its components#### Vector \(\vec{A}\) Magnitude: \(A = 60\) Angle: \(\theta_1 = 45^\circ\) Components: \[ A_x = A \cos(\theta_1) = 60 \cos(45^\circ) = 60 \left(\frac{\sqrt{2}}{2}\right) = 60 \times 0.7071 = 42.43 \] \[ A_y = A \sin(\theta_1) = 60 \sin(45^\circ) = 60 \left(\frac{\sqrt{2}}{2}\right) = 60 \times 0.7071 = 42.43 \] #### Vector \(\vec{B}\) Magnitude: \(B = 60\) Angle: \(\theta_2 = 20^\circ\) (below the horizontal axis, so it's negative) Components: \[ B_x = B \cos(\theta_2) = 60 \cos(20^\circ) = 60 \times 0.9397 = 56.38 \] \[ B_y = B \sin(\theta_2) = 60 \sin(20^\circ) = 60 \times 0.3420 = 20.52 \] Since \(\theta_2\) is below the horizontal axis, \(B_y\) is negative: \[ B_y = -20.52 \] #### Vector \(\vec{C}\) Magnitude: \(C = 60\) Angle: \(180^\circ\) (directly to the left) Components: \[ C_x = C \cos(180^\circ) = 60 \cos(180^\circ) = 60 \times (-1) = -60 \] \[ C_y = C \sin(180^\circ) = 60 \sin(180^\circ) = 60 \times 0 = 0 \] ### Step 2: Sum the components#### Sum of \(x\)-components: \[ R_x = A_x + B_x + C_x \] \[ R_x = 42.43 + 56.38 - 60 \] \[ R_x = 38.81 \] #### Sum of \(y\)-components: \[ R_y = A_y + B_y + C_y \] \[ R_y = 42.43 - 20.52 + 0 \] \[ R_y = 21.91 \] ### Step 3: Find the magnitude and direction of the resultant vector#### Magnitude: \[ R = \sqrt{R_x^2 + R_y^2} \] \[ R = \sqrt{(38.81)^2 + (21.91)^2} \] \[ R = \sqrt{1506.29 + 479.36} \] \[ R = \sqrt{1985.65} \] \[ R \approx 44.56 \] #### Direction (angle \(\theta\) with respect to the positive x-axis): \[ \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) \] \[ \theta = \tan^{-1}\left(\frac{21.91}{38.81}\right) \] \[ \theta \approx 29.4^\circ \] ### ResultThe resultant vector has a magnitude of approximately \(44.56\) and makes an angle of approximately \(29.4^\circ\) with the positive x-axis.