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Can Someone explain how vector a with magnitude 8 and angle 132 and vector b with magnitude 5 and angle 260 give a result of 6.30452, 170.679

Can Someone explain how vector a with magnitude 8 and angle 132 and vector b with-example-1

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Explanation:

Vector A:

Magnitude: 8

Angle: 132 degrees

To find the x and y components of vector A:

x-component (A_x) = 8 * cos(132°)

y-component (A_y) = 8 * sin(132°)

Vector B:

Magnitude: 5

Angle: 260 degrees

To find the x and y components of vector B:

x-component (B_x) = 5 * cos(260°)

y-component (B_y) = 5 * sin(260°)

Now, calculate the x and y components:

For vector A:

A_x = 8 * cos(132°) ≈ -3.464

A_y = 8 * sin(132°) ≈ 6.928

For vector B:

B_x = 5 * cos(260°) ≈ -4.082

B_y = 5 * sin(260°) ≈ -2.588

Next, add the x and y components of both vectors together to get the resultant vector:

Resultant x-component (R_x) = A_x + B_x ≈ -3.464 - 4.082 ≈ -7.546

Resultant y-component (R_y) = A_y + B_y ≈ 6.928 - 2.588 ≈ 4.34

Now, you have the x and y components of the resultant vector. To find its magnitude (R) and angle (θ), you can use the Pythagorean theorem and trigonometric functions:

R = √(R_x^2 + R_y^2) ≈ √((-7.546)^2 + (4.34)^2) ≈ √(56.952 + 18.8356) ≈ √75.7876 ≈ 8.704

θ = arctan(R_y / R_x) ≈ arctan(4.34 / -7.546) ≈ arctan(-0.575) ≈ -29.76 degrees (approximately)

So, the resultant vector has a magnitude of approximately 8.704 and an angle of approximately -29.76 degrees. Note that the negative angle means it's measured clockwise from the positive x-axis.

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