Vector B has a magnitude of 21.50. Its direction measured counterclockwise from the x axis is 125.0°. Find its vector components. (Assume that the +x axis is to the right and the +y axis is up along the page. Also, assume
B
is in the x-y plane.)
in quadrant 2 ---> x negative, y +
reference angle = 180 -125 = 55 above -x axis
Bx = -21.50 cos 55
By = +21.50 sin 55
B = 21.5[125o].
X = 21.5*Cos125. = -12.33.
Y = 21.5*sin125 = 17.61.
To find the vector components of vector B with a magnitude of 21.50 and a direction of 125.0° counterclockwise from the x-axis, we can use trigonometric functions.
The x-component of vector B can be found using the cosine function:
x-component = magnitude * cos(angle)
x-component = 21.50 * cos(125°)
x-component ≈ -9.452
The y-component of vector B can be found using the sine function:
y-component = magnitude * sin(angle)
y-component = 21.50 * sin(125°)
y-component ≈ 17.806
Therefore, the vector components of vector B are approximately:
x-component ≈ -9.452
y-component ≈ 17.806
To find the vector components of vector B, we can use trigonometry.
Given that the magnitude of vector B is 21.50 and its direction (angle) measured counterclockwise from the x-axis is 125.0°, we can use the following formulas:
Bx = B * cos(θ)
By = B * sin(θ)
where Bx is the x-component of vector B, By is the y-component of vector B, B is the magnitude of vector B, and θ is the angle measured counterclockwise from the x-axis.
Let's substitute the given values into the formulas:
Bx = 21.50 * cos(125.0°)
By = 21.50 * sin(125.0°)
Now, we can calculate the vector components:
Bx = 21.50 * (-0.5736)
By = 21.50 * 0.8192
Bx ≈ -12.33
By ≈ 17.61
Therefore, the vector components of vector B are approximately -12.33 in the x-direction and 17.61 in the y-direction.