Calculating the x-component of a vector with magnitude and direction

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SUMMARY

The discussion focuses on calculating the x-component of vector A, which has a magnitude of 35.1 units and is directed at an angle of 301° counterclockwise from the positive x-axis. The x-component can be derived using trigonometric functions, specifically the cosine function, as follows: vx = magnitude * cos(angle). Thus, vx = 35.1 * cos(301°). This straightforward application of trigonometry confirms the simplicity of resolving vector components using known angles and magnitudes.

PREREQUISITES
  • Understanding of basic trigonometry, specifically sine and cosine functions
  • Familiarity with vector components and their representation
  • Knowledge of degrees and radians in angle measurement
  • Ability to perform calculations involving square roots and trigonometric functions
NEXT STEPS
  • Learn how to calculate the y-component of a vector using sine functions
  • Explore vector addition and subtraction in two dimensions
  • Study the application of the Pythagorean theorem in vector analysis
  • Investigate the conversion between degrees and radians for angle measurements
USEFUL FOR

Students in physics or mathematics, engineers, and anyone involved in vector analysis or trigonometry applications.

ymehuuh
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The magnitude of vector A is 35.1 units and points in the direction 301° counterclockwise from the positive x-axis. Calculate the x-component of this vector.

So...35.1=sqrt(vx^2 + vy^2)

and 301°= vy/vx

How do I do this?
 
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You just need trig. It's a right angled triangle. You know an angle and the hypotenuse, so what can you do to find the x component?
 
Wow. It was that easy. Thanks.
 

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