Solving Homework Problems: Understanding Ax and Ay Equations | Get Help Now

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The discussion focuses on solving homework problems related to the Ax and Ay equations, specifically using trigonometric functions to find components of a vector B. The equations Ax = A cos(θ) and Ay = A sin(θ) are applied to determine the angles between vector B and the coordinate axes. The magnitude of vector B is calculated as √(6² + 4² + 3²) = √61, leading to angles of approximately 39.81 degrees for the x-axis, 59.19 degrees for the y-axis, and 67.41 degrees for the z-axis. There is confusion regarding the conversion between radians and degrees, which is clarified during the discussion. The thread emphasizes the importance of correctly applying trigonometric principles to resolve vector components.
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1. Homework Statement
http://gyazo.com/f4c02f2a57b24a39dfee575837e0a807.png


2. Homework Equations
Ax = A cos (?)
Ay = A sin(?)


3. The Attempt at a Solution
I tried putting the magnitude of x cos (?) but honestly have no idea what to do. Thanks
 
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Angle between B and x-axis: arccos(6/sqrt(61))
Angle between B and y-axis: arccos(4/sqrt(61))
Angle between B and z-axis: arccos(3/sqrt(61))
 
Where did you get those numberts?
 
jaytm2291 said:
Where did you get those numberts?

B = 6i + 4j + 3z

Bx = B * cos(angle between x-axis and vector B) ->>cos(angle) = Bx / B ->> angle = arccos(Bx/B)

By = B * cos(angle between y-axis and vector B) ->>cos(angle) = By / B ->> angle = arccos(By/B)

Bz = B * cos(angle between z-axis and vector B) ->>cos(angle) = Bz / B ->> angle = arccos(Bz/B)

B = sqrt(6^2 + 4^2 + 3^2) = sqrt(61)

angle between x-axis and B = arccos(6/sqrt(61))
angle between y-axis and B = arccos(4/sqrt(61))
angle between z-axis and B = arccos(3/sqrt(61))
 
This is wrong. ?
 
jaytm2291 said:
This is wrong. ?

You didnt convert radians to degrees did you, no you didn't

You done didn't, did ya..

x-axis and B = 39.8055 degrees
y-axis and B = 59.1930 degrees
z-axis and B = 67.4114 degrees
 
=) Thank you.
 
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