Respuesta :
Answer:
a) [tex] \theta = tan^{-1} (\frac{11}{11})= tan^{-1} (1)=45[/tex] degrees
b) [tex] \theta = tan^{-1} (\frac{11}{19})= tan^{-1} (0.579)=30.07[/tex] degrees
c) [tex] \theta = tan^{-1} (\frac{19}{11})= 59.93[/tex] degrees
Explanation:
If we have a vector [tex] A= (A_x ,A_y)[/tex] in a two dimensional space. The angle respect the x axis can be founded from the following expression:
[tex]tan (\theta)= \frac{A_y}{A_x}[/tex]
And then the angle is given by:
[tex] \theta = tan^{-1} (\frac{A_y}{A_x})[/tex]
Part a
Ax = 11 m and Ay = 11 m
For this case the angle would be:
[tex] \theta = tan^{-1} (\frac{11}{11})= tan^{-1} (1)=45[/tex] degrees
Part b
Ax = 19 m and Ay = 11 m
For this case the angle would be:
[tex] \theta = tan^{-1} (\frac{11}{19})= tan^{-1} (0.579)=30.07[/tex] degrees
Part c
Ax = 11 m and Ay = 19 m
For this case the angle would be:
[tex] \theta = tan^{-1} (\frac{19}{11})= 59.93[/tex] degrees
Answer:
(a) theta = arctan (Ay/Ax) = arc tan (11/11) = 45°
(b) theta = arctan (Ay/Ax) = arc tan (11/19) = 30.07°
(c) theta = arctan (Ay/Ax) = arc tan (19/11) = 59.93°
To find the angle a vector makes with the positive axis, one needs to know the components of the given vector. Then the arctan of the ratio of the component of the vector along the y-axis and the component along the x-axis is computed. This gives the angle the vector makes with the positive x-axis.
Explanation:
This angle is always measured from the positive x-axis in a counter clockwise direction towards the positive y-axis. The values of theta range from 0 - 360°.
The component vectors Ay and Ax are simply representing of the vector A along the respective axis. In terms of force, they represent the effect that force will have along the x- and y-axis respectfully. Mathematically they can be calculated using the formulas
Ax = Acos(theta)
Ay = Asin(theta)
Where theta is measured from the positive X-axis counter clock wisely
A = magnitude of the vector A and mathematically
A = squareroot(Ay² + Ax²)
