Answer:
The coordinates of the P' point are (-17.3,-10).
Step-by-step explanation:
We can use the components of the vector from 0 to P point. We know that the diameter is 40 meters, so the magnitude of the vector will be 20 m.
Now, the angle from the negative x-direction to the vector is 210°-180° = 30°.
The x-component of the vector will be:
[tex]V_{x}=-Vcos(30)[/tex]
[tex]V_{x}=-20cos(30)=-17.32\: m[/tex]
It is negative because the component is in the negative x-direction.
The y-component of the vector will be:
[tex]V_{y}=-Vsin(30)[/tex]
[tex]V_{y}=-20sin(30)=-10\: m[/tex]
It is negative too because the component is in the negative y-direction.
Therefore, the coordinates of the P' point are P'(-17.3,-10)
I hope it helps you!
