Please can you help me to resolve this vector force?

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The discussion focuses on resolving a vector force of 20N directed N 60° E into its components. The resolved parts of the force are calculated as 10N north and 17.3N east using trigonometric functions. For the direction N 30° E, the angle between the force and this direction is 30 degrees, leading to the calculation of the force component in that direction as 20 cos(30°). The participants confirm the accuracy of the calculations for the north and east components.

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slingboi
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A force P is 20N in a direction N 60(degrees) E. What is the resolved part of the force
(a) north
(b) east
(c) N 30(degrees) E

I have an idea of what to do for a and b but can you please show me your working aswell, thanks
 
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Sorry, you need to show your work first. ehild
 
north = 20Cos(60) = 10
east = 20Sin(60) = 17.3
What could (c) be, does N 30(degrees) E become the hypotenuse?
 
(a) & (b) are correct.
 
slingboi said:
north = 20Cos(60) = 10
east = 20Sin(60) = 17.3
What could (c) be, does N 30(degrees) E become the hypotenuse?

See attached figure. You need the projection of the force on the direction N 30° E. What is the angle between the force and the given direction?

ehild
 

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    resolving.JPG
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30 degrees, oh I see now so would the force N 30(degrees) E be 20cos(30)?
 
I hope so ...

ehild
 

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