How to Find the X Component of a Vector | Decomposing Vectors Homework

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To find the x component of a vector with a magnitude of 6 m pointing 35 degrees north of west, the cosine function can be applied using two methods. One approach involves calculating -(6 m)cos(35), resulting in -4.9 m, while the other uses (6 m)cos(145), also yielding -4.9 m. Both methods are valid, but it's suggested to use the first method and apply the negative sign directly when necessary. Understanding cardinal angles can help clarify the direction, but memorizing them isn't essential. Ultimately, consistency in approach is key for solving similar problems.
Mr Davis 97
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Homework Statement


The magnitude of the vector is 6 m, and points 35 degrees north of west. Find the x component of the vector.

Homework Equations



x = |R|cos(theta)

The Attempt at a Solution



I am confused about what degrees to input into the cosine function. I know that the vector is pointing 35 degrees north of west, which means that the x component will be negative, but there are two ways I could do it am I am not sure which one I should do. First, I could tack on a negative and calculate -(6 m)cos35 = -4.9 m. Another way is not tack on a negative and calculate (6 m)cos145 = -4.9 m. Which method is better, and which one should I use on a regular basis?
 
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Both ways are fine, as long as you don't try to memorize a rule and blindly apply it. Personally, I like the first approach, and just tack on the negative sign when I know there should be one.
 
Mr Davis 97 said:

Homework Statement


The magnitude of the vector is 6 m, and points 35 degrees north of west. Find the x component of the vector.

Homework Equations



x = |R|cos(theta)

The Attempt at a Solution



I am confused about what degrees to input into the cosine function. I know that the vector is pointing 35 degrees north of west, which means that the x component will be negative, but there are two ways I could do it am I am not sure which one I should do. First, I could tack on a negative and calculate -(6 m)cos35 = -4.9 m. Another way is not tack on a negative and calculate (6 m)cos145 = -4.9 m. Which method is better, and which one should I use on a regular basis?
Well, you could learn the angles for the cardinal points of the compass, but why do that when you can just make something up?

East = 0°
North = 90°
West = 180°
South = 270°
 
SteamKing said:
Well, you could learn the angles for the cardinal points of the compass, but why do that when you can just make something up?

East = 0°
North = 90°
West = 180°
South = 270°

Huh?
 
Mr Davis 97 said:
Huh?
You've never seen the following diagram (or a similar one)?:
ucad.gif
It should have been used when you studied trigonometry.

After all,
cos (0°) = 1.0
sin (0°) = 0.0

cos (90°) = 0.0
sin (90°) = 1.0

cos (180°) = -1.0
sin (180°) = 0.0

cos (270°) = 0.0
sin (270°)= -1.0

etc., etc.
 

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