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Hi

Take this question: A velocity vector is described as 400 km/hr in a direction of 150 degrees (i.e., 30 degrees north of west). Separate this into its north and west vectors.

I know this is hard without a diagram, but to do this I made a right angled triangle from the vector with 400 as the hypotenuse and the right angle 'at the top right' between north and west. I took the angle between the north vector and the hypotenuse as 60 (150-90) and did:

cos(60) = adj (VNorth)/hyp =

cos(60) = VNorth/ 400 = 200km/hr

The solution on the website was Vnorth = sine(150) * 400. There is no diagram. I don't understand how you could make a right angled triangle with the north vector the opposite side to a 150 degree angle.

In this case you get the same answer but in other examples on the same site they get a different answer to me using their approach. The quesiton is actually q1 on this page about the plane

http://www.physicsclassroom.com/Class/vectors/U3l1g.cfm

thankls

Take this question: A velocity vector is described as 400 km/hr in a direction of 150 degrees (i.e., 30 degrees north of west). Separate this into its north and west vectors.

I know this is hard without a diagram, but to do this I made a right angled triangle from the vector with 400 as the hypotenuse and the right angle 'at the top right' between north and west. I took the angle between the north vector and the hypotenuse as 60 (150-90) and did:

cos(60) = adj (VNorth)/hyp =

cos(60) = VNorth/ 400 = 200km/hr

The solution on the website was Vnorth = sine(150) * 400. There is no diagram. I don't understand how you could make a right angled triangle with the north vector the opposite side to a 150 degree angle.

In this case you get the same answer but in other examples on the same site they get a different answer to me using their approach. The quesiton is actually q1 on this page about the plane

http://www.physicsclassroom.com/Class/vectors/U3l1g.cfm

thankls

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```
0 degrees is East ==> 3 o'clock
90 degrees is North ==> 12 o'clock
150 degrees is NNE ==> 10 o'clock
I would say the North vector is
sin(150)*400 km/hr = (1/2)*400 km/hr = 200 km/hr
The vector can be drawn from the horizontal center line of clock
to the tip of hour hand at 10 o'clock. Then scale the magnitude.
A vector is a magnitude and a direction. Once you have calculated it,
you can move it back to the origin.
```

You are making this too hard. (perhaps because you are trying to get method with out understanding what is going on).

You just want the x and the y components of the vector.

more later

Let's say you have a N vector and W vector both drawn from the origin.

There are two ways to add them graphically:

Move the tail of the W vector to the head of the N vector.

OR Move the tail of the N vector to the head of the W vector.

The resultant is the same in both cases.

You need to be comfortable moving the vectors around.

And you really need to make drawing to solve these sort of problems.

V.North = Vsin(a)

V.East = Vcos(a)

So here in your case V.North = 400sin(150)

V.East = Vcos(150) . But you need V.West and since V.West = - V.North so V.West = -400cos(150)

If you need V.South for ex, note that V.South= -V.North so it's -400sin(150) . Forget your 60 or it will be easy to make mistakes.

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Start your 7-day free trialI believe i am comfortable in moving the vector around. I solved the problem in 2 separate ways as shown in the attached image. The resultant was indeed the same in both cases.

I imagine the problem is with my trigonometry because my 2 solutions only involve acute angles. I don't know how you can use those basic cos/sine/tan ratios other than in right angled triangles. The solution given uses them with obtuse angles.

I have never seen the formula given by arabia in the last comment.

img008.jpg

Arabia's formulas are basically the definitions of the sine and cosine. Usually you see them defined on a circle of radius 1 with 0 degrees being to the east (that's why I said you measure the 150 degrees from East). Except they usually use radians instead of degrees in the definitions of the trigonometric functions.

I'm aware of the unit circle and that 0 degrees is to the east but I don't seem to be making the connection about how I could apply that here. I could just simply use the formulae given by arabia, but i do prefer to understand where they came from.

Could you perhaps recommend a webpage explaining the maths behind why

V.North = Vsin(a)

V.East = Vcos(a)

Thanks

Math / Science

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