basic physics - vector resolution


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

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If you measure the east-west leg of the triangle like it sounds like you're describing what you originally did, that would be the cosine, and the north-south leg would be the sine.  That would be from the point on the 400-radius circle straight down to the east-west axis, with the right angle in the center of the circle.  Either way would work.  You could think of the website's solution as the sine of 30 degrees; that would be the normal way to measure the angle of a triangle, but it is also the same as the supplementary angle from the east-pointing ray to the hypotenuse which is 150 degrees.

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.

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"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."
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.
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>> 150 degrees is  approximately  WNW   ==>  10 o'clock exactly

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.
for what example do  they get a different answer to you?
First, for easy to remember use the value that is given. Keep this always in your mind and you don't need anything else
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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andiejeAuthor Commented:

I 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.
Your diagram makes it easier.  As I tried to describe in my original comment, if you sook at your last diagram (solution 2), you would just measure the angle from the East axis to your hypotenuse, instead of the angle to the west access.  That would be your 150 degrees.

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.
andiejeAuthor Commented:
Superdave, as i said, it's probably a maths issue as i don't know very much maths either and I'm teaching myself that along the way. I only know how to use the cos/sine/tan ratios when the angle is inside a right angled triangle as I have done in my solutions. The 150 degree angle is not is 'inside' any triangle on my diagrams.

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)

andiejeAuthor Commented:
it's ok - i found one now
andiejeAuthor Commented:
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