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January 8th, 2015, 02:10 PM   #1
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The anticlockwise angle a vector makes with the positive x axis

I tried all possible answers and its says I'm wrong...

vector b = 6i - 8j So I drew the angle and found its magnitude:


I need to find the angle it makes with the positive x axis.
I found the angle by Tan^-1(6/-8 ) = -37 (2sf)

To find the angle it makes with the positive x axis, I thought it was 143 since id imagine it would be 180 - 37 = 143. But apparently it's not. Any help?
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January 8th, 2015, 05:18 PM   #2
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vector b = 6i - 8j is in quad IV, not quad II
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January 8th, 2015, 10:39 PM   #3
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Math Focus: सामान्य गणित
your graph should look like this.
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File Type: jpg graph.jpg (35.6 KB, 0 views)
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January 8th, 2015, 10:49 PM   #4
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This graph shows the magnitude and direction of $\displaystyle \underset{i}{\rightarrow}$ and $\displaystyle \underset{j}{\rightarrow}$
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File Type: jpg component.jpg (17.2 KB, 2 views)
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January 9th, 2015, 03:13 AM   #5
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Quote:
Originally Posted by MATHEMATICIAN View Post
This graph shows the magnitude and direction of $\displaystyle \underset{i}{\rightarrow}$ and $\displaystyle \underset{j}{\rightarrow}$
Okay that makes sense! 360-53 = 307 anticlockwise from the positive x axis, it's right, thank you!

Last edited by Tangeton; January 9th, 2015 at 03:16 AM.
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