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Applied Maths

aarond280 · 2012-10-18 18:18:37

I was wondering if Northαwest does this mean that the angle is in the 2nd quadrant and that the angle α is towards the positive y (or j) - axis. If this is correct does this mean that the if I want to find the angle towards the horizontal I could use tan(90- α)= (tan90-tanα)/(1+tan90tanα) = -tanα as tan90 is undefined. The…

18-10-2012 7:18pm

#1

aarond280

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Join Date: May 2011

Posts: 238

I was wondering if Northαwest does this mean that the angle is in the 2nd quadrant and that the angle α is towards the positive y (or j) - axis. If this is correct does this mean that the if I want to find the angle towards the horizontal I could use tan(90- α)= (tan90-tanα)/(1+tan90tanα) = -tanα as tan90 is undefined.

The reason I ask this is because its the way I seem to be able to get the answer in question 2(b) 2008 http://www.examinations.ie/archive/markingschemes/2008/LC020ALP000EV.pdf

Also in the question for Vwm=(x-a)i-3j so what I did was let -tanα = (x-a)/3=(2). So x-a=6
Then -tanb=(x)/(3+a)=3/2, 2x=9+3a,(2x-9)/3 . Subbing these values for a gives x=9. This is the answer but I want to know if this right thanks

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