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I have to find to component of vector DE having magnitude 1 m .now the vector is in 4th quad making angle 90 degree with postive x axis The solution that my teacher showed is ax=1cos(270) .and ay = 1sin(270) ... Now my confusion is that if it makes 90 degree with x axis ....why not ax=1cos90 and ay=1sin90 refrence shows ax=0 and ay=-1

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closed as off-topic by innisfree, Kyle Kanos, Kyle Oman, rob, John Rennie Jun 16 '15 at 5:38

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$\cos 90^o = \cos 270^o =0$. – gonenc Jun 15 '15 at 18:23
But sin90 is not equal to sin270 – saiman qureshi Jun 15 '15 at 18:25
Note that the angles are measured counter-clockwise with respect to the positive $x$ axis. – gonenc Jun 15 '15 at 18:26
Draw a picture - think why the angle might in fact be $-90^\circ$ or equivalently $270^\circ$. Something to do with those quadrants – innisfree Jun 15 '15 at 18:26
@innisfree: 4. quadrant $90^o$ means a unit vector pointing downwards in the $-y$ direction :) – gonenc Jun 15 '15 at 18:28
up vote 0 down vote accepted

Draw a picture...

enter image description here

From here the answer should be obvious. Don't focus just on the math - look at what is actually going on.

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Thanx ,but i still dont understand how it works ,say we have to resolve a vector with magnitude 1 which lies in fourth quad (3,-4)...which is parallal to y axis ....what will be its angle with x axis – saiman qureshi Jun 15 '15 at 20:08
By convention we measure angles as I drew in my diagram: start at X axis and work counter clockwise. That makes the angle of the vector I drew 270°, and you get (0,-1) in vector notation $(\cos 270°, \sin 270°$) . The result is the same if you use an angle of -90° – Floris Jun 15 '15 at 20:11
Oh thanxs a lot ...i got you ..thanx a ton – saiman qureshi Jun 15 '15 at 20:17

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