Find the magnitude and direction angle of the vector (-2,5). Round to tenths
square, add, sqrt .... usual distance process
and since this is in quadrant the arctan will represent in quadrant 4 so adjust by adding pi
*since its in quadrant 2 ... i said it in my head, but my fingers refused to type it :)
or just do arccos of x/r
If we're arc-ing anyway, might as well just take the arctan of y/x ...
thats what i normally do, but that only pops out q1 or q4 results
oh... right -.-
if arctan pops out say: -85, then adding 180 will put it into q2 if it pops out a radian measure, adding pi will accomplish the same thing but since we can determine r, arccos will give it without any adjustments required
You had me at "that's what I normally do, but" XD
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