Trig problem, can't figure out left and right.
#1

Ok so if I use this code:

pawn Код:
x += ( distance * floatsin( -angle, degrees ) );
y += ( distance * floatcos( -angle, degrees ) );
It will give me the point infront or behind the x/y based on the distance variable (pos/neg value passed).

How do I do left or right though ?
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