Help Please

Discussion in 'Math' started by Mr Walker, Aug 15, 2007.

  1. Mr Walker

    Thread Starter New Member

    Aug 15, 2007
    6
    0
    http://sub.allaboutcircuits.com/images/12113.png

    Can some one help me out with this equation i need a full break down if possible. I know i have to co sine the degrees but can you please explain to me how i'm using a scientific caculator
     
  2. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    for addition,subtraction of complex nos in R<phi always transform to rectangular form
    i;e A+jB form.
    where A=Rcos(phi)
    and
    B=Rsin(phi)
    then add imaginary parts separately and real separately.
    then if u wish u can convert to polar form again by
    sqrt(A^2 +b^2)=R
    and tan(phi)=(B/A)

    in case of calculator it migh depend on ur calculator model.
    try solving in complex mode (be sure to express the nos properly)

    did that help?
     
  3. Mr Walker

    Thread Starter New Member

    Aug 15, 2007
    6
    0
    thank you very much but can you break down the equation....its complicated ut if you solve it i'm sure i can follow your steps back and figure it from there
     
  4. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
    120<0 - 120<120
    =120cos0 + j.120sin0 -(120cos120 + j.120sin120)
    =120-120*-0.5 + j(0-103.92)
    =180 - j.103.92
    180=A;
    (-)103.92=B---note there is a minus sign before 103.92 this means B is to be taken with sign if it wud have been '+' in front then B wud have been +ve
    R= sqrt(sq(180) + sq(103.92))=207.84
    phi= tan inverse of (B/A)=tan inverse of (-0.577)=-30 degrees
    thus answer = R<phi
    =207.84<-30
    anything else?
     
  5. Mr Walker

    Thread Starter New Member

    Aug 15, 2007
    6
    0
    can you explain the steps between 1 to 2 to 3....like how you got from point one to point b
    never mind i firgured it out thanks
     
  6. recca02

    Senior Member

    Apr 2, 2007
    1,211
    0
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