#### Mr Walker

Joined Aug 15, 2007
6

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

#### recca02

Joined Apr 2, 2007
1,212
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?

#### Mr Walker

Joined Aug 15, 2007
6
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

#### recca02

Joined Apr 2, 2007
1,212
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
=207.84<-30
anything else?

#### Mr Walker

Joined Aug 15, 2007
6
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