Question

The direction Of three forces 1N,2N,3N are acting at a point are parallel to the sides of an equilateral Triangle taken in order. The magnitude of their resultant??

Answer 1

the three forces will be along three directions separated by 120 deg. from each other drawn at a point P.  Let them be  PA of 1 N.  PB = 2 N  and then  PC = 3 N..

Let us draw PA  horizontally along  positive x axis.
 PB is at 30 deg. anticlockwise from + y axis.
PC will be at   30 deg. clockwise from  - y axis...
   
   Resolving components along x axis and adding:
        = 1 N +  2 N * Cos 120 + 3 N * Cos 240
        =  1 - 2 * 1/2 - 3 /2  =  - 3/2 Newtons
  
   Resolving components along y axis and adding
      = 0 N + 2 N * √3 /2  -  3 N * √3/2   =  - √3/2  Newtons
  

  So resultant is    R =  -1.5 i  - √3 /2  j  where   i and j  are unit vectors.
   magnitude =     √[ 1.5² + 3/4 ]   = √3  Newtons.

======
The calculations are much simpler,  if you take  2 Newtons force/vector along x axis and the other two at 120 deg. on either side to that....  

Answer 2

Answer:

three directions separated by 120 deg. from each other drawn at a point P.  Let them be  PA of 1 N.  PB = 2 N  and then  PC = 3 N..

Let us draw PA  horizontally along  positive x axis.

 PB is at 30 deg. anticlockwise from + y axis.

PC will be at   30 deg. clockwise from  - y axis...

   

   Resolving components along x axis and adding:

        = 1 N +  2 N * Cos 120 + 3 N * Cos 240

        =  1 - 2 * 1/2 - 3 /2  =  - 3/2 Newtons

  

   Resolving components along y axis and adding

      = 0 N + 2 N * √3 /2  -  3 N * √3/2   =  - √3/2  Newtons

  

  So resultant is    R =  -1.5 i  - √3 /2  j  where   i and j  are unit vectors.

   magnitude =     √[ 1.5² + 3/4 ]   = √3  Newtons.

======

The calculations are much simpler,  if you take  2 Newtons force/vector along x axis and the other two at 120 deg. on either side to that....