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I Parallelogram law of vector addition

  1. Jul 20, 2016 #1
    download4.jpg
    • Acording to this diagram vector P = vector BC and vector Q = vector OB(their magnitudes are also respectively equal.)
    • Therefore acoording to the congruency of triangles angle alpha = (theta)/2. But this is not right( whats wrong )
    • {Tan(alpha) = Qsin(theta)/(P + Qcos(thets)) ]
    • Why resultant vector of P vector and Q vector touches C to form OC vectors.
    • Does the length of vector represents its magnitude.
     
  2. jcsd
  3. Jul 20, 2016 #2
    you have found something wrong....
    so why not draw a good diagram say P=Q and angle theta=60 degree and check !
     
  4. Jul 20, 2016 #3

    Simon Bridge

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    You are observing that OAC and OBC are similar (congruent) triangles.
    If |OA|=|OB| (ie P and Q have the same magnitudes) then you have similar isosceles triangles.
    What is the relationship between the angles then?
     
  5. Jul 20, 2016 #4
    You seem to be confused into thinking that angles AOC and angle BOC are equal because the two triangles are congruent.But its actually the angles ACO and BOC;AOC and BCO that are equal.
     
  6. Jul 20, 2016 #5
    Ohh sorry . It was just basic congurency .
    ,
     
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