Parallelogram law of vector addition

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parshyaa
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  • 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( what's 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.
 
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parshyaa said:
Therefore acoording to the congruency of triangles angle alpha = (theta)/2. But this is not right( what's wrong )

you have found something wrong...
so why not draw a good diagram say P=Q and angle theta=60 degree and check !
 
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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?
 
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.
 
Ohh sorry . It was just basic congurency .
Daymare said:
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.
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