I Parallelogram law of vector addition

[parshyaa]

• 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.

 

[drvrm]

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 !

 

[Simon Bridge]

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?

 

[Daymare]

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.

 

[parshyaa]

Ohh sorry . It was just basic congurency .

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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