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Question Number 162344 by stelor last updated on 28/Dec/21
Commented by mr W last updated on 29/Dec/21
twoequationsforthreeunknowns⇒nouniquesolution!
Answered by aleks041103 last updated on 29/Dec/21
cosa+cosb+cosc=0(1)sina+sinb+sinc=0(2)⇒eia+eib+eic=0⇒ei(a−c)+ei(b−c)+1=0⇒ei(a−c)=(ei(b−c))∗=ei(c−b)a−c=c−ba+b=2calso−12=cos(a−c)⇒a−c=2π3and4π3⇒a=c±2π3⇒b=c±(−2π3)
Commented by aleks041103 last updated on 29/Dec/21
thishasacutegeometricprooftoo.thecomplexvectorseia,eib,eicmustsumupto0andhavethesamelength.Thereforethetreevectorsmustbesymmetric⇒theanglebetweenanytwoofthemmustbe360°3=120°.
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