At what angle two forces A+B and A-B act, so that their resultant is squqre root of 3A2 + B2.
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the angle on the one side :)
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Answer:
root 3a^2 + b^2 = root (a+b)^2 + (a-b)^2 + 2 (a+b) (a-b) cos teta
3a^2 + b^2 = (2a^2 + b^2) + 2(a^2 - b^2 ) cos teta
a^2 - b^2 = 2 (a^2 - b^2) cos teta
cos teta = 1/2
teta= 60
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