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Question

Five forces AB, AC, AD, AE and AF act at the vertex of a regular hexagon ABCDEF. Prove that the resultant is
6 AO, where O is the centre of hexagon.

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Solution



AB+AC+ AD+AE+AF

Consider ∆ADE,

AD+DE+EA=0AD+DE=AE2AO+AB=AE AD=2AOand ED ||ABDE=ABAE+AB=2AO ..... (1)

Now, consider ∆ADC

AC+CD+DA=0AC+CD=AD CD=AFAC+AF=2AO .....2

Using (1) and (2),

AB+AE+AC+AF+AD2AO+2AO+2AO=6AO

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