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hi,
at point B,
T2*cosβ = mg .....(in vertical dirxn).............[1]
T2*sinβ = mg........(in horizontal dirxn)........[2]
dividing these 2 eqns, we get
tanβ = 1 ,.i.e β = ∏/4......
now at pont A,
T1*cosα = T2*cosβ + mg........(in vertical dirxn).................[3]
=> T1*cosα = 2T2*cosβ........(using [1])
in horizontal dirxn
T1*sinα = T2*sinβ ........................[4]
dividing 4 by 3 ,we get
tanα = 0.5*tanβ
=> 2tanα = tanβ
so correct option is (a)........
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