Tension

this is the correct figure

and the ans is mg/2tantheta

i'm getting the ans mg/2tan(theta)

but how tanθ it should be sinθ

2Tsinθ = mg ??[7]

@philip i gave u the wrong angle sry

@prateek ur ans is correct can u post the method

@ subhash
i don't think the θ given by u earlier was wrong....

if we consider the right portion of the wire....
the vertical component of tension at the point where the string is hanged by the wire is Tsinθ and the horizontal component is Tcosθ(towards right).....
where T=tension on the wire due to wall.....

NOW,
Let, the tension at the middle point of wire will be T'....
and the direction of tension will be horizontal towards left....
since it is equilibrium....

Tsinθ=mg/2

Tcosθ=T'
Solving the two eqⁿ we can calculate T'.....

check this out.....
is this not to be pinked....
guys many of my posts are not been pinked though they r useful....