Answer
$ tan^{-1}\frac{ILB}{mg}$
Work Step by Step
We know that
$\Sigma F_y=0$
$\implies Tcos\theta-mg=0$
$\implies T=\frac{mg}{cos\theta}$....eq(1)
Similarly $\Sigma F_x=0$
$\implies Tsin\theta-ILB=0$
$\implies T=\frac{ILB}{sin\theta}$...eq(2)
comparing eq(1) and eq(2), we obtain:
$\frac{mg}{cos\theta}=\frac{ILB}{sin\theta}$
$\implies tan\theta=\frac{ILB}{mg}$
$\implies \theta=tan^{-1}\frac{ILB}{mg}$
