Answer
$\theta=56.0^o$
Work Step by Step
$\sum F_Y=F_{T}\cos(\theta)-mg=0$
$F_T=\frac{mg}{\cos(\theta)}$
$F_{T}\sin(\theta)=\frac{mv^2}{r}$
$F_{T}=\frac{mv^2}{r\sin(\theta)}$
$\frac{mg}{\cos(\theta)}=\frac{mv^2}{r\sin(\theta)}$
$\frac{g}{\cos(\theta)}=\frac{v^2}{r\sin(\theta)}$
$\theta=\tan^{-1}(\frac{v^2}{rg})$
$r=l\sin(\theta)$
$v=\frac{2\pi r}{T}=\frac{2\pi l\sin(\theta)}{T}$, where T is the period
$\frac{g}{\cos(\theta)}=\frac{\Big(\frac{2\pi l\sin(\theta)}{T}\Big)^2}{l\sin(\theta)\sin(\theta)}$
$\frac{g}{\cos(\theta)}=\frac{4\pi^2l}{T^2}$
$\theta=\arccos\Big(\frac{T^2g}{4\pi^2l}\Big)=\arccos\Big(\frac{(0.75s)^2\big(9.8\frac{m}{s^2}\big)}{4\pi^2(0.25m)}\Big)=56.0^o$
