Answer
$\frac{T_{B}}{T_{A}}$ =$ cos^{2}$$\beta$
Work Step by Step
Before the horizontal string is cut, the ball is in equilibrium, and the vertical component of the tension force must balance the weight,
so $T_{A}$cos$\beta$ = W
$T_{A}$=$\frac{W}{cos\beta}$
At point B, the ball is not in
equilibrium; its speed is instantaneously 0, so there is no radial acceleration, and the tension force must balance the radial component of the weight so,
$T_{B}$=$Wcos\beta$
Hence, $\frac{T_{B}}{T_{A}}$ =$ cos^{2}$$\beta$
