Answer
a) $T=229N$
b) $F_x=216N$
c) $F_y=78.5N$
Work Step by Step
(a) We know that
$T_y=\frac{mg}{2}$
We plug in the known values to obtain:
$T_y=\frac{(16Kg)(9.81m/s^2)}{2}=78.5N$
and $T_x=\frac{T_y}{tan 20^{\circ}}$
$\implies T_x=\frac{78.5N}{tan 20^{\circ}}$
$T_x=216N$
Now $T=\sqrt{T_x^2+T_y^2}$
$\implies T=\sqrt{(216N)^2+(78.5N)^2}$
$T=229N$
(b) and (c)
We know from part (a) that the horizontal component of the force is $F_x=229\times cos(20^{\circ})=216N$
and the vertical component of the force is $F_y=229\times sin(20^{\circ})=78.5 N$.