Need help with physics

A 3.45 kg object is attached to a vertical rod by two strings as in the figure below. The object rotates in a horizontal circle at constant speed 7.25 m/s.

(a) Find the tension in the upper string (in N).

(b) Find the tension in the lower string (in N).

Diagram: http://www.webassign.net/pse/p6-11.gif

based on drawing radius = 1.323m

let tension in the upper string be T and tension in the lower string be F

centripetal force = mv^2/r
= 3.45 * 7.25^2 / 1.323
= 137.07N
resolving force vertically,
T sin A = F sin A + mg
where A is the angle between string and horizontal circle sin A = 1.5 / 2 = 0.0.75
0.75 T = 0.75 F + 3.45 * 9.81
T = F + 45.13 ----(1)

resolving force horizontally,
T cos A + F cos A = 137.07
where A is the angle between string and horizontal circle cos A = 1.323 / 2 = 0.6615
0.6615 T + 0.6615 F = 137.07
T + F = 207.21 ----(2)

sub (1) into (2)
F + 45.13 + F = 207.21
F = 81.04 N ---answer
from (1)
T = 126.17 N ----answer