A mass m = 9.300 kg is suspended from a string of length L = 1.510 m. It revolves in a horizontal circle (see

A mass m = 9.300 kg is suspended from a string of length L = 1.510 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 2.660 m/s. What is the angle theta between the string and the vertical (in degrees)?

figure http://gryphon.phy.umassd.edu/msuphysicslib/Graphics/Gtype09/prob33_flymass.gif

centripetal force
= mv^2/ r
= 9.3 * 2.66^2 / 1.51 sin A
= T sin A
where T = tension in string; A = angle theta between the string and the vertical (in degrees)
9.3 * 2.66^2 / 1.51 sin A = T sin A
43.578 = T sin ^2 A ---(1)

resolving vertically,
T cos A = mg
T = 9.3 * 9.81 / cos A
sub into (1),
43.578 = [9.3 * 9.81 / cos A] * sin ^2 A
0.4777 cos A = sin^2 A
0.4777 cos A = 1 - cos ^2 A
cos^2 A + 0.4777 cos A -1 =0
cos A = { -0.4777 + sq root [ 4.2282] } /2 = 0.7893
A = 37.88 deg
answer
cos A1 = { -0.4777 - sq root [ 4.2282] } /2 = -2.5339 not valid