This is a force problem.  So let's go:
Vertical:
We have the weight of the light (30 kg)(9.8 N/kg) = 294 N down (-), and the vertical components of the tensions in the cords.  I will call the leftmost cord T_L and the rightmost T_R. The vertical component of T_L is T_Lsin(53^o), and of T_R, T_Rsin(37^o) and these are both upward (+) so our equilibrium looks like:
 T_Lsin(53^o) + T_Rsin(37^o) - 294 N= 0
Which is not solvable, as it has two unknowns.  So let's set up Horizontal.
In the x direction, we have the horizontal components of the two cords acting in the opposite directions.  The horizontal component of T_L, T_Lcos(53^o), acts to the left (-), and the horizontal component of T_R, T_Rcos(37^o) acts to the right (+) so we have
T_Rcos(37^o) - T_Lcos(53^o) = 0

Which means that 
T_R = T_Lcos(53^o)/cos(37^o)

Popping this into our first equation, we get:
T_Lsin(53^o) + T_Rsin(37^o) - 294 N= 0
T_Lsin(53^o) + {T_Lcos(53^o)/cos(37^o)}sin(37^o) - 294 = 0
So
T_Lsin(53^o) + T_Lcos(53^o)tan(37^o) = 294 N
T_L(sin(53^o) + cos(53^o)tan(37^o)) = 294 N
T_L = (294 N)/(sin(53^o) + cos(53^o)tan(37^o)) = 234.79884 N = 230 N (the book says 240??)
And since
T_R = T_Lcos(53^o)/cos(37^o)
T_R = (234.79884 N)cos(53^o)/cos(37^o) = 176.9336 N = 180 N