How do they derive at the 202.6 degrees on the vector table below (first column, second cell)? This is for F₂₁ (3.83, 202.6 degrees). How do they get 202.6 degrees from 3.83?

Example 16.3 Suppose three point charges are arranged as shown in the figure. A charge q1 = + 1.2 \"C is located at the origin of an (x, y) coordinate system; a second charge q2 = 0.60 [AC is located at (1.20 m, 0.50 m) and the third charge q3 = + 0.20 uC is located at (1.20 m, 0). What is the force on q2 due to the other two charges? .\\~ 0.5 m \fVectors (mN, () X - components Y - components F21 (3.83, 202.6.) 3.83 cos 202.6 = -3.53 3.83 sin 202.6 = -1.47 F23 (4.32, -90' ) 4.32 cos (-90 ) = 0 4.32 sin (-90 ) = -4.32 F2 -3.53 -5.79 F, = V3.53 2 + 5.792 = 6.78 mN y tan O = Ry / RX tan O = -5.79/-3.53 = 1.64 92 0 = tan-1 (1.64) = 58.6 F 2 ( = 90 - 58.60 = 31 a = inv tan (1.2/0.5 ) 91 + X + 93 (The correct angle is 270 - 31 = 239) 38