Consider the forces at A.
T_{1h} = T_{1} Sin 35°
T_{1v} = T_{1} Cos
35° 
Recall Sin θ = ^{opposite}/_{hypotenuse}
and Cos θ = ^{adjacent}/_{hypotenuse}
F can be resolved into F Sin θ and F Cos θ 
(b)
T_{3h} + [(T_{2R})] = 0
T_{3h} = T_{2R}
The tension on the horizontal string T_{2R} = T_{2L}
T_{3h} = T_{2L}
T_{3h} = 28N
T_{3} Sin θ = 28N
T_{3} = 28N / Sin θ
From Eq2
50N / Cos θ = 28N / Sin θ
Sin θ / Cos θ = 28N / 50N
Tan θ = 28N / 50N
θ =Tan^{1} (28N / 50N)
θ = 29°
T_{3} = 28N / Sin θ
T_{3} = 28N / Sin θ
T_{3} = 57N 
Alternate Solution for T_{3}.
using Pythagoras
T_{3}^{2} = T_{3v}^{2} + T_{3h}^{2}
T_{3}^{2} = 50^{2} _{} + 28^{2}
T_{3} = 57N
Recall tan θ = ^{Sin θ}/_{Cos θ}
