Answer:
 T₁ = 93.6 N
, T₂ = 155.6 N
, T₃ = 200 N
Explanation:
This is a balance exercise where we must apply the expressions for translational balance in the two axes
      ∑  F = 0
Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º
decompose the tension of the two upper cables
           cos 41 = T₁ₓ / T1
           sin 41 = T₁y / T1
           T₁ₓ = T₁  cos 41
           T₁y= T₁  sin 41
for cable gold
            cos 63 = T₂ / T₂
            sin 63 =  / T₂
 / T₂
We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.
Let's start by analyzing the point where the traffic light meets the vertical cable
               T₃ - W = 0
               T₃ = W
               T₃ = 200 N
now let's write the equations for the single point of the three wires
X axis
    - T₁ₓ + T₂ₓ = 0
   T₁ₓ = T₂ₓ
    T1 cos 41 = T2 cos 63
    T1 = T2 cos 63 / cos 41                (1)
y Axis 
        + T_{2y} - T3 = 0
 + T_{2y} - T3 = 0
        T₁ sin 41 + T₂ sin 63 = T₃          (2)
to solve the system we substitute equation 1 in 2
         T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W
          T₂ (cos 63 tan 41 + sin 63) = W
          T₂ = W / (cos 63 tan 41 + sin 63)
We calculate
           T₂ = 200 / (cos 63 tan 41 + sin 63)
           T₂ = 200 / 1,2856
            T₂ = 155.6 N
we substitute in 1
             T₁ = T₂ cos 63 / cos 41
              T₁ = 155.6 cos63 / cos 41
              T₁ = 93.6 N
therefore the tension in each cable is
             T₁ = 93.6 N
              T₂ = 155.6 N
              T₃ = 200 N