Answer:
The tensions in 
 is approximately 4,934.2 lb and the tension in 
 is approximately  6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC = 
The tension in rope segment BD = 
The tension in rope segment AB = 
 = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
 × cos(26.0°) + 
 × cos(21.0°) = 
 = 1200 lb
 × cos(26.0°) + 
 × cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
 × sin(26.0°) + 
 × sin(21.0°) = 0...........................(2)
Which gives;
 × sin(26.0°) = - 
 × sin(21.0°)
 = - 
 × sin(21.0°)/(sin(26.0°))  ≈ - 
 × 0.8175
Substituting the value of, 
, in equation (1), gives;
- 
 × 0.8175 × cos(26.0°) + 
 × cos(21.0°) = 1200 lb
- 
 × 0.7348  + 
 ×0.9336 = 1200 lb
 
 ×0.1988 = 1200 lb
 
 ≈ 1200 lb/0.1988 = 6,035.6938 lb
 
 ≈ 6,035.6938 lb
 ≈ - 
 × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
 ≈ -4934.1733 lb
From which we have;
The tensions in 
 ≈ -4934.2 lb and  
 ≈ 6,035.7 lb.