Answer:
33.3° south of east
Step-by-step explanation:
You want the angle a smaller tug must pull if a larger tug with 30% greater force is pulling 25° north of east and the desired resultant is due east.
The net force in the northerly direction must be zero.
The northerly component of the force from the larger tug is ...
1.3·sin(25°)
We want that to be the opposite of the northerly component from the smaller tug:
1.3·sin(25°) +1.0·sin(x°) = 0
sin(x°) = -1.3sin(25°)
x° = arcsin(-1.3·sin(25°)) ≈ -33.3°
The smaller tug must pull at an angle of 33.3° south of east.