Assume the arrow is notched at the center of the bowstring. Thatis where the forces are in equilibrium and all forces add up to zero.
The tensions in each half of the bow string each have a horizontal and a vertical component.
Since the arrow is in the center and both haves are mirror images of each other, the tensions in each half are equal.
Since the point of equilibrium is in the center it is also true that both horizontal components are equal, and since there are no other horizontal forces acting on the arrow and it is not accelerating they must add to zero.
∑ F hor = 0
All the forces in the vertical at that point must also add to zero since there is no acceleration in that direction either. I just assumed the arrow was still drawn and not loosed.
There are 3 forces there. The two equal vertical components of the 2 tensions (1 for each half of the bowstring) and the downward 90 N pull the shooter makes.
The 2 sides of the bowstring make a right triangle with the arrow and the hypotenuse is the tension in the string and is T sin(35). The vertical forces add to zero so you have:
∑ F vert = 2•T•sin(35) - 90 N = 0
solving for T you get:
2•sin(35) = 90 N /T ==>> T = 90 N / (2 sin(35) = 45 N / sin(35)
Just don't forget; what goes up straight, comes back down straight ... on your head;)