First a diagram of the field and the point where the shortstop throws to first and second base.

Now add given dimensions and some calculated dimensions. I used black ink for given dimensions and blue ink for derived dimensions, but it is hard to tell the difference. Oh well.

Note that the derived dimension of F6's throw to first is 157.56 feet. This was calculated using two decimal precision. Using full precision the actual distance is 157.62969 feet (157.63').

My method:

1) First used Pythagorean to find distance from 1st to 3rd base. 90² + 90² = sqrt(16200) = 127.279 feet.
2) Next found the angle between the foul line and the 55' line: arcsin A = 20/55 = 21.3237°
3) Next found the angle between the line from 1st to 3rd base and the 55' line. 180 - 45 - 21.32 = 113.68°
4) Because an angle and two adjacent sides of the triangle formed by the line from 1st to 3rd base, the 55' line, and F6's throw are known, F6's throw can be calculated:

c = sqrt(55² + 127.28² - 2(55)(127.28)(-0.40)) = 157.56 feet  (with better decimal precision the real answer is 157.63 feet)

Now the throw to 2nd base.

I'm going to use the triangle formed by F6's throw to first, his throw to 2nd, and the 90' baseline from 1st base to 2nd base. There are two knowns, the length of the throw to 1st and the baseline. I'm going to find an angle. To do that, I need to use the triangle formed by the line from 1st to 3rd base, 3rd base to where F6 is, and F6's throw to 1st base.

sinA/a = sinC/c ===> 0.915803/157.56 = sinC/55 ===> 0.005812(55) = sinC ===> 0.319682 = sinC ===> arcsin(0.319682) = 18.64°

Subtracting 18.64 from 45 is 26.36°. Now we have 3 knowns and can solve the problem.

c² = 90² + 157.56² -  2(90)(157.56)(0.896022) ===> c = sqrt(58337.1) ===> c = 86.67

With better decimal precision  the real answer is 86.75.