Let x be the first slaned distance (hypotenuse) travelled in 15 degree triangle. Let y be horizontal distance

x = 69.546 ft

y = 67.177 ft

Let m be the second slaned distance (hypotenuse) travelled in 21 degree triangle. Let n be horizontal distance of triangle.

m = 50.228 ft

n = 46.892 ft

The middle horizontal distance travelled = 200 - 67.177 - 46.892 = 85.931 ft

Total distance travelled on bridge = 69.546 + 85.931 + 50.228 = 205.705 ft