So Coco,

1) Have you drawn a diagram for this problem yet? If not, you should do that first, because it will give you an idea of what to do next.

2) Now that you've drawn a diagram, you see that you have an isosceles triangle defined, with the third side length 2500 ft., and the central angle of 35" -- it it really " = seconds?? I think you meant ^{o} , degrees, right? I'll assume you meant degrees, which would be reasonable; (if you really meant seconds, then you'll solve in the same general way, the radius of the circle is going to be enormous however).

3) So you have a pair of matched right triangles then defined (each taking half the isosceles area), each with an angle 35/2 degrees opposite a side of 2500/2 ft.

4) Solve by trig for the hypotenuse length: sin (35/2)^{o }= 1250 ft./h

5) Use h and the original central angle (35^{o}) to determine the arc length s (proportion to the entire circle C=2πr, where r = h you just found, and the 35^{o} is a proportion of 360^{o} = 2π)

6) Check your answer for reasonableness: it should be slightly more than 2500 ft.

7) If that angle really was 35 seconds, the arc will be almost flat, so of length only a tiny bit more than 2500 ft.

Hope this walk-through helped you.

-- Stanton D.