output_description: Print the answer without units. Your output will be accepted when its absolute
  or relative error from the correct value is at most 10^{-9}.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      A B H M
attempt: """
  ・なんとか自力AC。大変だった。これ本当に300点か。。
  ・三角形、角度、三角関数、余弦定理
  ・2辺の長さが分かっていてもう1辺を知りたいので、角度が分かればいいなーって気持ちになる。
  ・時刻が与えられるが、短針の微妙な位置とか分からんだろ、ってなる。
  ・ここで先にD解いたり、Eしばらく考えたりして、E厳し目っぽかったので戻ってきた。
  ・もう1回、300点でそんな高度な要求がある訳ないという前提の元、考え直してみる。
  ・ここでやっと、1分毎に動く量を出せば経過時間から割り出せると分かる。
  ・ちょっとゴニョゴニョして無事角度が出て、さて、2辺と角度が揃ったらどうするんだっけな、とググる。
  ・余弦定理の公式が出てきて、あぁこんなんあったな、と当てはめる。無事AC。
  ・こういうのを300点っていう低難度で出せちゃうのは、やっぱり競プロは数学ゲーなんだな、と改めて思わされる所。
  """
  
  import sys
  from math import sin, cos, radians, hypot
  
  
  def input():
      return sys.stdin.readline().strip()
  
  
  def list2d(a, b, c):
      return [[c] * b for i in range(a)]
  
  
  def list3d(a, b, c, d):
      return [[[d] * c for j in range(b)] for i in range(a)]
  
  
  def list4d(a, b, c, d, e):
      return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
  
  
  def ceil(x, y=1):
      return int(-(-x // y))
  
  
  def INT():
      return int(input())
  
  
  def MAP():
      return map(int, input().split())
  
  
  def LIST(N=None):
      return list(MAP()) if N is None else [INT() for i in range(N)]
  
  
  def Yes():
      print("Yes")
  
  
  def No():
      print("No")
  
  
  def YES():
      print("YES")
  
  
  def NO():
      print("NO")
  
  
  sys.setrecursionlimit(10**9)
  INF = 10**18
  MOD = 10**9 + 7
  
  a, b, h, m = MAP()
  
  hm = h * 60 + m
  
  heach = hm / 720
  meach = m / 60
  hdig = heach * 360
  mdig = meach * 360
  
  hx = a * cos(radians(hdig))
  hy = a * sin(radians(hdig))
  mx = b * cos(radians(mdig))
  my = b * sin(radians(mdig))
  
  ans = hypot(hx - mx, hy - my)
  print(ans)
  
problem_description: Statement
  
  Consider an analog clock whose hour and minute hands are A and B centimeters
  long, respectively.
  
  An endpoint of the hour hand and an endpoint of the minute hand are fixed at
  the same point, around which each hand rotates clockwise at constant angular
  velocity. It takes the hour and minute hands 12 hours and 1 hour to make one
  full rotation, respectively.
  
  At 0 o'clock, the two hands overlap each other. H hours and M minutes later,
  what is the distance in centimeters between the unfixed endpoints of the
  hands?
samples: [{"input": "3 4 9 0", "output": "5.00000000000000000000\n    \n\nThe two hands will be in the positions shown in the figure below, so the\nanswer is 5 centimeters.\n\n![The clock at <var>9</var>\no'clock](https://img.atcoder.jp/ghi/when_a_nameless_star_falls_into_the_sky.png)\n\n* * *"}, {"input": "3 4 10 40", "output": "4.56425719433005567605\n    \n\nThe two hands will be in the positions shown in the figure below. Note that\neach hand always rotates at constant angular velocity.\n\n![The clock at\n<var>10:40</var>](https://img.atcoder.jp/ghi/when_flower_petals_flutter.png)"}]