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An aircraft took off point A(xºN, 15ºE) at 0720h, local time. It flew due West to another point B(xºN, 75ºW) a distance of 5005 km from A.
After a stopover of 1 hour 30 minutes at point B, the aircraft took off and flew for 3 hours 40 minutes due South to a point C. The aircraft maintained an average speed of 910 km/h for the journey from A to B and also from B to C
(Take π = 22/7and the radius of the earth to be 6370 km)

  1. Calculate the:
    • position of point B. 
    • position of point C 
  2. Determine the local time at point C when the aircraft arrived.            

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  1.  
    1. θ= 75 + 15 = 90
       90   x 2 x 22 x 6,370 cos a = 5005
      360            7
      cos a= 0.500
      a= 60
      B(60 N, 75W)
    2. speed x time= (θ/360) x 2πR
      910 x 3hrs 40 mins = (θ/360)2 x 22/7 x 6370
      θ= 30
      new latitude = 60N − 30 = 30N
      C(30 N, 75 W)
  2. local time C when departing from A(0720hr)
    10   = 4 min
    900 =?
    90/1 x 4 min = 6hrs
    time in c when departing from A
    = 0720 hrs − 6 hrs
    = 0120 hrs
    arrival time = departure time + travelling time
    = 0120hrs + 33mins + 1 1/2 + 3hrs 40mins
    = 0703 hrs or 7. 03 am

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