Solution Q.1.

Dlat  = 23o.25S – 20o.4S= 2o.85S = 171’S

Mean Lat    = 20o.4S + 1o.425  = 21o.825S

Dlong          = 120o.5W – 118o.3W  = 2o.2W = 132’W

Dep   = d.longxcos(mean lat)  = 132 cos(21o.825)  = 122’.5E

Tan(course) =   Dep / D.Lat = 22’.5 /171’ = 0.716o

0.716-1 = 35o.6  = 36o  Therefore, course to steer = S36oW  = 216o

Dist   =   Dep /  Sin (course)  = 122.5 /  Sin(216)  = 208’.41

Therefore, distance to new position = 208.41 n.m.

Time to reach SOS position = Dist / Speed

= 208.41 ÷ 25 = 8.3 hrs. = 8 hrs. 18 mins.

ETA = 1230 + 0818  = 2048 GMT

Solution Q.2.

Dist.      = 60 n.m.

Dep.         = 60.Sin(3o)  = 30’ E.

Dlat          = 60.Cos(3o) = 51’.96 N.

New Lat   = 31o 12’.96 N.

M.lat        = 3o 46’.98 N.  = 3o.78N.

DLong.    = 30.Sec(3o.78) = 34’.92E

New Long = 15o 54’ – 34’.92 = 15o 19’.08W.

New Pos. = 31o 12’.96N.  15o 19’.08W.

Solution Q.3.

Dist.      = 2.47 hrs @ 28 knots= 69.16nm

Dep.         = 69.16.Sin(148)  = 36’.65 E.

Dlat          = 69.16.Cos(148) = 58’.65 S.

New Lat   = 49o 17’.4 S.

M.lat        = 48o 48’.08 S.

DLong.    = 36.65.Sec(48.8) = 55’.6E

New Long = 30o 23’.9E

New Pos. = 49o 17’.4 S. 30o 23’.9E

Solution Q.4

Dist.      = 24 n.m.

Dep.         = 24.Sin(75)  = 23’.18 E.

Dlat          = 24.Cos(75) = 6’.21 N.

New Lat   = 36o 29’.61 N.

M.lat        = 36o 26’.5 N.

DLong.    = 23.18.Sec(36.44) = 28’.8E

New Long = 08o 46’.6W.

New Pos.  36o 29’.61 N.  08o 46’.6W.