Where is the ship with respect to the starting point (0,0)
I assume we do 20 knots the whole time.
East 40 - 20 sin 54 = 40 - 16.2 = 23.8 East of start
North 20 cos 54 = 11.8 North of start
tan angle East of North = 23.8/11.8
bearing angle East of North = 63.6 deg. Note that is compass bearing, not on math xy system
distance = sqrt (11.8^2 + 23.8^2) nautical mile
A ship leaves port at noon at heads due east at 20 nautical miles/hour (20 knots). At 2PM the ship changes course to N 54° W.
From the port of departure towards the ship at 3 PM, find the following:
a) the bearing to the ship (to the nearest degree)
b) the distance to the ship (round to 1 decimal place)
2 answers
I bet no math teacher wrote that question. It is worded correctly by someone who knows how to operate a sextant.
1 answer