(a) 52.475nm many nautical miles south and 29.09nm many nautical miles west will the ship have traveled by 6:00 p.m.
(b) at 6:00pm the ship changes course to due west when 36.68° the ship's bearing and distance from port at 7:00 p.m.
Given that,
A ship sets out from port at noon with a bearing of s 23° west and a speed of 10 knots.
We have to find
(a) By the time it gets dark at six o'clock, how far south and west will the ship have sailed? The ship alters its course to head due west at 6:00 p.m. miles south miles west
(b). at 7:00 p.m., determine the ship's bearing and distance from the port. nautical miles in s.
We know that,
(a) For 6:00 p.m
Sin(θ)=opposite/ hypothesis
θ=29
Opposite=west
Hypothesis =60
Sin(29°)=w/60
w=29.09nm
West= 29.09nm
Cos(θ)=adjacent/ hypothesis
θ=29
Adjacent= south
Hypothesis =60
Cos(29°)= south/60
south=52.475nm
Therefore, 52.475nm many nautical miles south and 29.09nm many nautical miles west will the ship have traveled by 6:00 p.m.
(b) tan(θ)=opposite/adjacent
Opposite=29.09nm
Adjacent= 52.475nm
Tan(θ)=29.09nm/52.475nm
θ=36.68°
Therefore, at 6:00pm the ship changes course to due west when 36.68° the ship's bearing and distance from port at 7:00 p.m.
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