Asked by Anonymous
A ship sails from a port,P, on a course of 060 degrees for 15km to a buoy ,Q. It then changes course in a direction 120 degrees from Q for 25 km to a rig R.
Answers
Answered by
Henry
D = 15km @ 60deg + 25km @ 120deg.
X = hor. = 15cos60 + 25cos120 = -5km.
Y = ver. = 15sin60 + 25in120 = 34.64km.
tanA = Y/X = 34.64 / -5 = -6.9282,
A = -81.8 Deg., CW.
A = -81.8 + 180 = 98.2 Deg., CCW.
D = -5 / cos98.2 = 35km @ 98.2 Deg.,CCW
X = hor. = 15cos60 + 25cos120 = -5km.
Y = ver. = 15sin60 + 25in120 = 34.64km.
tanA = Y/X = 34.64 / -5 = -6.9282,
A = -81.8 Deg., CW.
A = -81.8 + 180 = 98.2 Deg., CCW.
D = -5 / cos98.2 = 35km @ 98.2 Deg.,CCW
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