Asked by john williams

Bob has a helicopter from the launch pad he flies the following path. First he travels from the launch pad a distance of 24 kilometers at heading of 60 degrees East of South. Then he flies 47 degrees heading 77 degrees West of North. After this he flies 34 kilometers heading 75 degrees West of North. Now he is ready to return to the launch pad. What is the displacement vector that he needs to take to return directly to the launch pad, from his present location (for the heading give the number degrees north of east - your answer may be greater than 90 degrees)?
Answered by Henry
All angles are measured CCW from +x-axis.

Disp. = 24km[330o] + 47[167] + 34[165].
Disp. = (20.78-12i) + (-45.8+10.57i) + (-32.84+8.80i).
Disp. = -57.86 + 7.37i = 58.33km [-7.26o] = 58.33km[172.74o].
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