Asked by Dr.evidence
An air craft travelled from calabar to kano as follows, it flew first to illori covering a distance of 300km, 30° west of North, and then flew 400km, 60° east of North to kano. What is the resultant displacement? It's urgent
Answers
Answered by
Damon
distance north = N = 300 cos30+400 cos 60
distance east = E = -300 sin 30+400 sin 60
magnitude of displacement = sqrt(N^2+E^2)
angle east of north =tan^-1(E/N)
distance east = E = -300 sin 30+400 sin 60
magnitude of displacement = sqrt(N^2+E^2)
angle east of north =tan^-1(E/N)
Answered by
Henry
All angles are measured CCW from +x-axis.
Disp. = 300km[120o] + 400km[30o].
Disp. = (300*Cos120+300*sin120)+(400*Cos30+400*sin30),
Disp. = (-150+260i) + (346.4+200i).
Combine like-terms and convert to polar form.
Disp. = 300km[120o] + 400km[30o].
Disp. = (300*Cos120+300*sin120)+(400*Cos30+400*sin30),
Disp. = (-150+260i) + (346.4+200i).
Combine like-terms and convert to polar form.
Answered by
idris babangida labaran
thanks for the answer
Answered by
henry2,
Disp. = 196 + 460i = 500km[66.9o].
Answered by
Unknown
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