Asked by Abigail
a man walks 8km North and then 5km in a. direction 60degrees. find its distance from his starting point
Answers
Answered by
MathMate
Use the component method to break down each vector into x,y components.
8km North = <0,8>
5km at 60° = <5cos60°,5sin60°>
Assuming angle is in Cartesian plane (counterclockwise from east) and not bearing (clockwise from north).
Add the two vectors and calculate the magnitude of the resultant.
8km North = <0,8>
5km at 60° = <5cos60°,5sin60°>
Assuming angle is in Cartesian plane (counterclockwise from east) and not bearing (clockwise from north).
Add the two vectors and calculate the magnitude of the resultant.
Answered by
Anonymous
56
Answered by
dorcas
5km
Answered by
henry2,
d = 8km[90o] + 5km[60o],
d = X + Yi = (8*Cos90+5*Cos60) + (8*sin90+5*sin60)I = 2.5 + 12.3i,
d = sqrt(X^2+Y^2) = 12.6 km.
Tan A = Y/x.
d = X + Yi = (8*Cos90+5*Cos60) + (8*sin90+5*sin60)I = 2.5 + 12.3i,
d = sqrt(X^2+Y^2) = 12.6 km.
Tan A = Y/x.
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