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A Man Moves in an open field such that after moving 10 m on a straight line , he makes a sharp turn of 60 degree to his left . The Total Displacement just at the start of 8th Turn is equal to ........ ????

A Man Moves in an open field such that after moving 10 m on a straight line , he makes a sharp turn of 60 degree to his left . The Total Displacement just at the start of 8th Turn is equal to ........ ????
 

Grade:12th pass

1 Answers

Arun
25758 Points
3 years ago
Dear student
 
 
No of sides of the polygon is 360/60=6
We get a isosceles triangle inside a hexagon during his eighth turn. Interior angle of hexagon is 120°.
Leta perpendicular be dropped which divides his displacement into 2 halves. Let his displacement be d.Then sin60°=d/2÷10  Then d/2=10*sin 60°
d=2*10*sin 60°=17.32(approx)

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