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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 ........ ????

Tanvir Alam , 7 Years ago
Grade 12th pass
anser 1 Answers
Arun
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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Last Activity: 7 Years ago
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