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Grade 12th passAIIMS MBBS

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 agoGrade 12th pass

1 Answer

Arun

Approved Tutor Answer7 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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