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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 after 8 such turns is equal to

sid pal , 12 Years ago

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3 Answers

kumar adarsh

Last Activity: 12 Years ago

try drawing a fig. for the problem.

so,after 6 turns the person reaches the same spot from which he had started aftet travelling along a reg. hexagon of side 10m and angle 150°.

then he again covers 10m and takes a turn and covers 10m.so we get an iso. triangle with 2 sides of 10m and the angle between them = 150° (60+90).

now solve using trigo.

1/2disp.÷10=sin75°

so disp.= 20sin75°

=20(√3+1/2√2)

=19.4m approx

Divyansh Singhvi

Last Activity: 12 Years ago

10*8=80

Debjit De

Last Activity: 7 Years ago

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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Mechanics> kinematics...

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 after 8 such turns is equal to

sid pal , 12 Years ago

kumar adarsh

Divyansh Singhvi

Debjit De

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