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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

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

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3 Answers

kumar adarsh
18 Points
10 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
20 Points
10 years ago

10*8=80

Debjit De
15 Points
5 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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