Points $(2 ,2 )$ and $(8 ,1 )$ are $(5 pi)/6$ radians apart on a circle. What is the shortest arc length between the points?

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Answer 1 · 2018-01-27T11:54:20

Shorter arc length between the points is $8.25$ unit.

Distance between two points $(2,2) and (8,1)$ is

$D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2) =sqrt ((2-8)^2+(2-1)^2$ or

$D=sqrt37 ~~ 6.08$ unit. So, chord length is $L_c=6.08$ unit.

Formula for the length of a chord is $L_c= 2r sin (theta/2)$

where $r$ is the radius of the circle and $theta$ is the angle

subtended at the center by the chord.

$theta =(5pi)/6=(5*180)/6=150^0 ; theta/2=75^0$

$6.08 = 2*r *sin75 or r = 6.08/(2*sin75)= 3.15$ unit.

Arc length is $L_a= 2* pi * r*theta/360=2*pi*3.15*150/360$ or

$L_a ~~ 8.25$ unit. Shorter arc length between the points is

$8.25$ unit. [Ans]

Points (2 ,2 )(2 ,2 ) and (8 ,1 )(8 ,1 ) are (5 pi)/6 (5 pi)/6 radians apart on a circle. What is the shortest arc length between the points?