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

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Points $(2, 2)$ $(2 ,2 )$ and $(8, 1)$ $(8 ,1 )$ are $\frac{5 π}{6}$ $(5 pi)/6$ 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 $8.25$ $8.25$ unit.

Explanation:

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

$D = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}} = \sqrt{{(2 - 8)}^{2} + {(2 - 1)}^{2}}$ $D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2) =sqrt ((2-8)^2+(2-1)^2$ or

$D = \sqrt{37} \approx 6.08$ $D=sqrt37 ~~ 6.08$ unit. So, chord length is $L_{c} = 6.08$ $L_c=6.08$ unit.

Formula for the length of a chord is $L_{c} = 2 r sin (\frac{θ}{2})$ $L_c= 2r sin (theta/2)$

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

subtended at the center by the chord.

$θ = \frac{5 π}{6} = \frac{5 \cdot 180}{6} = 150^{0}; \frac{θ}{2} = 75^{0}$ $theta =(5pi)/6=(5*180)/6=150^0 ; theta/2=75^0$

$6.08 = 2 \cdot r \cdot sin 75 or r = \frac{6.08}{2 \cdot sin 75} = 3.15$ $6.08 = 2*r *sin75 or r = 6.08/(2*sin75)= 3.15$ unit.

Arc length is $L_{a} = 2 \cdot π \cdot r \cdot \frac{θ}{360} = 2 \cdot π \cdot 3.15 \cdot \frac{150}{360}$ $L_a= 2* pi * r*theta/360=2*pi*3.15*150/360$ or

$L_{a} \approx 8.25$ $L_a ~~ 8.25$ unit. Shorter arc length between the points is

$8.25$ $8.25$ unit. [Ans]

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

1 Answer

Explanation:

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Points (2 ,2 )(2,2)(2 ,2 ) and (8 ,1 )(8,1)(8 ,1 ) are (5 pi)/6 5π6(5 pi)/6 radians apart on a circle. What is the shortest arc length between the points?

1 Answer

Explanation:

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