Points $(8, 5)$ $(8 ,5 )$ and $(7, 6)$ $(7 ,6 )$ are $\frac{π}{3}$ $( pi)/3$ radians apart on a circle. What is the shortest arc length between the points?

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Points $(8, 5)$ $(8 ,5 )$ and $(7, 6)$ $(7 ,6 )$ are $\frac{π}{3}$ $( pi)/3$ radians apart on a circle. What is the shortest arc length between the points?

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Answer 1 · 2016-11-19T18:50:17

$s = \sqrt{2} \frac{π}{3}$ $s = sqrt(2)pi/3$

Explanation:

The length of the chord is:

$c = \sqrt{{(7 - 8)}^{2} + {(6 - 5)}^{2}}$ $c = sqrt((7 - 8)^2 + (6 - 5)^2)$

$c = \sqrt{2}$ $c = sqrt(2)$

Two radii, each drawn from the center to its respective end of the chord, form a triangle with the chord, therefore, we can use a variant of the Law of Cosines where $a = b = r$ $a = b = r$ :

$c^{2} = r^{2} + r^{2} - 2 (r) (r) cos (θ)$ $c^2 = r^2 + r^2 - 2(r)(r)cos(theta)$

Substitute $c = \sqrt{2} and θ = \frac{π}{3}$ $c = sqrt(2) and theta = pi/3$ :

NOTE: At this point, we should realize that an isosceles triangle with the third angle equal to $\frac{π}{3}$ $pi/3$ is an equilateral triangle but, let's proceed as an example of how to solve the problem, when it is not this special case:

${(\sqrt{2})}^{2} = 2 r^{2} - 2 r^{2} cos (\frac{π}{3})$ $(sqrt(2))^2 = 2r^2 - 2r^2cos(pi/3)$

$r^{2} = \frac{{(\sqrt{2})}^{2}}{2 (1 - cos (\frac{π}{3}))}$ $r^2 = (sqrt(2))^2/(2(1 - cos(pi/3))$

$r^{2} = 2$ $r^2 = 2$

$r = \sqrt{2}$ $r = sqrt(2)$

The arc length is, $s = r θ$ $s = rtheta$

$s = \sqrt{2} \frac{π}{3}$ $s = sqrt(2)pi/3$

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Points $(8, 5)$ $(8 ,5 )$ and $(7, 6)$ $(7 ,6 )$ are $\frac{π}{3}$ $( pi)/3$ radians apart on a circle. What is the shortest arc length between the points?

1 Answer

Explanation:

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Science

Math

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Points (8,5)(8 ,5 ) and (7,6)(7 ,6 ) are π3( pi)/3 radians apart on a circle. What is the shortest arc length between the points?

1 Answer

Explanation:

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Points $(8, 5)$ $(8 ,5 )$ and $(7, 6)$ $(7 ,6 )$ are $\frac{π}{3}$ $( pi)/3$ radians apart on a circle. What is the shortest arc length between the points?