Points $(4, 4)$ $(4 ,4 )$ and $(7, 3)$ $(7 ,3 )$ are $\frac{5 π}{4}$ $(5 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

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Points $(4, 4)$ $(4 ,4 )$ and $(7, 3)$ $(7 ,3 )$ are $\frac{5 π}{4}$ $(5 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

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Answer 1 · 2017-08-20T17:15:46

$S = \frac{3 π}{4} \sqrt{\frac{5}{3} (2 - \sqrt{2})}$ $S = (3pi)/4sqrt(5/3(2-sqrt2))$

Explanation:

Because the angle $\frac{5 π}{4}$ $(5pi)/4$ is greater than $π$ $pi$ , we know that this is NOT the smallest angle between the two points; the smallest angle is:

$∠ θ = 2 π - \frac{5 π}{4}$ $angle theta=2pi-(5pi)/4$

$∠ θ = \frac{3 π}{4}$ $angle theta = (3pi)/4$

This will be the angle between the two radii that connect the two points $(4, 4)$ $(4,4)$ and $(7, 3)$ $(7,3)$ . The two radii and the chord, c, between the two points form a triangle, therefore, we can write an equation, using the Law of Cosines:

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

We know that $c^{2}$ $c^2$ is the square of the distance between the two points and we know the value of $θ$ $theta$ :

${(7 - 4)}^{2} + {(3 - 4)}^{2} = r^{2} + r^{2} - 2 (r) (r) cos (\frac{3 π}{4})$ $(7-4)^2+(3-4)^2 = r^2 + r^2-2(r)(r)cos((3pi)/4)$

Remove a common factor of $r^{2}$ $r^2$ :

$3^{2} + {(- 1)}^{2} = r^{2} (2 - 2 cos (\frac{3 π}{4}))$ $3^2+(-1)^2= r^2(2-2cos((3pi)/4))$

Evaluate the cosine function:

$10 = r^{2} (2 - 2 (- \frac{\sqrt{2}}{2}))$ $10= r^2(2-2(-sqrt2/2))$

Simplify:

$10 = r^{2} (2 + \sqrt{2})$ $10= r^2(2+sqrt2)$

Multiply both sides by the conjugate:

$10 (2 - \sqrt{2}) = r^{2} (2 + \sqrt{2}) (2 - \sqrt{2})$ $10(2-sqrt2)= r^2(2+sqrt2)(2-sqrt2)$

$10 (2 - \sqrt{2}) = 6 r^{2}$ $10(2-sqrt2)= 6r^2$

Flip the equation and divide both sides by 6:

$r^{2} = \frac{5}{3} (2 - \sqrt{2})$ $r^2 = 5/3(2-sqrt2)$

Use the square root operation on both sides:

$r = \sqrt{\frac{5}{3} (2 - \sqrt{2})}$ $r = sqrt(5/3(2-sqrt2))$

We know that the arclength, S, is the radius multiplied by the radian measure of angle:

$S = θ r$ $S = thetar$

$S = \frac{3 π}{4} \sqrt{\frac{5}{3} (2 - \sqrt{2})}$ $S = (3pi)/4sqrt(5/3(2-sqrt2))$

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Points $(4, 4)$ $(4 ,4 )$ and $(7, 3)$ $(7 ,3 )$ are $\frac{5 π}{4}$ $(5 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

1 Answer

Explanation:

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Points (4,4)(4 ,4 ) and (7,3)(7 ,3 ) are 5π4(5 pi)/4 radians apart on a circle. What is the shortest arc length between the points?

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Points $(4, 4)$ $(4 ,4 )$ and $(7, 3)$ $(7 ,3 )$ are $\frac{5 π}{4}$ $(5 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?