Points $(2, 9)$ $(2 ,9 )$ and $(1, 3)$ $(1 ,3 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

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Points $(2, 9)$ $(2 ,9 )$ and $(1, 3)$ $(1 ,3 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

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Answer 1 · 2016-03-06T04:59:32

6.24 unit

Explanation:

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It is evident from the above figure that shortest $a r c A B$ $arcAB$ having end point A(2,9) and B (1,3) will subtend $\frac{π}{4}$ $pi/4$ rad angle at the center O of the circle. AB chord is obtained by joining A,B. A perpendicular OC is also drawn on it at C from center O.
Now the triangle OAB is isosceles having OA=OB=r (radius of circle)
Oc bisects $∠ A O B$ $/_AOB$ and $∠ A O C$ $/_AOC$ becomes $\frac{π}{8}$ $pi/8$ .
AgainAC= BC $= \frac{1}{2} A B = \frac{1}{2} \cdot \sqrt{{(2 - 1)}^{2} + {(9 - 3)}^{2}} = \frac{1}{2} \sqrt{37}$ $=1/2AB=1/2*sqrt((2-1)^2+(9-3)^2)=1/2sqrt37$

$:.AB=sqrt37$

Now $AB=AC+BC=rsin/_AOC+rsin/_BOC=2rsin(pi/8)$
$r=1/2AB*(1/sin(pi/8))=1/2sqrt37csc(pi/8)$

Now,
Shortest Arc length of AB = Radius $*/_AOB=r*/_AOB=r*(pi/4)=1/2sqrt37csc(pi/8)*(pi/4)=6.24$ unit

**More easily by properties of triangle **
$r/sin(3pi/8)=(AB)/sin(pi/4)$
$r=(AB)/sin(pi/4)*(sin(3pi/8))=sqrt2AB*sin(3pi/8)$
Now
Shortest Arc length of AB = Radius $*/_AOB=r*/_AOB=r*(pi/4)=sqrt2AB*sin(3pi/8)*pi/4=6.24$ unit

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Points $(2, 9)$ $(2 ,9 )$ and $(1, 3)$ $(1 ,3 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?

1 Answer

Explanation:

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Points $(2, 9)$ $(2 ,9 )$ and $(1, 3)$ $(1 ,3 )$ are $\frac{3 π}{4}$ $(3 pi)/4$ radians apart on a circle. What is the shortest arc length between the points?