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

1 Answer
P dilip_k
Mar 6, 2016

6.24 unit

Explanation:

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It is evident from the above figure that shortest #arcAB# having end point A(2,9) and B (1,3) will subtend #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 #/_AOB# and #/_AOC# becomes #pi/8#.
AgainAC= BC#=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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