A circle has a chord that goes from #( pi)/8 # to #(4 pi) / 3 # radians on the circle. If the area of the circle is #64 pi #, what is the length of the chord?

Radius:
Area#=64pi#
#pir^2=64pi#
#r^2=64#
#r=8#

Chord length:
The first angle, #pi/8#, is in quadrant I, and the second angle, #(4pi)/3#, is in quadrant III. Imagine a right triangle where the hypotenuse is the chord from polar coordinates #(8,pi/8)# to #(8,(4pi)/3)#. The horizontal and vertical distances can be found by adding the absolute values of the cosine and sine values, respectively.

(Drawing not to scale)

To solve for chord length:
#x^2+y^2=c^2" (Pythagorean theorem)"#

Substitute in values for x and y (shown in drawing above):
#c^2 = [|8cos(pi/8)|+|8cos((4pi)/3)|]^2+[|8sin(pi/8)|+|8sin((4pi)/3)|]^2#

Simplify (get rid of absolute values by determining their sign)
#c^2=[8cos(pi/8)-8cos((4pi)/3)]^2+[8sin(pi/8)-8sin((4pi)/3)]^2#

#c^2=[8cos(pi/8)-8(-1/2)]^2+[8sin(pi/8)-8(-sqrt3/2)]^2#

#c^2=(8cos(pi/8)+4)^2+(8sin(pi/8)+4sqrt3)^2#

#c=sqrt{(8cos(pi/8)+4)^2+(8sin(pi/8)+4sqrt3)^2}#

#c approx 15.15088#

This answer appears correct because if the radius is 8, then the diameter is 16, and this given chord is just less than 16 because it is not the diameter.