first of all theta + 2pie/3 = theta +120 =theta + 90 + 30 cos(theta + 30 +90) = -sin(theta +30) sin(theta + 30 +90) = cos(theta + 30)
using distance formula dist = sqrt of [{acos(theta + 30) - acos(theta + 60)}^2 + {asin(theta + 30) - asin(theta + 60)}^2] take a^2 common from both brackets also let theta +30 =A and theta + 60 =B
dist = a^2 * sqrt of [(cosA - cos B)^2 + (sinA - sin B)^2] = a^2 * sqrt of [2-2( sinAsinB + cosAcosB ) ] = a^2 * sqrt of[2 -2(cos (A - B))] now cos(A-B) = cos(-30)=cos30=(sqrt of 3)/2 put value and get the answer rate if u like my efforts
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The polar coordinates of the two points is (a, +2/3) and (a,+/3) d = r12 + r22 - 2r1r2cos(1-2) = a2+a2 - 2a2cos/3 = a
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If you are not familiar with polar coordinates and this formula, pick up SL Loney. The proof of this result is given somewhere in the begining.
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+2
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+
r12 + r22 - 2r1r2cos(
