A circular path of radius 20m …

Let Ankur be represented as A, Syed as S and David as D. The boys are sitting at an equal distance. Hence, △ASD is an equilateral triangle. Let the radius of the circular park be r meters. ∴OS=r=20m. Let the length of each side of △ASD be x meters. Draw AB⊥SD ∴SB=BD= 2 1 ​ SD= 2 x ​ m In △ABS,∠B=90 o By Pythagoras theorem, AS 2 =AB 2 +BS 2 ∴AB 2 =AS 2 −BS 2 =x 2 −( 2 x ​ ) 2 = 4 3x 2 ​ ∴AB= 2 3 ​ x ​ m Now, AB=AO+OB OB=AB−AO OB=( 2 3 ​ x ​ −20) m In △OBS, OS 2 =OB 2 +SB 2 20 2 =( 2 3 ​ x ​ −20) 2 +( 2 x ​ ) 2 400= 4 3 ​ x 2 +400−2(20)( 2 3 ​ x ​ )+ 4 x 2 ​ 0=x 2 −20 3 ​ x ∴x=20 3 ​ m The length of the string of each phone is 20 3 ​ m.