What is the length of the chord ?

8-6=2 and finally 8+2=10

draw another line from O to the chord at 60.now length of each side of the equilateral triangle thus formed is 6.The length left is 2.Hence chord length is 6+2+2=10

Wow, and here I go using Law of Cosines to find the radius when I clearly could have just did it your way. Nice and simple solution.

cos 60=1/2=x/6... hence x=3... 8-3=5... and finally ac=5-3=2... so AB=AC+BC= 2+8=10....

simple 8-6=2 becz equilateral triangle and 8+2=10

Good question

10

length of the Perpendicular will be 3 root(3) = 6 sin60 and the corresponding base will be 3 = 6*cos60 so half of the chord value 5 only [Perpendicular to chord divide it equally] ] so value is 10

10

BC-OC=AC AC+BC= AB AO=OB 8cm-6cm=2 and finally 8 +2= 10

>>solve"cos(60)=(8-x)/(2*6)";
>>x=
2
>>length=(8+x) ;
>>length =
10

there is a better approach, draw a line $\quad D =60\degree\quad$ from o such that it touches $\quad BC\quad$ and,now $\quad\triangle COD\quad$ is equatorial due to its $\quad 60,60,60\quad$ angles. now, $\quad CD=6cm\quad$ and $\quad DB =2 cm\quad$ we can say $\quad AC=DB=2cm$ $\therefore 6+2+2=\boxed{10}$

another simpler way, is to construct a line that is 60 degree from o and lies on bc, name it op so angle opc is also 60 degree that is equilateral triangle so pc = 6 cm. From that 8-6 = 2 so 8+2 = 10