Find the length of the chord which subtends an angle of `120^(@)` at the centre of the circle of radius 6 cm.

Explanation

let the chord be AB and O be the circle.

In the figure, D is the mid-point of AB and OD od perpendicular to AB.
` OA=OB=6 cm, angleAOB=120^(@)`.
Clearly, triangles AOD and BOD are congruent.
`:. angleAOD=angleBOD=1/2angleAOB=60^(@)`
In `DeltaAOD, sin60^(@)=(AD)/(AO)=(AD)/6`
`:. AD=6sin60^(@)=6xxsqer3/2=3sqrt3`
`:. AB=2xxAD=6sqrt3`