In a circle with centre O, AD is a diameter and AC is a chord. Point B is on AC such that OB = 7 cm and $\angle OBA = 60^\circ$. If $\angle DOC = 60^\circ$, then what is the length of BC?

7 cm

\(\angle\)DOC = 60

\(\angle\)DOC + \(\angle\)AOC = 180 (Sum of angles on a straight line is 180).

= 60 + \(\angle\)AOC = 180

= \(\angle\)AOC = 120

In \(\Delta \)AOC,

AO = OC  (Radius of the circle)

= \(\angle\)OAC = \(\angle\)OCA = \(\frac{180\;-\;120}{2}\) = 30

= \(\angle\)OBC = 180 - 60 = 120

= \(\angle\)BOC = 180 - 120 - 30 = 30

\(\angle\)BOC = \(\angle\)OCB = 30

In \(\Delta \)BOC,

OB = BC  (Isosceles triangle)

OB = 7 cm

BC = 7 cm

Therefore, BC is 7 cm.