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 ∠OBA = 60°. If ∠DOC = 60°, then what is the length of BC (in cm) ?

7

\(\angle\)DOC = \({60}^\circ\)

\(\angle\)DOC + \(\angle\)AOC = 180 (Sum of the 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 = \(\frac{(180\; - \;120)}{2}\) = \({30}^\circ\)

= \(\angle\)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.