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?
Answer & explanation
Correct answer: option 3
\(\angle\)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.