Question : 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 $\angle \mathrm{DOC}=60^{\circ}$, then what is the length of BC?
Option 1: $3 \sqrt{7} \mathrm{~cm}$
Option 2: $3.5 \mathrm{~cm}$
Option 3: $7 \mathrm{~cm}$
Option 4: $5 \sqrt{7} \mathrm{~cm}$
Correct Answer: $7 \mathrm{~cm}$
Solution :
Given,
AD is a diameter
AC is a chord
OB = 7
$\angle{OBA} = 60°$
$\angle{DOC} = 60°$
Now, $\angle{DOC} + \angle{AOC} = 180°$ (The sum of the angles on a straight line is 180°)
⇒ $60° + \angle{AOC} = 180°$
⇒ $ \angle{AOC}= 120° $
In ΔAOC
AO = OC (Radius of the circle)
⇒ $\angle{OAC} = \angle{OCA} = \frac{180° – 120°}{2}$
⇒ $\angle{OAC} = \angle{OCA} = 30° $
⇒ $\angle{OBC} = 180°- 60° = 120°$
⇒ $\angle{ BOC} = (180°- 120° - 30°) = 30°$
⇒ $\angle{ BOC} = \angle{OCB} = 30°$
In ΔBOC,
OB = BC (Isosceles triangle)
⇒ OB = 7 cm
⇒ BC = 7 cm
Hence, the correct answer is $7 \mathrm{~cm}$.
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