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If O is the in-centre of ∆ABC; if ∠BOC = 120°, then the measure of ∠BAC is
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- 30°
- 60°
- 150°
- 75°
- 30°
Correct Option: B
On the basis of given in question , we draw a figure triangle ABC ,
BO, CO and AO are internal bisectors of ∠B, ∠C and ∠A respectively.
Here , ∠BOC = 120°
∴ ∠BOC = 90° + | |
2 |
⇒ 120° = 90° + | |
2 |
⇒ | = 120° - 90° = 30° | |
2 |
∴ ∠A = 30 × 2 = 60°