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Internal bisectors of ∠B and ∠C of ABC intersect at O. If ∠BOC = 102°, then the value of ∠BAC is
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- 12°
- 24°
- 48°
- 60°
- 12°
Correct Option: B
As per the given in question , we draw a figure of a triangle ABC and the internal bisectors of ∠B and ∠C intersect at O 
Here , ∠BOC = 102°
In ∆ABC ,
∠A + ∠B + ∠C = 180°
| ⇒ | + | = 90° - | .........( 1 ) | |||
| 2 | 2 | 2 |
In ∆ BOC,
| ∠BOC + | + | = 180° | ||
| 2 | 2 |
| ⇒ 102° + 90° – | = 180° { ∴ Using ( 1 ) } | |
| 2 |
| ⇒ | = 102° + 90° - 180° = 12° | |
| 2 |
∴ ∠A = 24°