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					 The internal bisectors of the angles B and C of a triangle ABC meet at I. If ∠BIC = (∠A/2) + X, then X is equal to
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                        -  60° 
 
-  30° 
 
-  90° 
 
- 45°
 
-  60° 
Correct Option: C
On the basis of given in question , we draw a figure triangle ABC whose I is incentre ,
In ∆ ABC, 
We know that , ∠A + ∠B + ∠ = 180° 
∴ ∠B + ∠ = 180° – ∠A
| ∴ | (∠B + ∠C) = 90° - | |||
| 2 | 2 | 
In ∆ BIC,
| + | + ∠BIC = 180° | ||||
| 2 | 2 | 
| ∴ 90° - | + ∠BIC = 180° | ||
| 2 | 
| ⇒ ∠BIC = 180° - 90° + | ||
| 2 | 
| ∠BIC = 90° + | ||
| 2 | 
∴ X = 90°
 
	