Correct option is (A) 90° + 1/2 ∠A
\(\angle A+\angle B+\angle C\) \(=180^\circ\) (Sum of angles in \(\triangle ABC)\)
\(\Rightarrow\) \(\frac{\angle A}2+\frac{\angle B}2+\frac{\angle C}2\) \(=\frac{180^\circ}2\)
\(\Rightarrow\) \(\frac{\angle B}2+\frac{\angle C}2\) \(=90^\circ-\frac{\angle A}2\)
\(\Rightarrow\) \(\angle XBC+\angle XCB=90^\circ-\frac{\angle A}2\)
\((\because\) BX and CX are angle bisectors of \(\angle B\;\&\;\angle C)\)
\(\therefore\) \(\angle BXC\) \(=180^\circ-(\angle XBC+\angle XCB)\)
\(=180^\circ-(90^\circ-\frac{\angle A}2)\)
= \(90^\circ+\frac{\angle A}{2}\)