question_answer

The bisectors BI and CI of \[\angle B\]and \[\angle C\]of a \[\Delta ABC\]meet in I. What is \[\angle BIC\]?

A) \[90{}^\circ -\frac{A}{A}\]

B) \[90{}^\circ +\frac{A}{A}\]

C) \[90{}^\circ -\frac{A}{2}\]

D) \[90{}^\circ +\frac{A}{2}\]

Correct Answer: D

Solution :

[d] Now, in \[\Delta \Beta \Iota C\] \[x{}^\circ +\frac{B}{2}+\frac{C}{2}=180{}^\circ \] \[\Rightarrow \]                           \[x{}^\circ =180{}^\circ -\frac{1}{2}(B+C)\] \[\Rightarrow \]                           \[x{}^\circ =180{}^\circ -\frac{1}{2}(180{}^\circ -A)\] \[\Rightarrow \]                           \[x{}^\circ =180{}^\circ -90{}^\circ +\frac{A}{2}\] \[\therefore \]                              \[\angle BIC=x{}^\circ -90{}^\circ +\frac{A}{2}\]