question_answer

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

A) \[90{}^\circ -\frac{A}{4}\]                    

B) \[90{}^\circ +\frac{A}{4}\]

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

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

Correct Answer: D

Solution :

Given that, \[BI\]and \[CI\]are angle bisectors of\[\angle B\]and \[\angle C\]respectively.

Now, in \[\Delta BIC\]

\[x{}^\circ +\frac{B}{2}+\frac{C}{2}=180{}^\circ \]

[let \[\angle BIC=x{}^\circ \]]

\[\Rightarrow \]\[x{}^\circ =180{}^\circ -\frac{1}{2}(B+C)\]

\[\Rightarrow \]\[x{}^\circ =180{}^\circ -\frac{1}{2}(180{}^\circ -A)\]

\[[\because \text{in }\Delta ABC,A+B+C=180{}^\circ ]\]

\[\Rightarrow \]\[x{}^\circ =180{}^\circ -90{}^\circ +\frac{A}{2}\]

\[\therefore \]\[\angle BIC=x{}^\circ =90{}^\circ +\frac{A}{2}\]