question_answer

The external bisector of \[\angle B\]and \[\angle C\]of \[\Delta ABC\] (where AB and AC extended to E and F, respectively) meet at point P.  If \[\angle BAC=100{}^\circ ,\]then the measure of \[\angle BPC\]is                                                                        [SSC (FCI) 2012]

A) \[50{}^\circ \]                          

B) \[80{}^\circ \]

C) \[40{}^\circ \]  

D) \[100{}^\circ \]

Correct Answer: C

Solution :

In \[\Delta ABC\]side AB and AC are produced to E and F, respectively and the external bisector \[\angle EBC\]and \[\angle FCB\]intersect at P.

\[x+y+z=180{}^\circ \]

\[y+z=180-x\]

\[=180-100=80{}^\circ \]

Now,     \[2\angle 1+y=180{}^\circ \]

and       \[2\angle 2+z=180{}^\circ \]

\[\therefore \]      \[2\,\,(\angle 1+\angle 2)=360{}^\circ -(y+z)\]

\[=360{}^\circ -80{}^\circ =280{}^\circ \]

and       \[\angle BPC=180{}^\circ -(\angle 1+\angle 2)\]

\[=180{}^\circ -140{}^\circ =40{}^\circ \]