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 \[\angle BPC=180{}^\circ -(\angle 1+\angle 2)\] |
\[=180{}^\circ -140{}^\circ =40{}^\circ \] |
You need to login to perform this action.
You will be redirected in
3 sec