A) 4
B) 5
C) 6
D) 8
Correct Answer: B
Solution :
In \[\Delta \]ABC, \[\angle A=x,\angle B=y:\angle C=z\] In \[\Delta \] PBC \[\angle PBC+\angle PCB+\angle BPC=180{}^\circ \] \[\Rightarrow \]\[\frac{1}{2}\]\[\angle EBC+\angle FCB+2\angle BPC=360{}^\circ \] = \[180{}^\circ \] \[\Rightarrow \]\[\angle EBC+\angle FCB+2\angle BPC=360{}^\circ \] \[\Rightarrow \]\[\text{(180}{}^\circ -y)\text{ +}\left( 180{}^\circ -z \right)+2\text{ }\angle BPC\] =\[360{}^\circ \] \[\Rightarrow \]\[360{}^\circ -\left( y+z \right)+2\] \[\angle BPC=360{}^\circ \] \[\Rightarrow \]\[2\angle BPC=y+z\] \[\Rightarrow \]\[2\angle BPC=180{}^\circ -x\] \[=180{}^\circ -\angle BAC\] \[\therefore \]\[\angle BPC=90{}^\circ -\frac{1}{2}\angle BAC\] \[=90{}^\circ -50{}^\circ =40{}^\circ \]You need to login to perform this action.
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