A) \[{{50}^{{}^\circ }}\]
B) \[{{80}^{{}^\circ }}\]
C) \[{{30}^{{}^\circ }}\]
D) \[{{100}^{{}^\circ }}\]
Correct Answer: C
Solution :
(c): In \[\Delta ABC\], \[\angle A=x,\angle B=y;\angle C=z\] In \[\Delta PBC\], \[\angle PBC+\angle PCB+\angle BPC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\frac{1}{2}\angle EBC+\frac{1}{2}\angle FCB+\frac{1}{2}\angle BPC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle EBC+\angle FCB+2\angle BPC={{360}^{{}^\circ }}\] \[\Rightarrow \] \[\left( {{180}^{{}^\circ }}-y \right)+\left( {{180}^{{}^\circ }}-z \right)+2\angle BPC={{360}^{{}^\circ }}\] \[\Rightarrow \]\[{{360}^{{}^\circ }}-\left( y+z \right)+2\angle BPC={{360}^{{}^\circ }}\] \[\Rightarrow \]\[2\angle BPC=y+z\] \[\Rightarrow \] \[2\angle BPC={{180}^{{}^\circ }}-x\] \[=180{}^\circ -\angle BAC\] \[\therefore \] \[\angle BPC={{90}^{{}^\circ }}-\frac{1}{2}\angle BAC\] \[={{90}^{{}^\circ }}-{{60}^{{}^\circ }}={{30}^{{}^\circ }}\]You need to login to perform this action.
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