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9th Class Mathematics Triangles Question Bank Triangle

  • question_answer
    In the given figure below, the external bisector of \[\angle \mathbf{B}\] and \[\angle C\] of \[\Delta \mathbf{ABC}\] (where AB and AC extended to E and F respectively) meet at point P. If \[\angle \mathbf{BAC}=\mathbf{12}{{\mathbf{0}}^{{}^\circ }}\], then the measure of \[\angle \mathbf{BPC}\] is

    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 \[\Delta PBC\], \[\angle PBC+\angle PCB+\angle BPC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\frac{1}{2}\angle EBC+\frac{1}{2}\angle FCB+\frac{1}{2}\angle BPC={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle EBC+\angle FCB+2\angle BPC={{360}^{{}^\circ }}\] \[\Rightarrow \] \[\left( {{180}^{{}^\circ }}-y \right)+\left( {{180}^{{}^\circ }}-z \right)+2\angle BPC={{360}^{{}^\circ }}\] \[\Rightarrow \]\[{{360}^{{}^\circ }}-\left( y+z \right)+2\angle BPC={{360}^{{}^\circ }}\] \[\Rightarrow \]\[2\angle BPC=y+z\] \[\Rightarrow \] \[2\angle BPC={{180}^{{}^\circ }}-x\] \[=180{}^\circ -\angle BAC\] \[\therefore \] \[\angle BPC={{90}^{{}^\circ }}-\frac{1}{2}\angle BAC\] \[={{90}^{{}^\circ }}-{{60}^{{}^\circ }}={{30}^{{}^\circ }}\]                            


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