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9th Class Mathematics Triangles Question Bank Triangle

  • question_answer
    In the given figure below, 0 and C are respectively the orthocentre and circum centre of an acute angled triangle PQR. The points P and 0 are joined and produced to meet the side QR at S. If \[\angle \mathbf{PQS}=\mathbf{6}{{\mathbf{0}}^{{}^\circ }}\]and \[\angle \mathbf{QCR}=\mathbf{13}{{\mathbf{0}}^{{}^\circ }}\], then \[\angle \mathbf{RPS}=\]

    A)  \[{{30}^{{}^\circ }}\]                        

    B)  \[{{35}^{{}^\circ }}\]

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

    D)  \[{{60}^{{}^\circ }}\]

    Correct Answer: B

    Solution :

    (b): \[\angle PQS={{60}^{{}^\circ }}\] \[\angle QCR={{130}^{{}^\circ }}\] \[\therefore \]\[\angle QPR=\frac{1}{2}\times {{130}^{{}^\circ }}={{65}^{{}^\circ }}\] \[\Rightarrow \]\[\angle QRP={{180}^{{}^\circ }}-{{60}^{{}^\circ }}-{{65}^{{}^\circ }}={{55}^{{}^\circ }}\] \[\Rightarrow \]\[\angle PCQ={{110}^{{}^\circ }}\] \[\therefore \]In \[\Delta QCR,\] \[QC=CR\] \[\Rightarrow \]\[\angle CQR=\angle CRQ={{25}^{{}^\circ }}\] \[\left[ \therefore \angle CQR+\angle CRQ={{50}^{{}^\circ }} \right]\] \[\therefore \]\[\angle PQC+\angle QPC={{35}^{{}^\circ }}\] \[\left[ \therefore \angle PQC+\angle QPC={{70}^{{}^\circ }} \right]\] Similarly, \[\angle CPR={{30}^{{}^\circ }}\] \[\therefore \]\[\angle RPS={{35}^{{}^\circ }}\]


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