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Circles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSCircles1 Mark
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Answer

  1. 92º
    Solution:

    Since AD acts as a chord also, So, $\angle\text{ABD}=\angle\text{ACD}=58^\circ$
    Again as CD also acts as a chord also, therefore,
    $\angle\text{DBC}=\angle\text{DAC}$
    Now, $\angle\text{ABC}=\angle\text{ABD}+\angle\text{DBC}$
    Also, $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
    $\Rightarrow\angle\text{ABC}=180^\circ-77^\circ=103^\circ$
    And therefore
    $\angle\text{DBC}=103^\circ-58^\circ=45^\circ$
    Hence, $\angle\text{DAC}=45^\circ$
    Since,
    $\angle\text{DAC}=45^\circ$
    So, $\angle\text{CAB}=75^\circ-45^\circ=30^\circ$
    But, $\angle\text{CAB}=\angle\text{BDC}$
    $\Rightarrow\angle\text{BDC}=30^\circ$
    Now, In triangle CPD,
    $\angle\text{C}+\angle\text{P}+\angle\text{D}=180^\circ$
    $\Rightarrow58^\circ+\angle\text{P}+30^\circ=180^\circ$
    $\Rightarrow\angle\text{P}=180^\circ-30^\circ-58^\circ=92^\circ$

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