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Triangles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSTriangles1 Mark
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Answer

  1. 120º
    Solution:

    In $\triangle\text{ABD}$
    $\angle\text{A}+\angle\text{B}+\angle\text{D}=180^\circ$
    $\Rightarrow55^\circ+\angle\text{DBA}+25^\circ=180^\circ$
    $\Rightarrow\angle\text{DBA}=180^\circ-55^\circ-25^\circ$
    $=180^\circ-80^\circ$
    $\Rightarrow\angle\text{DBA}=100^\circ$
    So, $\angle\text{DBC}=180^\circ-\angle\text{DBA}$
    $=180^\circ-100^\circ$
    $\Rightarrow\angle\text{DBC}=80^\circ$
    Now, in $\triangle\text{EBC}$
    $\angle\text{E}+\angle\text{EBC}+\angle\text{C}=180^\circ$
    $\Rightarrow\angle\text{E}+80^\circ+40^\circ=180^\circ$ $\big(\angle\text{DBC}=\angle\text{EBC}\big)$
    $\Rightarrow\angle\text{E}=180^\circ-120^\circ=60^\circ$
    Also, $\text{x}=180^\circ-\angle\text{E}=180^\circ-60^\circ$
    $\Rightarrow\text{x}=120^\circ$

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