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Triangles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSTriangles1 Mark
Question
In $\triangle\text{ABC},$ if $\angle\text{A}=100^\circ,\text{AD}$ bisects $\angle\text{A}$ and $\text{AD}\perp\text{BC}.$ Then, $\angle\text{B}=$
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Answer

  1. 40º
    Solution:

    $\text{AD}\perp\text{BC}$ and AD bisects $\angle\text{A}.$
    $\Rightarrow\angle\text{BAD}=\angle\text{CAD}=50^\circ$
    In Right $\triangle\text{ADB}$
    $\angle\text{BAD}=50^\circ,\angle\text{ADB}=90^\circ$
    Also sum of all interior angles = 180º
    $\Rightarrow\angle\text{BAD}+\angle\text{ADB}+\angle\text{B}=180^\circ$
    $\Rightarrow\angle\text{B}=180^\circ-50^\circ-90^\circ$
    $\Rightarrow\angle\text{B}=40^\circ$

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