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Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsTriangles1 Mark
MCQ
Line segments AB and CD intersect at O such that AC || DB. If $\angle\text{CAB}=45^\circ$ and $\angle\text{CDB}=55^\circ,$ then $\angle\text{BOD}=$
  • A
    100º
  • B
    80º
  • C
    90º
  • D
    135º
✓

Answer

  1. 80º
    Solution:

    AC || BD
    And, AB is transverse to these parallal lines
    So $\angle\text{CAB}=\angle\text{ABD}$ (Alternate angles)
    $\Rightarrow\angle\text{ABD}=45^\circ$
    Now In $\triangle\text{BOD}$
    $\angle\text{BOD}+\angle\text{ODB}+\angle\text{DBA}=180^\circ$
    $\angle\text{DBA}=\angle\text{ABD}=45^\circ,\angle\text{ODB}=55^\circ$
    So $\angle\text{BOD}=180^\circ-45^\circ-55^\circ$
    $=80^\circ$

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