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  • In Fig. 6.35, Delta ODC ~ Delta OBA, angle BOC = 125° and angle CDO = 70°. Find angle DOC, angle DCO and angle OAB.

Q2 In Fig. 6.35, \Delta ODC \sim \Delta OBA, \angle BOC = 125 \degree  and \angle CDO = 70 \degree. Find

      \angle DOC , \angle DCO , \angle OAB

      

Answers (1)

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Given :  \Delta ODC \sim \Delta OBA, \angle BOC = 125 \degree  and \angle CDO = 70 \degree

 

\angle DOC+\angle BOC=180 \degree            (DOB is a straight line)

\Rightarrow \angle DOC+125 \degree=180 \degree

\Rightarrow \angle DOC=180 \degree-125 \degree

\Rightarrow \angle DOC=55 \degree

In \Delta ODC ,

\angle DOC+\angle ODC+\angle DCO=180 \degree

\Rightarrow 55 \degree+ 70 \degree+\angle DCO=180 \degree

\Rightarrow \angle DCO+125 \degree=180 \degree

\Rightarrow \angle DCO=180 \degree-125 \degree

\Rightarrow \angle DCO=55 \degree

Since ,\Delta ODC \sim \Delta OBA , so

\Rightarrow\angle OAB= \angle DCO=55 \degree   ( Corresponding angles are equal in similar triangles).

 

 

Posted by

seema garhwal

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