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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

$\angle DOC , \angle DCO , \angle OAB$

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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).

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