Question

In figure $2.35$, $\Delta ODC\sim \Delta OBA,\angle BOC={125}^{\circ }$ and $\angle CDO={70}^{\circ }$. Find $\angle DOC,\angle DCO$ and $\angle OAB$.

Answer 1

$\angle DOC+\angle BOC={180}^{\circ }$(linear pair)

$\Rightarrow \angle DOC+{125}^{\circ }={180}^{\circ }$

$\Rightarrow \angle DOC={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

In $△DOC,$ we have

$\angle DOC+\angle ODC+\angle DCO={180}^{\circ }$ by angle sum property

$\Rightarrow {55}^{\circ }+{70}^{\circ }+\angle DCO={180}^{\circ }$

$\Rightarrow {125}^{\circ }+\angle DCO={180}^{\circ }$

$\Rightarrow \angle DCO={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

Given:$△ODC\sim △OBA$

$\therefore \angle OCD=\angle OAB$

$\Rightarrow \angle DCO=\angle OAB$

$\Rightarrow \angle OAB=\angle DCO$

But $\angle DCO={55}^{\circ }$

$\Rightarrow \angle OAB={55}^{\circ }$