Question

Question 2
In the figure given below,
$\Delta ODC\sim \Delta OBA,\angle BOC={125}^{\circ }and\angle CDO={70}^{\circ }$
$Find\angle DOC,\angle DCOand\angle OAB$.

Answer 1

DOB is a straight line.
$Therefore,\angle DOC+\angle BOC={180}^{\circ }\left(linearpairangles\right)$
$\Rightarrow \angle DOC={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

In $\Delta DOC$,
$\angle DCO+\angle CDO+\angle DOC={180}^{\circ }$ (Sum of the measures of the all angles in a triangle is ${180}^{\circ }$)
$\Rightarrow \angle DCO+{70}^{\circ }+{55}^{\circ }={180}^{\circ }$
$\Rightarrow \angle DCO={55}^{\circ }$

It is given that $\Delta ODC\sim \Delta OBA$.
$\therefore \angle OAB=\angle OCD$ [Corresponding angles are equal in similar triangles.]
$\Rightarrow \angle OAB={55}^{\circ }$