Question

In the given figure $∆ODC~∆OBA,\angle BOC=115°\mathrm{and}\angle CDO=70°.$ Find


(i) ∠DOC
(ii) ∠DCO
(iii) ∠OAB
(iv) ∠OBA.

Answer 1

(i)
It is given that DB is a straight line.
Therefore,
$$\begin{gathered} \angle DOC+\angle COB=180° \\ \angle DOC\hspace{0.17em}=180°-115°=65° \end{gathered}$$

(ii)
In $△DOC$, we have:
$$\begin{gathered} \angle ODC+\angle DCO+\angle DOC=180° \\ \mathrm{Therefore}, \\ 70°+\angle DCO+65°=180° \\ \Rightarrow \angle DCO=180-70-65=45° \end{gathered}$$

(iii)
It is given that $△ODC~△OBA$
Therefore,
$\angle OAB=\angle OCD=45°$

(iv)
Again, $△ODC~△OBA$
Therefore,
$\angle OBA=\angle ODC=70°$