Question

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

• A. $\angle DOC={55}^o;\angle DCO={55}^o;\angle OAB={55}^o$
• B. $\angle DOC={70}^o;\angle DCO={55}^o;\angle OAB={55}^o$
• C. $\angle DOC={70}^o;\angle DCO={70}^o;\angle OAB={55}^o$
• D. $\angle DOC={55}^o;\angle DCO={55}^o;\angle OAB={70}^o$

Answer 1

The correct option is A $\angle DOC={55}^o;\angle DCO={55}^o;\angle OAB={55}^o$

$\angle DOC+{125}^o={180}^o$ (linear pair)
$\Rightarrow$ $\angle DOC={180}^o-{125}^o={55}^o$
In $△DOC$
$\angle DCO+\angle CDO+\angle DOC={180}^o$ (sum of three angles of $△ODC$)
$\Rightarrow$ $\angle DCO+{70}^o+{55}^o={180}^o$
$\Rightarrow$ $\angle DCO+{125}^o={180}^o$
$\Rightarrow$ $\angle DCO={180}^o-{125}^o={55}^o$
Now we are given that $△ODC\sim △OBA$
$\Rightarrow$ $\angle OCD=\angle OAB$ (Corresponding angles of similar triangles)
$\Rightarrow$ $\angle OAB=\angle OCD=\angle DCO={55}^o$
i.e., $\angle OAB={55}^o$
Hence we have,
$\angle DOC={55}^o;\angle DCO={55}^o;\angle OAB={55}^o$