Question

In the given figure, $ ∆OBA ~ ∆ODC, \angle BOC = 125°$ and $ \angle CDO = 70°$. Find the value of $ \angle OAB$.

Answer 1

Compute the required value:

Given: $∆OBA~∆ODC$

Thus, $\angle ODC=\angle OBA=70°$ [ corresponding angles of similar triangles ]

From the figure, $AOC$ is a straight line. Therefore,

$\begin{array}{rcl}\angle \mathrm{AOB}& =& 180°-125°\\ & \Rightarrow & \angle \mathrm{AOB}=55°\end{array}$ [ Linear pair ]

$\begin{array}{rcl}& & \angle \mathrm{AOB}=\angle \mathrm{DOC}\\ & & \Rightarrow \angle \mathrm{DOC}=55°\end{array}$ [ vertically opposite angles ]

In $∆ODC$,

$\begin{array}{rcl}\angle \mathrm{OCD}& =& 180°-70°-55°\\ \angle \mathrm{OCD}& =& 55°\end{array}$ [ Sum of interior angles of a triangle is $180°$ ]

$\begin{array}{rcl}\angle \mathrm{OAB}& =& \angle \mathrm{OCD}\\ \angle \mathrm{OAB}& =& 55°\end{array}$ [ corresponding angles of similar triangles ]

Hence, the value of $\angle OAB$ is $55°$.