Ex.6.3 Q2 Triangles Solution - NCERT Maths Class 10

Go back to  'Ex.6.3'

Question

In Figure  \(\Delta ODC\sim \Delta OBA\) ,\(\angle BOC = 125°\)and \(\angle CDO = 70°\). Find ,\(\angle \text DCO, \angle \text{ DCO}\) and \(OAB\). 

Diagram

Text Solution

Reasoning:

As we are aware if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred as \(AA\) criterion for two triangles.

Steps:

In the given figure.

\[\begin{align}  \angle DOC&={{180}^{\circ }}-\angle COB \\ & \qquad \left[ \because \,\,\angle DOC\,\,\rm{and}\,\,\angle COB\,\,\text{from}\,\text{a}\,\,\text{linear}\,\,\text{pair} \right] \\  \angle DOC&={{180}^{\circ }}-{{125}^{\circ }} \\ \angle DOC& ={{55}^{\circ }} \\ \end{align}\]

In \(\Delta ODC\) 

\[\begin{align}\angle DCO&={{180}^{\circ }}-(\angle DOC+\angle ODC) \\ & \qquad [\because \,\text{angle sum property}] \\  \angle DCO&={{180}^{\circ }}-({{55}^{\circ }}+{{70}^{\circ }}) \\  \angle DCO &={{55}^{\circ }} \\ \end{align}\]

In \(\Delta ODC\) and \(\Delta OBA\) 

\[\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta ODC\sim \Delta OBA \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \angle DCO=\angle OAB\,\,\, \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\angle DCO={{55}^{\circ }} \\ \end{align}\]