# 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

Video Solution
Triangles
Ex 6.3 | Question 2

## 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 \\& \left[ \begin{array} & \because \angle DOC\text{ and }\angle \text{COB}\\ \text{form a linear pair} \\\end{array} \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) \\ &[\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 \\ &\quad\Rightarrow \angle DCO=\angle OAB \\ &\qquad\quad\;\; \angle DCO={{55}^{\circ }} \\ \end{align}

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