# Similarity of Triangles CBSE NCERT Solutions Chapter 6 Triangles Exercise 6.3 Question 2

CBSE NCERT Solutions Chapter 6 Triangles Exercise 6.3 Question 2

2.   In fig 6.35, $\triangle ODC$ ~ $\triangle OBA$, $\angle BOC$ = 125$^\circ$ and $\angle CDO$ = 70$^\circ$. Find $\angle DOC$, $\angle DCO$ and $\angle$ OAB.

Fig 6.35

Solution:

Given: $\triangle ODC$ ~ $\triangle OBA$, $\angle BOC$ = 125$^\circ$ and $\angle CDO$ = 70$^\circ$

$\angle BOC$$\angle DOC$ = 180$^\circ$    {Linear Pair}

$\Rightarrow \angle DOC$ = 180 - 125 = 55$^\circ$

In $\triangle ODC$

$\angle ODC$$\angle DCO$$\angle COD$ = 180$^\circ$   {Sum of angles of Triangle}

$\Rightarrow 70 + 55 + \angle DCO$ = 180

$\Rightarrow \angle DCO = 180 - 70 - 55 = 55^\circ$

It is given that $\triangle ODC$ ~ $\triangle OBA$.

Therefore, corresponding angles of $\triangle ODC$ and $\triangle OBA$ are equal.

By, AAA similarity criterion, $\angle OCD = \angle OAB$

Therefore, $\angle OAB = 55^\circ$

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