In the figure, \(\triangle\) ODC ~ \(\triangle\) OBA, \(\angle\) BOC = 125 and \(\angle\) CDO = 70. Find the \(\angle\) DOC, \(\angle\) DCO and \(\angle\) OAB.

Solution :

Since BD is a line and OC is a ray on it.similar triangles

\(\angle\) DOC + \(\angle\) BOC = 180

So,  \(\angle\) DOC + 125 = 180

\(\angle\) DOC = 55

In triangle CDO, we have :

\(\angle\) CDO + \(\angle\) DOC + \(\angle\) DCO = 180

70 + 55 + \(\angle\) DCO = 180

\(\angle\) DCO = 55

Given that \(\triangle\) ODC ~ \(\triangle\) OBA

\(\angle\) ODC = \(\angle\) OBA, \(\angle\) OCD = \(\angle\) OAB

\(\angle\) OBA = 70 and \(\angle\) OAB = 55

Hence, \(\angle\) DOC = 55, \(\angle\) DCO = 55 and \(\angle\) OAB = 55 

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