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. \(\angle\) 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 = …