A) \[65{}^\circ \]
B) \[55{}^\circ \]
C) 75
D) \[45{}^\circ \]
Correct Answer: B
Solution :
(b):\[\angle DOC+125{}^\circ =180{}^\circ \] (\[\because \] DOC is a straight line) \[\Rightarrow \]\[~\angle DOC=180{}^\circ -125{}^\circ =55{}^\circ \] (Sum of three angles of\[\Delta \,ODC\]) \[\angle DCO+\angle CDO+\angle DOC=180{}^\circ \] \[\Rightarrow \]\[\angle DCO+70{}^\circ +55{}^\circ =180{}^\circ \] \[\Rightarrow \]\[\angle DCO+125{}^\circ =180{}^\circ \] \[\Rightarrow \]\[\angle ~DCO=180{}^\circ -125{}^\circ =55{}^\circ \] Now, we are given that, \[\Delta \,ODC\tilde{\ }\Delta \,OBA.\] \[\Rightarrow \]\[\angle OCD=\angle OAB\] \[\Rightarrow \]\[\angle OAB=\angle OCD=55{}^\circ \] i.e., \[\angle OAB=55{}^\circ \]
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