| In the given figure,\[\Delta ODC\tilde{\ }\Delta OBA\]. \[\angle BOC=125{}^\circ \]and\[\angle CDO=70{}^\circ \], then the value of \[\angle OAB\]is |
 |
A) \[70{}^\circ \]
B) \[125{}^\circ \]
C) \[65{}^\circ \]
D) \[55{}^\circ \]
Correct Answer: D
Solution :
| \[\angle DOC+\angle COB=180{}^\circ\] [linear pair] |
| \[\Rightarrow \,\,\,\,\angle DOC+125{}^\circ =180{}^\circ\] |
| Hence, \[\angle DOC=180{}^\circ -125{}^\circ =180{}^\circ\] |
| Again, \[\angle DCO+\angle CDO+\angle DOC=180{}^\circ\] |
| [angles of a triangle] |
| \[\Rightarrow \,\,\,\,\angle DCO+70{}^\circ +55{}^\circ =180{}^\circ\] |
| Hence, \[\angle DCO=180{}^\circ -125{}^\circ =55{}^\circ\] ...(i) |
| \[\because \Delta ODC\tilde{\ }\Delta OBA\] [given] |
| \[\therefore \angle OAB=\angle OCD={{55}^{{}^\circ }}\] [from Eq. (i)] |
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