A) 1 : 2
B) 1 : 3
C) 1:1
D) 2 : 3
Correct Answer: C
Solution :
Since \[\angle OAC=\angle OBC={{90}^{o}}\] and \[\angle AOB={{120}^{o}}\] \[\therefore \] \[\angle ACB={{60}^{o}}\] Also \[\angle CAB={{90}^{o}}-\angle OAM={{60}^{o}}\] and \[\angle CBA={{90}^{o}}-\angle OBM={{60}^{o}}\] Hence \[\Delta \,ABC\] is equilateral \[CA=CB=AB=2AM=\sqrt{3}r\] Then \[CM=CA\,\cos \,30\] \[=\sqrt{3}\,\,r.\,\frac{\sqrt{3}}{2}=\frac{3}{2}r\] Further \[OM=OA\,\,\cos \,\,{{60}^{o}}=\frac{r}{2}\] \[\therefore \] \[DM=OD-OM=r-\frac{r}{2}=\frac{r}{2}\] and \[CD=CM-DM=\frac{3}{2}r-\frac{1}{2}r=r\] \[\therefore \] s \[OD:DC=r:r:1:1\]You need to login to perform this action.
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