question_answer

OA, OB are the radii of a circle with 0 as centre, the angle AOB = \[120{}^\circ \]. Tangents at A and B are drawn to meet in the point C. If OC intersects the circle in the point D, then D divides OC in the ratio of

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 \[OD:DC=r:r:1:1\]