Question

OA, OB are the radii of a circle with 0 as the center, the $\angle AOB={120}^o$. 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

Answer 1

Answer: (C)

1 : 1

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 $CM=CAcos30=\sqrt{3}r.\frac{\sqrt{3}}{2}=\frac{3}{2}r$
Further $OM=OAcos{60}^o=\frac{r}{2}$
$\therefore DM=OD-OM=r-\frac{r}{2}=\frac{r}{2}$
and $CD=CM-DM=\frac{3}{2}r-\frac{1}{2}r=r$
$\therefore OD:DC=r:r:1:1$