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Nikhil Manohar

· started a discussion

· 1 Months ago

Simple solution (Question: 113116.0 )

Consider triangle OBC,
OC = hypotenuse
=2 x 60°_adjacent
=2 x r
OD = r
So DC =2r-r=r
So OD:DC = 1:1

Question:
OA, OB are the radii of a circle with O as centre, the angle AOB = 120°. 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:
Options:
A) 1 : 2
B) 1 : 3
C) 1 : 1
D) 2 : 3
Solution:
Ans: (c) 


Since \(\angle \)OAC = \(\angle \)OBC = 90°

and \(\angle \)AOB = 120°

\(\therefore\) \(\angle \)ACB = 60°

\(\angle \)CAB = 90° - \(\angle \)OAM

= 60°

and \(\angle \)CBA = 90° - \(\angle \)OBM

= 60°

\(\therefore\) \(\triangle\)ABC is equilateral

and CA = CB = AB = 2AM = \(\sqrt[]{3r}\)

Then CM = CA cos 30° = \(\sqrt[]{3r}.\cfrac{\sqrt[]{3}}{2}=\cfrac{3}{2}r\)

Further \(OM=OA\ cos\ 30^o=\cfrac{r}{2}\)

\(\therefore DM=OD-OM=r\cfrac{r}{2}=\cfrac{r}{2}\)

and \(CD=CM-DM=\cfrac{3}{2}r-\cfrac{1}{2}r=r\)

\(\therefore OD:DC=r:r=1:1\)

vivek tripathi

· commented

· 1 Months ago

Yet to be approved!

surender kumar

· commented

· 1 Months ago

thnx..

Vijay Sharma

· commented

· 1 Months ago

thanks bro...

harshit

· commented

· 1 Months ago

nice one.

Mohit yadav

· commented

· 1 Months ago

nice bro...good solution.... inko kuch nhi aata
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