'O' is the centre of the inscribed circle in a 30°-60°-90° triangle ABC with right angled at C. If the circle is tangent to AB at D then the angle COD is?

Dear Student,

Please find below the solution to the asked query:

From given information we form our diagram , As :

Here , tangent BC and AC meet radius at E and F respectively .

We know : A tangent to a circle is perpendicular to the radius at the point of tangency. So

ODA  =  ODB  =  OEB  =  OEC  =  OFA =  OFC = 90°                       --- ( 1 )

From angle sum property in quadrilateral ADOF we get

ODA +  OFA +  DAF +  DOF = 360°  , Now substitute values from equation 1 and given value and get

90° + 90° + 30°  +   DOF = 360° 

  DOF = 150°                                                                                                         ---- ( 2 )

In quadrilateral CEOF 

  OEC  =  OFC =  ACB = 90°    , From equation 1 and given angle C at right angle .                 ---- ( 3 )

From angle sum property in quadrilateral CEOF we get

OEC +  OFC +  ACB +  EOF = 360°  , Now substitute values from equation 3 and get

90° + 90° + 90°  +   EOF = 360° 

  EOF = 90°   and OF  =  OE (  Radius of circle )  Hence

CEOF is a square and we know diagonals of square bisect the angles of square , So

  COF =    COE  = 45°  

And

DOF + COF +  COD  = 360°              (  Sum of angles on a point )

150° + 45° + COD  = 360°     

COD  = 165°                                             ( Ans )  

Hope this information will clear your doubts about topic.

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