'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

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

From angle sum property in quadrilateral ADOF we get

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

90$°$ + 90$°$ + 30$°$  +   $\angle$ DOF = 360$°$ 

  $\angle$ DOF = 150$°$                                                                                                         ---- ( 2 )

In quadrilateral CEOF 

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

From angle sum property in quadrilateral CEOF we get

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

90$°$ + 90$°$ + 90$°$  +   $\angle$ EOF = 360$°$ 

  $\angle$ 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

  $\angle$ COF =    $\angle$ COE  = 45$°$  

And

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

150$°$ + 45$°$ + $\angle$COD  = 360$°$     

$\angle$COD  = 165$°$                                             ( Ans )  

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