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Question

In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B.  BDC = 65o. Find BAO.


Solution

AB is a straight line.

ADE+BDE=1800

ADE+650=1800

ADE=1150......(i)

AB i.e. DB is tangent to the circle at point B and BC is the diameter.

DBO=900

In BDC,DBC+BDC+DCB=1800900+650+DCB=1800

DCB=250

Now,OE=OC (radii of the same circle)

DCB or OCE=OEC=250

OEC=DEA=250.....(ii) (vertically opposite angles)

In triangle ADE ,

∠ADE+∠DEA+ ∠DAE =180

from (i) and (ii),

1150+250+DAE=1800DAE=BAO=400

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