Question

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

Answer 1

AB is a straight line.

$\angle ADE+\angle BDE={180}^0$

$\angle ADE+{65}^0={180}^0$

$\angle ADE={115}^0$......(i)

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

$\angle DBO={90}^0$

$In△BDC,\angle DBC+\angle BDC+\angle DCB={180}^0{90}^0+{65}^0+\angle DCB={180}^0$

$\angle DCB={25}^0$

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

$\angle DCBor\angle OCE=\angle OEC={25}^0$

$\angle OEC=\angle DEA={25}^0$.....(ii) (vertically opposite angles)

In triangle ADE ,

∠ADE+∠DEA+ ∠DAE =180

from (i) and (ii),

${115}^0+{25}^0+\angle DAE={180}^0\angle DAE=\angle BAO={40}^0$