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

From figure,

⇒ ∠ADE + ∠BDE = 180° [Linear pairs]

⇒ ∠ADE + 65° = 180°

⇒ ∠ADE = 180° - 65°

⇒ ∠ADE = 115° …………..(1)

∠DBO = 90° [As, DB is tangent and BC is diameter.]

In △BDC,

⇒ ∠BDC + ∠DBC + ∠DCB = 180° [Angle sum property of triangle]

⇒ 65° + 90° + ∠DCB = 180°

⇒ ∠DCB = 180° - 155° = 25°.

From figure,

OE = OC [Radius of same circle.]

⇒ ∠OCE = ∠OEC [As, angles opposite to equal sides are equal.]

⇒ ∠OEC = ∠DCB = 25°. [As, ∠OCE = ∠DCB]

In △ADE,

⇒ ∠ADE + ∠DEA + ∠DAE = 180° [Angle sum property of triangle]

⇒ 115° + 25° + ∠DAE = 180° [From figure, ∠DEA = ∠OEC [Vertically opposite angles are equal]]

⇒ ∠DAE = 180° - 140° = 40°.

From figure,

∠BAO = ∠DAE = 40°.

Hence, ∠BAO = 40°.