#### Question

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

#### Solution

AB is a straight line.

∴ `∠`ADE + `∠`BDE = 180°

⇒ `∠`ADE + 65° = 180°

⇒ `∠`ADE = 115° ………..(i)

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

∴`∠`DB = 90°

In ΔBDC,

`∠`DBC + `∠`BDC + `∠`DCB = 180°

⇒ 90° + 65° + `∠`DCB = 180°

⇒ `∠` DCB = 25°

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

∴ `∠`DCB or `∠`OCE = `∠`OEC = 25°

Also,

`∠`OEC = `∠`DEC = 25°

(vertically opposite angles)

In ΔADE,

`∠`ADE + `∠`DEA + `∠`DAE = 180°

From (i) and (ii)

115° + 25° + `∠` DAE = 180°

⇒ `∠`DAE or `∠`BAO = 180° - 140° =40°

∴`∠`BAO = 40°

Is there an error in this question or solution?

Solution In the Adjoining Figure, O is the Centre of the Circle and Ab is a Tangent to It at Point B. ∠Bdc = 65° Find ∠Bao. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.