#### 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 1

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°

#### Solution 2

As AB is a tangent to the circle at B and OB is radius, OB + AB ⇒ ∠ CBD = 90°.

In ΔBCD,

∠ BCD + ∠ CBD + ∠ BDC = 180°

∠ BCD + 90° + 65° = 180°

∠ BCD + 155° = 180°

∠ BCD = 180° - 155°

∠ BCD = 25°

∠ BOE = 2∠ BCE ....(angle at centre = double the angle at the remaining part of circle)

∠ BOE = 2 x 25°

∠ BOE = 50°

∠ BOA = 50°

In ΔBOA,

∠ BAO + ∠ ABO + ∠ BOA = 180°

∠ BAO + 90° + 50° = 180°

∠ BAO + 140° = 180°

∠ BAO = 180° - 140°

∠ BAO = 40°