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.

Answer 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°

Answer 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 × 25°

∠BOE = 50°

∠BOA = 50°

In ΔBOA,

∠BAO + ∠ABO + ∠BOA = 180°

∠BAO + 90° + 50° = 180°

∠BAO + 140° = 180°

∠BAO = 180° – 140°

∠BAO = 40°