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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 × 25°
∠BOE = 50°
∠BOA = 50°
In ΔBOA,
∠BAO + ∠ABO + ∠BOA = 180°
∠BAO + 90° + 50° = 180°
∠BAO + 140° = 180°
∠BAO = 180° – 140°
∠BAO = 40°
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