Mathematics
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.
Answer
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°.
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