QQuestion
In the given figure, O is the centre of the circle. CE is a tangent to the circle at A. If ∠ABD = 26°, find:
(a) ∠BDA
(b) ∠BAD
(c) ∠CAD
(d) ∠ODB
✓Answer
(a) Since BD is a diameter, the angle in a semicircle is a right angle.
Therefore, ∠BDA = 90°.
(b) In △BAD:
∠BDA + ∠BAD + ∠ABD = 180°
90° + ∠BAD + 26° = 180°
∠BAD + 116° = 180°
∠BAD = 180° − 116°
Therefore, ∠BAD = 64°.
(c) CE is tangent at A, so CE is perpendicular to radius OA.
∠CAD + ∠BAD = 90°
∠CAD + 64° = 90°
∠CAD = 90° − 64°
Therefore, ∠CAD = 26°.
(d) Join OD.
OD = OB, since both are radii.
Therefore, ∠ODB = ∠OBD.
Since O lies on AB, ∠OBD = ∠ABD = 26°.
Therefore, ∠ODB = 26°.
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