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प्रश्न
In the given figure O, is the centre of the circle. CE is a tangent to the circle at A. If ∠ABD = 26° find:
- ∠BDA
- ∠BAD
- ∠CAD
- ∠ODB
उत्तर
Given ∠ABD = 26°
a. Angle in semi-circle is a right angle
∴ ∠BDA = 90°
b. Sum of all angles in a triangle = 180°
∴ ∠BAD + ∠BDA + ∠ABD = 180°
∠BAD + 90° + 26° = 180°
∠BAD = 180° – 116°
= 64°
c. Angles in alternate segments are equal
∴ ∠CAD = ∠ABD
= 26°
d. In ΔODB,
OB = OD ...(Radii of same circle)
∴ ∠ODB = ∠OBD
= ∠ABD
= 26°
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