Mathematics
In the given figure, O is the centre of the circle. CE is a tangent to the circle at A. If ∠ABD = 26°, then find :
(i) ∠BDA
(ii) ∠BAD
(iii) ∠CAD
(iv) ∠ODB
Circles
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Answer
(i) We know that,
Angle in a semi-circle is a right angle.
∴ ∠BDA = 90°.
Hence, ∠BDA = 90°.
(ii) In △ BAD,
By angle sum property of triangle,
⇒ ∠BDA + ∠BAD + ∠ABD = 180°
⇒ 90° + ∠BAD + 26° = 180°
⇒ ∠BAD + 116° = 180°
⇒ ∠BAD = 180° - 116° = 64°.
Hence, ∠BAD = 64°.
(iii) From figure,
CE is tangent to the circle.
Angle between tangent and radius of the circle is 90°.
From figure,
⇒ ∠CAB = 90°
⇒ ∠CAD + ∠BAD = 90°
⇒ ∠CAD + 64° = 90°
⇒ ∠CAD = 90° - 64° = 26°.
Hence, ∠CAD = 26°.
(iv) In △ OBD,
⇒ OD = OB (Radius of same circle)
We know that,
Angles opposite to equal sides are equal.
⇒ ∠ODB = ∠OBD = 26°.
Hence, ∠ODB = 26°.
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