In Fig.. O is the Center of the Circle and Bcd is Tangent to It at C. Prove that ∠Bac + ∠Acd = 90° - Mathematics

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In fig.. O is the center of the circle and BCD is tangent to it at C. Prove that ∠BAC +
∠ACD = 90°

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Solution

Given

O is center of circle

BCD is tangent.

Required to prove: ∠BAC + ∠ACD = 90°

Proof: OA = OC [radius]

In ΔOAC, angles opposite to equal sides are equal.

∠OAC = ∠OCA …. (i)

∠OCD = 90° [tangent is radius are perpendicular at point of contact]

∠ACD + ∠OCA = 90°

∠ACD + ∠OAC = 90° [∵ ∠OAC = ∠BAC]

∠ACD + ∠BAC = 90° ⟶ Hence proved

 

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Chapter 8: Circles - Exercise 8 [Page 35]

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RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8 | Q 4 | Page 35

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