Question

If O is the center of the circle and ∠ACB=40, then find ∠OAB.
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Answer 1

Step-by-step explanation:

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Answer 2

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If O is the center of the circle and ∠ACB=40, then find ∠OAB.

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$\huge\green{\tt{\underline{\underline{Answer:-}}}}$

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In ΔOAB,

we have

OA=OB (radii of the same circle).

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∴ΔOAB is isosceles.

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∴∠OAB=∠OBA

⇒ ∠OAB+∠OBA=2∠OAB

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So, ∠AOB+∠OAB+∠OBA

= ∠AOB+2∠OAB=180degree

(by angle sum property of triangles).

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∴∠OAB= half (180 −∠AOB) .(eq. i)

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Now ∠AOB = 2∠ACB = 2×40 = 80degree

(The angle subtended by a chord of a circle at the centre is double of that at the cicumference)

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From (eq. i), we have

∠OAB = 1/2 × ( 180 - 80 )=50degree

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Hence, ∠OAB = 50 degree

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