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Question
In the following figure, ∠ACB = 40º. Find ∠OAB.
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Solution
Given, ∠ACB = 40°
We know that, a segment subtends an angle to the circle is half the angle subtends to the centre.
∴ ∠AOB = 2∠ACB
⇒ `∠ACB = (∠AOB)/2`
⇒ `40^circ = 1/2`∠AOB
⇒ ∠AOB = 80° ...(i) [Both are the radius of a circle]
In ΔAOB, AO = BO
⇒ ∠OBA = ∠OAB ...(ii) [Angles opposite to the equal sides are equal]
We know that, the sum of all three angles in a triangle AOB is 180°.
∴ ∠AOB + ∠OBA + ∠OAB = 180°
⇒ 80° + ∠OAB + ∠OAB = 180° ...[From equations (i) and (ii)]
⇒ 2∠OAB = 180° – 80°
⇒ 2∠OAB = 100°
∴ ∠OAB = `100^circ/2` = 50°
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