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Question
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
Solution
£ADE = 95°(Given) Since,
OA = OB, so
∠OAB = ∠OBA
∠OAB = 30°
∠ADE + ∠ADC = 180°
(Linear pair)
95° + ∠ADC = 180°
∠ADC = 85°
We know that,
∠ADC = 2∠ADC
∠ADC = 2 x 85°
∠ADC = 170°
Since,
AO = OC
(Radii of circle)
∠OAC = ∠OCA
(Sides opposite to equal angle) ... (i)
In triangle OAC,
∠OAC + ∠OCA + ∠AOC = 180°
∠OAC + ∠OAC + 170° = 180°
[From (i)]
2∠OAC = 10°
∠OAC = 5°
Thus,
∠OAC = 5°
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