Question

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Answer 1

Answer:

JOIN AC

LET ∠OAC = ∠COA = X

∴ ∠AOC = 180 - 2X

ALSO,

∠BAC = 30 + X

IN ΔABC

∠ABC = 180 - ∠BAC - ∠BCA

180 - [ 30 + X ] - [ 40 + X ]

⇒ 110 - 2X

NOW ,

∠AOC  = 2∠ABC

[ ANGLE AT THE CENTRE IS DOUBLE THE ANGLE AT THE CIRCUMFERENCE SUBTENDED BY THE SAME CHORD ]

180 - 2X = 2[ 110 - 2X ]

2X = 40

X = 40/20

X = 20

∴∠AOC = 180 - 2 × 20 = 140

Step-by-step explanation:

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