Answered

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in the figure above

In The Figure Above class=

Sagot :

Since AB is tangent to the circle, the angle BAO equals π/2.

The same happens to BC, so the angle BCO also equals π/2.

Now, for any quadrilateral, the sum of the internal angles is 2π. Therefore:

ABC + AOC + BAO + BCO = 2π

ABC + 3π/7 + π/2 + π/2 = 2π

ABC = 2π - 3π/7 - π/2 - π/2 = π - 3π/7 = (7π - 3π)/7

ABC = 4π/7

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