2007 AMC 12B Problem 3

Attempt Problem 3 of the 2007 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

3.

The point OO is the center of the circle circumscribed about △ABC,\triangle ABC, with ∠BOC=120∘\angle BOC=120^\circ and ∠AOB=140∘,\angle AOB=140^\circ, as shown. What is the degree measure of ∠ABC?\angle ABC?
3535
4040
4545
5050
6060
Answer: D
Concepts:inscribed angleangle sum
Small hint:
The three central angles at OO add up to 360∘360^\circ
Big hint:
The inscribed angle ∠ABC\angle ABC is half of the central angle ∠AOC\angle AOC
Solution:
The angles around OO sum to 360∘,360^\circ, so ∠AOC=360∘−140∘−120∘=100∘. \begin{aligned} \angle AOC&=360^\circ-140^\circ-120^\circ \\ &=100^\circ. \end{aligned}
By the inscribed angle theorem, ∠ABC\angle ABC subtends the same arc ACAC as the central angle ∠AOC,\angle AOC, so ∠ABC=12∠AOC=50∘. \angle ABC=\tfrac12\angle AOC=50^\circ.
Thus, the correct answer is D.
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