AMC 10B 2025 (Problem 5)In △ABC\triangle ABC△ABC, AB=10AB=10AB=10, AC=18AC=18AC=18, and ∠B=130∘\angle B=130^\circ∠B=130∘. Let OOO be the center of the circle containing AAA, BBB, and CCC. What is the degree measure of ∠CAO\angle CAO∠CAO?(A) 20\text{(A)}\;20(A)20(B) 30\text{(B)}\;30(B)30(C) 40\text{(C)}\;40(C)40(D) 50\text{(D)}\;50(D)50(E) 60\text{(E)}\;60(E)60Related TopicsCoreAngles in CirclesCheck AnswerYour answer:ABCDECheckHints (5)Hint 1Hint 2Hint 3Find ∠AOC\angle AOC∠AOC.Hint 4∠ABC=130∘\angle ABC=130^\circ∠ABC=130∘∠ABC\angle ABC∠ABC is an inscribed angle.By Angles in Circles:ADC^=2⋅130∘=260∘\widehat{ADC}=2\cdot130^\circ=260^\circADC=2⋅130∘=260∘⇒ABC^=360∘−260∘=100∘\Rightarrow \widehat{ABC}=360^\circ-260^\circ=100^\circ⇒ABC=360∘−260∘=100∘∠AOC\angle AOC∠AOC is a central angle.By Angles in Circles:∠AOC=ABC^=100∘\angle AOC=\widehat{ABC}=100^\circ∠AOC=ABC=100∘Hint 5△OAC\triangle OAC△OAC is an isosceles triangle.⇒∠CAO=180∘−100∘2=40∘\Rightarrow \angle CAO=\frac{180^\circ-100^\circ}{2}=40^\circ⇒∠CAO=2180∘−100∘=40∘Final Answer(C) 404040
