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• 2007 AMC 12B Problems/Problem 3
  <math>\angle AOC = 100^{\circ} \implies \angle ABC =\frac{\angle AOC}{2} =50^{ \circ},</math> or <math>\mathrm{(D)}.</math>
  905 bytes (130 words) - 10:39, 27 February 2022
• Isogonal conjugate
  <cmath>= \frac{180^\circ - \angle AOC}{2} = \frac{180^\circ - 2\angle ABC}{2} = \angle BAH.</cmath>
  54 KB (9,416 words) - 08:40, 18 April 2024
• 2008 AMC 12B Problems
  ...are on the same side of diameter <math>\overline{AB}</math>, <math>\angle AOC = 30^\circ</math>, and <math>\angle DOB = 45^\circ</math>. What is the rati
  14 KB (2,199 words) - 13:43, 28 August 2020
• 2008 AMC 12B Problems/Problem 4
  ...are on the same side of diameter <math>\overline{AB}</math>, <math>\angle AOC = 30^\circ</math>, and <math>\angle DOB = 45^\circ</math>. What is the rati <math>\angle COD = \angle AOB - \angle AOC - \angle BOD = 180^\circ - 30^\circ - 45^\circ = 105^\circ</math>.
  1 KB (183 words) - 22:35, 10 June 2017
• Spiral similarity
  <math>\triangle AOC = \triangle BOC \sim \triangle AXY \sim \triangle BZY \implies </math> <cmath>\angle AOC = \angle AO'C' = 2 \alpha \implies</cmath>
  28 KB (4,863 words) - 00:29, 16 December 2023
• 1995 USAMO Problems/Problem 3
  ...ngle B</math> which equals <math>\angle AOC</math>. Since <math>\triangle AOC</math> is isosceles we have that <math>\angle OAC = \angle OCA = 90 - \angl
  2 KB (364 words) - 01:42, 19 April 2024
• 2009 AMC 10B Problems/Problem 16
  ..., hence <math>\angle OAC = \angle OCA = 30^\circ</math>. Then <math>\angle AOC = 120^\circ</math>, and from symmetry we have <math>\angle AOB = \angle COB
  4 KB (576 words) - 19:59, 25 November 2023
• 2010 AMC 10B Problems/Problem 6
  ...ngle AOC = 180^\circ - 50^\circ = 130^\circ</math>. Because triangle <math>AOC</math> is isosceles, <math>\angle CAB = (180^\circ - 130^\circ)/2 = \boxed{
  2 KB (260 words) - 17:00, 1 August 2022
• Mock Geometry AIME 2011 Problems
  ...o <math>OB,OC,</math> and arc <math>BC</math>. It is known that <math>\tan AOC=\frac{24}{7}</math>. The ratio <math>\frac{r_2} {r_1}</math> can be express
  8 KB (1,349 words) - 19:10, 14 June 2022
• 2007 AMC 10B Problems/Problem 4
  \angle BOC + \angle AOB + \angle AOC &= 360\\ 120 + 140 + \angle AOC &= 360\\
  928 bytes (140 words) - 12:19, 4 July 2013
• Projective geometry
  &= \frac{\sin(\angle{AOC})\sin(\angle{BOD})}{\sin(\angle{AOD})\sin(\angle{BOC})},
  4 KB (720 words) - 15:03, 26 January 2015
• 2002 AMC 10B Problems/Problem 24
  ...\triangle OBC</math> by SAS, so <math>AC</math> is 20, and <math>\triangle AOC</math> is equilateral, and <math>\angle BOC = 60^\circ</math>
  3 KB (559 words) - 02:44, 8 February 2024
• Mock Geometry AIME 2011 Problems/Problem 11
  ...o <math>OB,OC,</math> and arc <math>BC</math>. It is known that <math>\tan AOC=\frac{24}{7}</math>. The ratio <math>\frac{r_2} {r_1}</math> can be express ...th> and <math> OC </math>, so it is on the angle bisector of <math> \angle AOC </math>.
  3 KB (432 words) - 14:12, 2 January 2012
• 2013 AIME II Problems/Problem 10
  <cmath>\triangle ABD \sim \triangle AOC \implies \frac {h}{H} = \frac {4}{4 + \sqrt{13}}.</cmath>
  5 KB (823 words) - 22:42, 15 June 2024
• 2014 AIME I Problems/Problem 7
  ...to maximize <math>\tan^2(\theta)</math>, we need to maximize <math>\angle{AOC}</math>. This angle is maximized when <math>AO</math> is tangent to <math>\
  5 KB (782 words) - 20:25, 10 October 2023
• 2015 AIME II Problems/Problem 15
  ...}{5}</math> and <math>\frac{2}{5}</math> respectively. Assign <math>\angle AOC=a,\angle COE=c,\angle DPB=d,\angle BPA=b.</math> We can figure out that <ma <cmath>[ACE]=[AOC]+[COE]-[AOC]=\dfrac{32}{5}-8\sin c-8\sin \theta,</cmath>
  31 KB (5,086 words) - 19:15, 20 December 2023
• 2017 USAJMO Problems/Problem 5
  ...ot AO=AN'\cdot AD.</cmath> Notice that <math>\triangle ABA' \sim \triangle AOC</math>, so that <cmath>\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=
  6 KB (968 words) - 12:05, 7 June 2024
• 1971 AHSME Problems/Problem 31
  ...e of that angle is 7/8. We use the double cosine angle to find angle <math>AOC</math>:
  3 KB (425 words) - 10:28, 18 August 2022
• 2008 iTest Problems/Problem 56
  ...{5})(\frac{4}{5})</math> or <math>\frac{24}{25}</math>. Angle <math>\angle AOC = 2\pi - 2\theta</math>. Let <math>E</math> be the altitude of <math>AOE</m
  7 KB (1,092 words) - 19:05, 17 December 2021
• 1973 AHSME Problems/Problem 15
  ...math>C</math> to <math>O</math>. By HL Congruency and CPCTC, <math>\angle AOC = \angle BOC = \theta /2</math>.
  2 KB (306 words) - 14:00, 20 February 2020

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