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• 2015 AIME II Problems/Problem 11
  ...circle at a point <math>K</math>. Then, note that <math>\angle{KAC}=\angle{CBK}=\angle{QBO}=90^\circ-\alpha</math>. But since <math>BK</math> is a diamete
  9 KB (1,249 words) - 21:59, 26 November 2023
• 2015 AMC 8 Problems/Problem 21
  ...} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}</math>, so <math>\Delta KBC</math> is a right triangle with le
  3 KB (471 words) - 20:36, 17 January 2024