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• Mock AIME 2 Pre 2005 Problems/Problem 11
  ...lpha) + \tan^{-1}(\beta) + \tan^{-1} (\gamma).</cmath> The value of <math>\tan(\omega)</math> can be written as <math>\tfrac{m}{n}</math> where <math>m</m ...ma - \gamma\alpha}</math>. Substituting Vieta's formulas, we obtain <math>\tan(\omega) = \frac{\frac{799}{4} - \frac{1}{4}}{1 - (-50)} = \frac{\frac{798}{
  1 KB (192 words) - 18:03, 13 September 2020
• 2023 AIME II Problems
  .../math> less than or equal to <math>1000</math> such that <math>\sec^n A + \tan^n A</math> is a positive integer whose units digit is <math>9.</math>
  8 KB (1,370 words) - 21:34, 28 January 2024
• 2022 AMC 12B Problems
  .../math> and the <math>y</math>-axis at point <math>B</math>. What is <math>\tan(\angle ABC)</math>?
  15 KB (2,233 words) - 13:02, 10 November 2023
• 2021 Fall AMC 10B Problems/Problem 25
  Hence, <math>\tan \theta = \frac{1}{3}</math>.
  16 KB (2,274 words) - 09:02, 10 December 2022
• 2021 AIME II Problems
  ...the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where <
  8 KB (1,429 words) - 14:31, 26 February 2024
• 2021 AMC 10A Problems/Problem 24
  Let <math>\tan A=a.</math> The area of the rectangle created by the four equations can be
  10 KB (1,662 words) - 19:31, 18 November 2022
• 1978 AHSME Problems/Problem 15
  If <math>\sin x+\cos x=1/5</math> and <math>0\le x<\pi</math>, then <math>\tan x</math> is Since <math>\tan x = \frac{\sin x}{\cos x}</math>, we have
  1 KB (207 words) - 21:10, 13 February 2021
• 2021 AIME II Problems/Problem 12
  ...the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where < ...and <math>\cos \theta</math> separately and use their values to get <math>\tan \theta</math>. We can start by drawing a diagram. Let the vertices of the q
  10 KB (1,669 words) - 17:33, 12 January 2024
• 2022 AIME II Problems/Problem 15
  <cmath>\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.</ The radius of the inscribed circle is <math>r = (s – B'D) \tan \alpha.</math>
  14 KB (2,217 words) - 00:28, 29 June 2023
• 2021 Fall AMC 12B Problems/Problem 15
  ...^\circ</math>, so <math>\angle{EAB} = 30^{\circ}</math>. Then <math>EB=AE\tan 30^\circ = \sqrt{3}</math>; therefore <math>BC=EC-EB=3-\sqrt{3}</math>. Thu line A'B' <math> \frac{y-(1/2 + \sqrt{3}/2)}{x - ( \sqrt{3}/2 - 1/2)} = Tan(90\circ+30\circ) = -\sqrt{3} </math>
  5 KB (805 words) - 01:45, 22 May 2024
• 2021 Fall AMC 12B Problems/Problem 22
  ...d <math>MP\perp AB</math>. Let <math>\theta = \angle BAC</math>; so <math>\tan\theta = \tfrac 68 = \tfrac 34</math>. Then <cmath>OP=MP-MO=AM\cot\theta - BM\tan\theta = 5(\tfrac 43 - \tfrac 34) = \boxed{\textbf{(C)}\ \tfrac{35}{12}}.</c
  6 KB (943 words) - 00:41, 6 August 2023
• 2021 Fall AMC 12B Problems/Problem 9
  ...erp</math> <math>\overline{AB}</math> and <math>AY = 3</math>. Thus <math>\tan(30^\circ) = \frac{OY}{3}</math> so <math>OY = \sqrt{3}</math>.
  5 KB (765 words) - 22:33, 10 July 2023
• 2021 Fall AMC 10A Problems/Problem 19
  ~MathFun1000 (Inspired by Way Tan)
  3 KB (475 words) - 17:45, 30 October 2023
• 2021 Fall AMC 10B Problems/Problem 11
  ...could find the length of the apothem by the formula <math>\frac{s}{2\text{tan}(\frac{180}{n})},</math> where <math>s</math> is the side length and <math>
  6 KB (1,021 words) - 19:40, 1 November 2023
• 2021 Fall AMC 10B Problems/Problem 17
  ...th the positive x axis. Thus, line <math>m</math> makes an angle of <math>\tan^{-1}(5) - 45^\circ</math> with the positive x axis. Thus, the slope of line <cmath> \tan (\tan^{-1}(5) - 45^\circ) = \frac{5 - 1}{1 + 5\cdot 1} = \frac{2}{3},</cmath>
  8 KB (1,331 words) - 22:44, 16 December 2023
• 2021 Fall AMC 12A Problems/Problem 13
  ...th>\theta_{3}</math>. Note that <math>\tan(\theta_{1})=1</math> and <math>\tan(\theta_{3})=3</math>, and this is why we named them as such. Let the angle \tan(\theta_k-\theta_1)&=\tan(\theta_3-\theta_k)\\
  16 KB (2,526 words) - 00:53, 6 May 2023
• 2023 AIME I Problems/Problem 5
  ...rom <math>P</math> to line <math>\overline{BD}</math>. We know that <math>\tan{\angle{POX}} = \frac{PX}{XO} = \frac{28}{45}</math> by triangle area and gi \frac{(2 \cdot \tan{\angle{OCP}})}{(1-\tan^2{\angle{OCP}})} = \tan{\angle{POX}} = \frac{28}{45}
  19 KB (3,107 words) - 23:31, 17 January 2024
• 2023 AIME I Problems
  ...erty that <cmath>\angle AEP = \angle BFP = \angle CDP.</cmath> Find <math>\tan^2(\angle AEP).</math>
  7 KB (1,154 words) - 12:54, 20 February 2024
• Kimberling’s point X(24)
  <cmath>HB' = 2 HD \sin (\alpha - 90^\circ) = - 2 CD \tan(90^\circ- \beta) \cos \alpha = - 2 AC \cos \gamma \frac {\cos \beta}{\sin \ <cmath>LM = \frac {R}{2} (\tan \beta + \tan \gamma) = \frac {R \sin (\beta + \gamma)}{2 \cos \beta \cdot \cos \gamma} \
  10 KB (1,751 words) - 15:34, 25 November 2022
• 2022 AMC 10B Problems/Problem 20
  By solving this equation, we get <math>\tan \angle FBC = \frac{\sin 46^\circ}{2 + \cos 46^\circ}</math>. \tan \angle BFC & = \frac{\sin \angle FBC}{\cos \left( 46^\circ - \angle FBC \ri
  11 KB (1,742 words) - 20:06, 8 September 2023

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