Search results

Create the page "Tan" on this wiki! See also the search results found.

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

4 KB (686 words) - 18:38, 12 July 2023
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
2022 AMC 12B Problems/Problem 25
<cmath>\frac 14\mathrm{base length}^2\cdot\tan(\mathrm{base angle})=\frac 14((\sqrt3-1)\sqrt2)^2\cdot\tan15^\circ=\frac 14

20 KB (2,980 words) - 18:17, 2 January 2024
2022 AMC 12B Problems/Problem 14
.../math> and the <math>y</math>-axis at point <math>B</math>. What is <math>\tan(\angle ABC)</math>? Thus, <cmath>\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.</c

7 KB (1,013 words) - 22:23, 27 October 2023
Gossard perspector
...\tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.</math> ...e O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.</math>

18 KB (3,046 words) - 06:44, 19 January 2023
2023 AIME I Problems/Problem 8
Which also gives us <math>\tan{\angle{OWX}}=\frac{1}{2}</math> and <math>OW=\frac{25\sqrt{5}}{2}</math>. \frac{OX}{OW}&=\tan{\angle{OWX}} \\

17 KB (2,612 words) - 14:54, 3 July 2023
2023 AIME I Problems/Problem 13
Denote <math>\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}</math>. & = \pm d \cos 2 \alpha \tan \alpha .

13 KB (2,130 words) - 01:52, 31 January 2024
2023 AIME II Problems/Problem 13
.../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> Denote <math>a_n = \sec^n A + \tan^n A</math>.

5 KB (881 words) - 04:52, 19 December 2023
2023 AIME II Problems/Problem 12
<cmath>\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\tan(\angle{QBL_2}-\angle{QCL_2})</cmath> ...}{6}</math>, <math>\tan{CAL_1}=\frac{9}{12}=\frac{3}{4}</math>, and <math>\tan{L_1AB}=\frac{5}{12}</math>. By abusing the tangent angle addition formula,

12 KB (1,900 words) - 18:14, 28 January 2024
2023 AIME II Problems/Problem 14
<cmath>\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC

10 KB (1,574 words) - 16:42, 8 March 2024
2023 AMC 12A Problems
...a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x} </cmath>whenever <math>\tan 2023x</math> is defined. What is <math>a_{2023}?</math>

15 KB (2,168 words) - 05:11, 4 February 2024
2023 AMC 12A Problems/Problem 11
.../math>. <math>\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1</math>. Therefore, <math>\alpha-\beta=\

6 KB (907 words) - 23:48, 18 February 2024
2023 AMC 12A Problems/Problem 25
...a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x} </cmath>whenever <math>\tan 2023x</math> is defined. What is <math>a_{2023}?</math>

6 KB (960 words) - 01:27, 11 November 2023
2023 AMC 12A Problems/Problem 9
...theta</math> be the angle opposite the smaller leg. We want to find <math>\tan\theta</math>. ...ta = \frac{1}{2},</math> or <math>\theta=15^\circ</math>. Therefore <math>\tan \theta = \boxed{\textbf{(C) }2-\sqrt3}</math>

7 KB (1,106 words) - 09:10, 8 March 2024
2023 AMC 12A Problems/Problem 15
&= \frac{100 - \frac{90}{\tan\theta}}{120 - \frac{90}{\sin\theta}}\\

7 KB (1,074 words) - 21:22, 20 November 2023
2023 AMC 12B Problems/Problem 25
...\frac{s}{2} = \frac{s}{4} \cdot (sec^2(54^{\circ})-2) = \frac{s}{4} \cdot (tan^2(54^{\circ})-1)</math>. ...t in a right triangle. Thus, <math>FU = \frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))</math>

19 KB (2,967 words) - 16:56, 24 February 2024
Polya’s method for extremums
...t \sin \alpha = F_B \cdot BC \cdot \cos \alpha \implies \frac {BC}{AC} = \tan^2 \alpha,</cmath> ...h>\frac {y_C}{x_C} = \frac {BC \cdot \sin \alpha}{AC \cdot \cos \alpha} = \tan^3 \alpha \implies</cmath>

12 KB (2,104 words) - 14:11, 24 February 2024
2024 AIME I Problems/Problem 7
...the fourth quadrant (side lengths of <math>3, -4, 5</math>). Since <math>\tan\theta = -\frac{4}{3},</math> we quickly see that <math>\sin\theta = -\dfrac

12 KB (1,842 words) - 19:26, 23 February 2024
Bisector
...{MG}{MF} = \frac {AM \tan \alpha}{BM \tan \gamma} =\frac {\tan \alpha}{\tan \gamma} = \frac {CB''}{AC''}=\frac{a+b-c}{b+c-a}.</cmath> ...^2 \alpha} = \frac {\sin 2\gamma}{\sin 2\alpha} \cdot \frac {\tan \gamma}{\tan \alpha} = \frac {c}{a} \cdot \frac{b+c-a}{a+b-c}.</cmath>

10 KB (1,668 words) - 09:47, 26 March 2024
1991 OIM Problems/Problem 2
<math>A_1=x\frac{y-x.tan(\theta)+y}{2}+\frac{y^2tan(\theta)}{2}</math> <math>A_1=\frac{y^2-x^2}{2}tan(\theta)+xy=1</math>

3 KB (482 words) - 09:40, 23 December 2023
2023 SSMO Relay Round 1 Problems
<cmath>\sec^{N} (Nx) - \tan^{N}(Nx) = 1</cmath>

640 bytes (94 words) - 21:32, 15 December 2023
2023 SSMO Relay Round 1 Problems/Problem 3
<cmath>\sec^{N} (Nx) - \tan^{N}(Nx) = 1</cmath>

180 bytes (31 words) - 22:28, 15 December 2023
2024 AIME I Problems/Problem 5
Let <math>\angle{DHE} = \theta.</math> This means that <math>DE = 17\tan{\theta}.</math> Since quadrilateral <math>ADHG</math> is cyclic, <math>\ang ...a} = \frac{16}{DX} = \frac{17}{FX}</math> and <math>DX + FX = DE + EF = 17\tan{\theta} + 184.</math>

7 KB (1,144 words) - 17:43, 22 April 2024
Proofs of trig identities
<math>\tan = \frac{\sin}{\cos}</math> pair D = (1,tan(d));

16 KB (2,796 words) - 13:12, 21 January 2024
2024 AIME II Problems/Problem 15
Case 1: <math>k = 0, \tan 30^\circ, \tan 60^\circ</math>. ...th>. The number of solutions for <math>k = \tan 30^\circ</math> and <math>\tan 60^\circ</math> are the same.

8 KB (1,395 words) - 17:26, 9 February 2024
DVI exam
<cmath>\tan \alpha = \frac {SO}{OH} = \frac {GO}{QH} \implies OH \cdot GO = \frac {15} <cmath>h = 5 \sqrt {\frac {3}{2}} \implies \tan \alpha = \sqrt{2} \implies \cos \alpha = \frac {1}{\sqrt{3}}.</cmath>

32 KB (5,375 words) - 11:43, 5 May 2024
2024 AIME II Problems/Problem 12
...=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math> ...math>\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{x}=\frac{1}{8}</math>. This mean

10 KB (1,655 words) - 00:31, 11 April 2024
Sharygin Olympiads, the best
<cmath>\angle CBO = 90^\circ \implies \angle COB = \alpha, MB = b \tan \alpha,</cmath> Denote <math>Q = AB \cap DP \implies BQ = \frac {BP}{\cos \alpha} = b \tan \alpha = MB.</math>

23 KB (4,003 words) - 16:17, 21 April 2024
2005 iTest Problems
...ath> equal the correct answer from 3A. In triangle <math>WXY</math>, <math>tan \angle YWX= (A + 8) / .5A</math>, and the altitude from <math>W</math> divi

25 KB (3,738 words) - 10:53, 23 April 2024

Page

Toolbox

Search

Search results