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- * '''Tangent''': The tangent of angle <math>A</math>, denoted <math>\tan (A)</math>, is defined as the r .../math>, denoted <math>\cot (A)</math>, is defined as the reciprocal of the tangent of <math>A</math>. <cmath>\cot (A) = \frac{1}{\tan (x)} = \frac{\textrm{adj7 KB (1,124 words) - 07:36, 29 September 2021
- ...meter <math>\overline{AB}</math> is constructed inside the square, and the tangent to the semicircle from <math>C</math> intersects side <math>\overline{AD}</ ...h> and <math>C</math> are externally tangent to each other, and internally tangent to circle <math>D</math>. Circles <math>B</math> and <math>C</math> are con13 KB (1,953 words) - 10:01, 6 September 2021
- ...eometry)|tangent]] to the circle at <math>A</math> and <math>\angle AOB = \theta</math>. If point <math>C</math> lies on <math>\overline{OA}</math> and <mat ...}\ \frac{1}{1+\sin\theta} \qquad \textbf {(E)}\ \frac{\sin \theta}{\cos^2 \theta}</math>13 KB (1,948 words) - 13:08, 19 February 2020
- ...h>9</math>, respectively. The equation of a common external [[tangent line|tangent]] to the circles can be written in the form <math>y=mx+b</math> with <math> ...ath> and the x-axis, so <math>m=\tan{2\theta}=\frac{2\tan\theta}{1-\tan^2{\theta}}=\frac{120}{119}</math>. We also know that <math>L_1</math> and <math>L_2<2 KB (251 words) - 11:55, 28 April 2019
- ...rnally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> ...etween their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>. So we have12 KB (2,000 words) - 13:17, 28 December 2020
- Let <math>\angle CAD = \angle BAE = \theta</math>. Note by Law of Sines on <math>\triangle BEA</math> we have <cmath>\frac{BE}{\sin{\theta}} = \frac{AE}{\sin{B}} = \frac{AB}{\sin{\angle BEA}}</cmath>12 KB (1,929 words) - 23:42, 24 December 2020
- ...The circle of radius <math>9</math> has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord. ...ed by faces <math>OAB</math> and <math>OBC.</math> Given that <math>\cos \theta=m+\sqrt{n},</math> where <math>m_{}</math> and <math>n_{}</math> are intege6 KB (1,000 words) - 07:37, 7 September 2018
- ...ersection. By the problem condition, however, the circle <math>P</math> is tangent to <math>BC</math> at point <math>N</math>. ...Moreover, since <math>AD</math> is the only chord, <math>BC</math> must be tangent to the circle <math>P</math>.14 KB (2,516 words) - 22:37, 19 December 2020
- ...frac{\tan\alpha -\tan\beta}{1+\tan\alpha \tan\beta}</math>. Therefore, the tangent of the acute angle between the medians from A and B will be <math>\frac{2-1 By the tangent subtraction identity, <math>\frac{2r-\frac{r}{2}}{1+2r \cdot \frac{r}{2}}=\8 KB (1,319 words) - 13:00, 6 September 2020
- ...1}{10}</math>. Now to find <math>\tan{\theta}</math>, we find <math>\cos{2\theta}</math> using the Pythagorean Identity, and then use the tangent double angle identity. Thus, <math>\tan{\theta} = 10-3\sqrt{11}</math>. Substituting into the original sum,3 KB (537 words) - 17:49, 2 September 2020
- ...</math>, and then we found <math>AP</math>, the segment <math>OB</math> is tangent to the circles with diameters <math>AO,CO</math>. ...a} = 4\cos^3{\theta} - 3\cos{\theta}</math>, and since we have <math>\cos{\theta} = \frac {4}{5}</math>, we can solve for <math>a</math>. The rest then foll8 KB (1,270 words) - 23:25, 30 July 2021
- ...have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> where <math>m_{}</math> an real theta = 29.66115; /* arctan(168/295) to five decimal places .. don't know other w6 KB (978 words) - 22:31, 28 May 2021
- ...The x-axis and the line <math>y = mx</math>, where <math>m > 0</math>, are tangent to both circles. It is given that <math>m</math> can be written in the form ...me positive reals <math>a</math> and <math>b</math>. These two circles are tangent to the <math>x</math>-axis, so the radii of the circles are <math>a</math>5 KB (834 words) - 22:22, 8 July 2021
- ...es that <math>e^{i\theta}=\cos(\theta)+i\sin(\theta)</math> for all <math>\theta</math>. He also discovered the power series for the [[tangent function|arctangent]], which is3 KB (500 words) - 21:28, 15 September 2008
- ...o the extension of [[leg]] <math>CB</math>, and the circles are externally tangent to each other. The length of the radius either circle can be expressed as ...s. As <math>\overline{AF}</math> and <math>\overline{AD}</math> are both [[tangent]]s to the circle, we see that <math>\overline{O_1A}</math> is an [[angle bi11 KB (1,750 words) - 18:39, 15 March 2021
- Consider a cone of revolution with an inscribed sphere tangent to the base of the cone. A cylinder is circumscribed about this sphere so t Now, let <math>\theta</math> be the angle subtended by a diameter of the base of the cone at the7 KB (1,214 words) - 18:49, 29 January 2018
- ...</math> is tangent to the circle at <math>A</math> and <math>\angle AOB = \theta</math>. If point <math>C</math> lies on <math>\overline{OA}</math> and <mat label("$\theta$",(0.1,0.05),ENE);6 KB (978 words) - 11:28, 3 August 2021
- label("\(\sin \theta = \frac{3}{5}\)",B-(.2,-.1),W); ...<math>CD</math> tangent to <math>O_a</math> <math>M</math>, and the point tangent to <math>O_b</math> <math>N</math>. Since <math>\triangle CO_aM</math> and6 KB (951 words) - 16:31, 2 August 2019
- label("\(\theta\)",(7,.4)); Let <math>x = CA</math>. Then <math>\tan\theta = \tan(\angle BAF - \angle DAE)</math>, and since <math>\tan\angle BAF = \f3 KB (513 words) - 14:35, 7 June 2018
- ...<math>T</math> lies on <math>\omega</math> so that line <math>CT</math> is tangent to <math>\omega</math>. Point <math>P</math> is the foot of the perpendicul label("\(\theta\)",C + (-1.7,-0.2), NW);7 KB (1,134 words) - 22:07, 5 October 2020