Search results
Create the page "Tangent theta" on this wiki! See also the search results found.
- * '''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{adj8 KB (1,217 words) - 20:15, 7 September 2023
- ...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) - 00:31, 26 January 2023
- ...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) - 12:26, 1 April 2022
- ...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 (253 words) - 22:52, 29 December 2021
- ...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>13 KB (2,129 words) - 18:56, 1 January 2024
- ...triangle]] because <math>OB</math> is a radius and <math>BA</math> is a [[tangent line]] at point <math>B</math>. We use the [[Pythagorean Theorem]] to find ...ave that <math>\cos\theta=\frac{1}{5}</math> which implies that <math>\sin\theta=\frac{2\sqrt{6}}{5}.</math> Note that the portion of rope not on the tower4 KB (729 words) - 01:00, 27 November 2022
- ...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) - 00:25, 27 March 2024
- ...el to <math>BC</math>. This means that <math>BC</math> is parallel to the tangent to the given circle at <math>D</math>. ...ersection. By the problem condition, however, the circle <math>P</math> is tangent to <math>BC</math> at point <math>N</math>.19 KB (3,221 words) - 01:05, 7 February 2023
- ...math>. Then by the [[Trigonometric_identities#Angle_addition_identities | tangent angle subtraction formula]], <cmath> \tan \theta = \tan (\theta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \ta11 KB (1,722 words) - 09:49, 13 September 2023
- ...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,5 KB (838 words) - 18:05, 19 February 2022
- ...</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) - 23:36, 27 August 2023
- ...Trigonometry#Tangent|tangent]] function is <math>180^\circ</math>, and the tangent function is [[one-to-one]] over each period of its domain. ...math>. We can set <math>\alpha (\cos{96^{\circ}}+\sin{96^{\circ}}) = \sin{\theta}</math>.Note that if we have <math>\alpha</math> equal to both the sine and4 KB (503 words) - 15:46, 3 August 2022
- ...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 w7 KB (1,184 words) - 13:25, 22 December 2022
- ...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>7 KB (1,182 words) - 09:56, 7 February 2022
- ...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 (1,851 words) - 12:31, 21 December 2021
- var theta=15; ...tension(A,dir(75),B/2,bisectorpoint(A,B)), Cp=rotate(theta,A)*C, Bp=rotate(theta,A)*B, X=extension(A,Bp,B,C), Y=extension(B,C,Bp,Cp);10 KB (1,458 words) - 20:50, 3 November 2023
- 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 (979 words) - 12:50, 17 July 2022
- Let <math>\theta = \angle ACP = \angle BCQ, \Theta = \angle ACQ = \angle BCP.</math> ...= \frac {PC \sin \theta}{PC \sin \Theta} = \frac {QC \sin \theta}{QC \sin \Theta} = \frac {QD'}{QE'}. \blacksquare</cmath>54 KB (9,416 words) - 08:40, 18 April 2024
- 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);8 KB (1,333 words) - 00:18, 1 February 2024
- .... Circle <math>Q</math> is externally [[tangent]] to <math>P</math> and is tangent to <math>\overline{AB}</math> and <math>\overline{BC}</math>. No point of c ...t <math>\angle ACB = 2\theta</math>; then <math>\angle PCX = \angle QBX = \theta</math>. Dropping the altitude from <math>A</math> to <math>BC</math>, we re6 KB (1,065 words) - 20:12, 9 August 2022
- ...nd <math>3 \theta = \angle CAB</math>. Then, it is given that <math>\cos 2\theta = \frac{AC}{AD} = \frac{2}{3}</math> and ...tan 3\theta - \tan 2\theta)}{AC \tan 2\theta} = \frac{\tan 3\theta}{\tan 2\theta} - 1.</math></center>4 KB (662 words) - 00:51, 3 October 2023
- ...sible values of <math>\tan\theta</math> given that <math>\sin\theta + \cos\theta = \dfrac{193}{137}</math>. If <math>a+b=m/n</math>, where <math>m</math> an ...ists a circle, lying inside the quadrilateral and having center I, that is tangent to all four sides of the quadrilateral. Points <math>M</math> and <math>N</71 KB (11,749 words) - 01:31, 2 November 2023
- ...ath> is any point of <math>AB, \theta </math> is circle <math>CPx_0, C' = \theta \cap A'B'</math> is the image <math>C</math> under spiral symilarity center ...x_0 = \Omega \cap \omega, x_0 \neq B, C </math> is any point of <math>AB, \theta </math> is circle <math>CBx_0,</math>28 KB (4,863 words) - 00:29, 16 December 2023
- ...ough <math>A</math>, <math>B</math>, <math>C</math>, and <math>P</math> is tangent to the sphere through <math>A'</math>, <math>B'</math>, <math>C'</math>, an ...e that pass through <math>P</math> for the <math>2</math> spheres that are tangent to each other.5 KB (807 words) - 18:37, 25 June 2021
- ...\angle DAC</math>. Notice <math>\tan \theta = BD</math> and <math>\tan 2 \theta = 2</math>. By the double angle identity, <cmath>2 = \frac{2 BD}{1 - BD^2} Remarks: You could also use tangent half angle formula2 KB (371 words) - 15:34, 15 October 2023
- ...ch that <math>\overline{AI}</math> and <math>\overline{BI}</math> are both tangent to circle <math>\omega</math>. The ratio of the perimeter of <math>\triangl ...}}</math>. By the double-angle formula <math>\sin(2\theta)=2\sin\theta\cos\theta</math>, it turns out that <cmath>\sin(\angle BAI)=\sin(2\angle DAO)=\dfrac{12 KB (1,970 words) - 22:53, 22 January 2024
- ...le, which is itself centered on <math>(1,0)</math>. Let us define <math>f(\theta)</math> as the value of the length of the first circle that lies within the <cmath>f(\theta)=\frac{\theta}{2\pi}\cdot2\pi=\theta</cmath>6 KB (1,105 words) - 13:39, 9 January 2024
- ...ath>\angle ABC=180^{\circ}-\alpha</math> and <math>\angle BCD=180^{\circ}-\theta</math>. Let the circle have center <math>O</math> and radius <math>r</math> ...\theta</math>, <math>GOH=180^{\circ}-\alpha</math>, <math>EOH=180^{\circ}-\theta</math>, and <math>FOE=\alpha</math>.4 KB (753 words) - 18:58, 2 June 2022
- ...ven points are such that (i) lines <math>PB</math> and <math>PD</math> are tangent to <math>\omega</math>, (ii) <math>P</math>, <math>A</math>, <math>C</math> Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD.4 KB (717 words) - 17:00, 14 April 2024
- ...nnot be on <math>p</math>. This implies that <math>q</math> is exactly the tangent line to <math>p</math> at <math>P</math>, that is <math>q=\ell(P)</math>. S ...h> be the reflection of <math>F</math> across <math>q</math>. Then <math>2\theta=\angle FBH=\angle C'HB</math>, and so <math>\angle C'HB=\angle AHB</math>.15 KB (2,593 words) - 13:37, 29 January 2021
- ...math>OA=\sqrt{18}</math>, <math>\angle OXA = 90</math> because the line is tangent to the circle. Using the pythagorean theorem, we have <math>OX=\sqrt{12}</m ...th>\tan (\theta + 45)</math> where <math>\angle AOX = \theta</math>. Using tangent addition,4 KB (614 words) - 20:09, 12 September 2022
- ...the equator, then <math>C=(cos(\theta),sin(\theta),0)</math> where <math>\theta</math> is the angle on the <math>xy</math>-plane from the origin to <math>C ...CN}}</math> is the unit vector in the direction of arc <math>CN</math> and tangent to the great circle of <math>CN</math> at <math>C</math>6 KB (1,013 words) - 22:09, 21 November 2023
- ...misses the circle altogether. This means <math>3x + 4y = A</math> will be tangent to the circle. ...will be <math>\frac{4}{3}</math>, since the radius is perpendicular to the tangent line.9 KB (1,441 words) - 17:51, 22 October 2023
- The incircle contains the tangent points of the incircle with the sides: Denote <math>R = PQ \cap P'Q', \theta = \odot P'QR, F = \Omega \cap \theta \notin \odot PQ'R, D \in \Omega</math> is the point isogonal conjugate to l25 KB (5,067 words) - 22:15, 31 March 2024
- ...e has a circle inscribed within the circle so that the inscribed circle is tangent to all <math>3</math> sides. What is the ratio of the sum of the areas of a ...ouches the origin and has its center located on the y-axis. The circle is tangent to the line <math>x + 3y = 6</math>. Given that the radius of the circle c15 KB (2,444 words) - 21:46, 1 January 2012
- ...e the acute angle formed by that side and the median. What is <math>\sin{\theta}</math>? A square region <math>ABCD</math> is externally tangent to the circle with equation <math>x^2+y^2=1</math> at the point <math>(0,1)14 KB (2,197 words) - 13:34, 12 August 2020
- ...theta </math> with the positive <math> x </math> axis. Compute <math> \cos\theta </math>.6 KB (910 words) - 17:32, 27 May 2012
- <math>\textbf{(A) } \sin^2\theta\qquad \textbf{(B) } \cos^2\theta\qquad16 KB (2,451 words) - 04:27, 6 September 2021
- We construct a tangent to the the circle <math>\odot DEF,</math> at the point <math>G.</math> It i ...math> and <math>E.</math> Line <math>\ell</math> is the tangent for <math>\theta</math> at the point <math>I.</math>15 KB (2,549 words) - 08:36, 2 September 2023
- ...n\theta\qquad\textbf{(E)}\ \dfrac{5}{2}-\dfrac{1}{2}\sin\left(\dfrac{1}{2}\theta\right) </math> ...ond circle is tangent internally to the circumcircle at <math>T</math> and tangent to sides <math>AB</math> and <math>AC</math> at points <math>P</math> and <17 KB (2,633 words) - 15:44, 16 September 2023
- ...\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> an ...radius <math>5</math>. Let <math>A</math> be the point where the disks are tangent, <math>C</math> be the center of the smaller disk, and <math>E</math> be th9 KB (1,472 words) - 13:59, 30 November 2021
- ...\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> an We know that <math>\tan{\theta}</math> is equal to the imaginary part of the above expression divided by t5 KB (782 words) - 20:25, 10 October 2023
- ...,C</math> and <math>D</math>. If <math>AB=1</math> and <math>\angle{APB}=2\theta</math>, then the volume of the pyramid is <math>\textbf{(A) } \frac{\sin(\theta)}{6}\qquad14 KB (2,099 words) - 01:15, 10 September 2021
- In the configuration below, <math>\theta</math> is measured in radians, <math>C</math> is the center of the circle, ...h>BCD</math> and <math>ACE</math> are line segments and <math>AB</math> is tangent to the circle at <math>A</math>.17 KB (2,512 words) - 18:30, 12 October 2023
- ...n acute angle, and <math>\sin 2\theta=a</math>, then <math>\sin\theta+\cos\theta</math> equals In the adjoining figure, <math>AB</math> is tangent at <math>A</math> to the circle with center <math>O</math>; point <math>D</17 KB (2,835 words) - 14:36, 8 September 2021
- In the configuration below, <math>\theta</math> is measured in radians, <math>C</math> is the center of the circle, ...h>BCD</math> and <math>ACE</math> are line segments and <math>AB</math> is tangent to the circle at <math>A</math>.2 KB (301 words) - 18:50, 1 April 2018
- ...rc is the argument of <math>z</math>), because the <math>y</math>- axis is tangent to the circle at the origin. So <math>\text{arg}(z-1)=\frac{\pi+1}{2}</math We want the argument of the whole expression <math>-\pi<\theta<0</math>. This translates into <math>\frac{-\pi-1}{2}<\text{arg}\left(z^n-16 KB (1,034 words) - 21:29, 14 January 2024
- ...<math>\mathcal{Q}</math> so that line <math>BC</math> is a common external tangent of the two circles. A line <math>\ell</math> through <math>A</math> interse ...be the intersection of <math>\overline{BC}</math> and the common internal tangent of <math>\mathcal P</math> and <math>\mathcal Q.</math> We claim that <math31 KB (5,086 words) - 19:15, 20 December 2023
- ...ect at points <math>X</math> and <math>Y</math>. Line <math>\ell</math> is tangent to <math>\omega_1</math> and <math>\omega_2</math> at <math>A</math> and <m pair a=tangent(q,o1,R1,2);14 KB (2,427 words) - 17:12, 8 January 2024
- ...n(\arctan(x)) = \tan(\theta)</math>, and simplifying gives <math>x = \tan(\theta)</math>. So <math>a = \tan(\theta_a) = \frac{1}{2}</math> and <math>b = \ta ...given two tangent measures, it is natural for us to think about the sum of tangent measures (what else can we try? Remember: we are not allowed to use calcula2 KB (363 words) - 12:46, 10 May 2022
- ...er such that <math>\frac{2Z}{\overline{Z}i}</math> has an argument <math>(\theta)</math> equal to <math>\frac{3\pi}{4}</math> and <math>\log_3(2Z+2\overline ...ning of the movement of <math>C'</math> the point <math>P</math> is at the tangent point <math>(4,0)</math>, like in the figure a. After some movement, the an7 KB (1,127 words) - 18:23, 11 January 2018
- ...mcenter and incenter, respectively. A circle with center <math>M</math> is tangent to the legs <math>AC</math> and <math>BC</math> and to the circumcircle of ...that <math>P_1P_2=P_2P_3=P_3P_1</math> and line <math>P_iP_{i+1}</math> is tangent to <math>\omega_i</math> for each <math>i=1,2,3</math>, where <math>P_4 = P14 KB (2,118 words) - 15:36, 28 October 2021
- ...that <math>P_1P_2=P_2P_3=P_3P_1</math> and line <math>P_iP_{i+1}</math> is tangent to <math>\omega_i</math> for each <math>i=1,2,3</math>, where <math>P_4 = P real theta = 41.5;13 KB (2,080 words) - 19:09, 21 October 2023
- ...ath> \sin \frac12 \theta = \sqrt{\frac{x-1}{2x}}</math>, then <math> \tan \theta</math> equals .../math> is acute and <math>\cos \tfrac{\theta}{2} = \sqrt{1 - \sin (\tfrac{\theta}{2})^2}</math>,1 KB (184 words) - 14:00, 20 February 2020
- ...sible values of <math>\tan\theta</math> given that <math>\sin\theta + \cos\theta = \dfrac{193}{137}</math>. If <math>a+b=m/n</math>, where <math>m</math> an <cmath>\frac{\sqrt{2}}{2}\sin\theta + \frac{\sqrt{2}}{2}\cos\theta = \frac{\sqrt{2}}{2}x</cmath>2 KB (343 words) - 20:35, 4 August 2018
- ...f a line connecting a point on the circle to the origin. That line must be tangent to the circle. ...Identity yields <math>\tan 2\theta = \frac34</math>, so <math>\tan (90 - 2\theta) = \frac43</math>.4 KB (722 words) - 20:53, 27 March 2019
- ...le of <math>3</math>, and negative otherwise. The degree measure of <math>\theta</math> is <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math ....</math> Circle <math>\omega_2</math> passes through <math>C</math> and is tangent to line <math>AB</math> at <math>A.</math> Let <math>K</math> be the inters7 KB (1,254 words) - 14:45, 21 August 2023
- ...nt to <math>\omega</math>, and the other two excircles are both externally tangent to <math>\omega</math>. Find the minimum possible value of the perimeter of ...C=a</math>, <math>AB=b</math>, <math>s</math> be the semiperimeter, <math>\theta=\angle ABC</math>, and <math>r</math> be the inradius. Intuition tells us t21 KB (3,915 words) - 19:55, 10 October 2023
- ...theta </math> with the positive <math> x </math> axis. Compute <math> \cos\theta </math>. ...gin has an equation of <math>y=mx</math>. If the line <math>y=mx</math> is tangent to the hyperbola than the equation <math>(mx)^2=x^2-x+1</math> will have on1 KB (261 words) - 18:38, 14 January 2020
- ...e intersection of the altitudes of <math>\triangle ABC.</math> Suppose the tangent to the circumcircle of <math>\triangle HBC</math> at <math>H</math> interse var phi=75.5, theta=130, r=4.8;16 KB (2,678 words) - 22:45, 27 November 2023
- Find the slope of the tangent at the point of inflection of <math>y = x^3 - 3x^2 + 6x + 2000</math>. An envelope of a set of lines is a curve tangent to all of them. What is the envelope of the family of lines y = <math>\frac3 KB (413 words) - 13:10, 21 January 2020
- ...triangle <math>\bigtriangleup ADM</math>. Let <math>\measuredangle AMD = \theta</math>. ...ega</math>. Therefore, <math>\measuredangle POM = 180 - 2(90 - \theta) = 2\theta</math>.9 KB (1,328 words) - 16:14, 11 September 2023
- ...<math>(-5,20)</math>, and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. There ...ection of the ordinate axis <math>2\theta,</math> its length <math>\rho_1(\theta),</math> the angle between the vector connecting the focus of the second pa10 KB (1,742 words) - 02:31, 13 November 2023
- ...AB = AC = 3\sqrt6</math>, and a circle with radius <math>5\sqrt2</math> is tangent to line <math>AB</math> at <math>B</math> and to line <math>AC</math> at <m ...th>\odot O_1</math> be the circle with radius <math>5\sqrt2</math> that is tangent to <math>\overleftrightarrow{AB}</math> at <math>B</math> and to <math>\ove7 KB (1,026 words) - 21:23, 15 June 2023
- ...h>\ell</math> to the point where the sphere with radius <math>13</math> is tangent to plane <math>\mathcal{P}</math> is <math>\tfrac{m}{n}</math>, where <math ...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 <math>m<8 KB (1,429 words) - 14:31, 26 February 2024
- ...nd perpendicular lines have slopes negative inverses of each other). Using tangent double angle formula, the slope of <math>\overline{EC'}</math> is <math>\fr8 KB (1,252 words) - 23:49, 3 March 2024
- ...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 <math>m< ...</math>, <math>CX = c</math>, and <math>DX = d</math>. We know that <math>\theta</math> is the acute angle formed between the intersection of the diagonals10 KB (1,669 words) - 17:33, 12 January 2024
- ...entroid <math>G</math>. Let <math>X</math> be the intersection of the line tangent to the circumcircle of <math>\Delta ABC</math> at <math>A</math> and the li ...lies \beta + \theta = 11\alpha.</cmath> However, from before, <math>\beta+\theta = 180 - 17 \alpha</math>, so <math>11 \alpha = 180 - 17 \alpha \implies 18010 KB (1,526 words) - 16:50, 25 December 2022
- ...DAM = \alpha</math>, <math>\angle BAD = \beta</math>, <math>\angle BMA = \theta</math>, <math>\angle CMD = \phi</math>. Hence, <math>MP = MB</math> and <math>\angle AMP = \theta</math>.14 KB (2,254 words) - 18:26, 8 February 2024
- Two externally tangent circles <math>\omega_1</math> and <math>\omega_2</math> have centers <math> Denote <math>\angle O_1 O O_2 = 2 \theta</math>.14 KB (2,217 words) - 00:28, 29 June 2023
- ...and passes through <math>A</math>. A circle centered at <math>P</math> is tangent to line <math>AC</math> at <math>A</math> and passes through <math>B</math> ...bel("$\theta$", A, 7*dir(162)); label("$\theta$", B, 7*dir(-20)); label("$\theta$", P, 7*dir(-110)); label("$6$", B--C, left); label("$8$", A--C, down); lab6 KB (943 words) - 00:41, 6 August 2023
- ...tively, that meet at an angle <math>\theta</math> is a rotation by <math>2\theta</math> around the intersection of <math>l</math> and <math>m</math>. by the tangent addition formula. Since the slope of line <math>m</math> is <math>\frac{2}{8 KB (1,331 words) - 22:44, 16 December 2023
- Now, we can take the tangent and apply the tangent subtraction formula: Denote by <math>\theta</math> the acute angle formed by lines <math>y = x</math> and <math>y = 3 x16 KB (2,526 words) - 00:53, 6 May 2023
- Using the half-angle formula for tangent, Denote by <math>\theta</math> the argument of point <math>P</math> on the circle.19 KB (3,107 words) - 23:31, 17 January 2024
- Let <math>\Theta</math> be the circle with diameter <math>OQ.</math> ...\cdot QC = (QO – R) \cdot (QO + R) = QP^2</math> <cmath>\implies P \in \Theta, \Omega \perp \omega.</cmath>10 KB (1,751 words) - 15:34, 25 November 2022
- ...th> be the incenter of <math>ABCD</math> for which <math>\odot O</math> is tangent to <math>\overline{DA},\overline{AB},</math> and <math>\overline{BC}</math> We denote <math>\theta = \angle BAC</math>.17 KB (2,612 words) - 14:54, 3 July 2023
- Denote <math>\theta = \angle PBQ</math>. \frac{BQ}{\sin C} = \frac{PQ}{\sin \theta} . \hspace{1cm} (1)12 KB (1,900 words) - 18:14, 28 January 2024
- ...als in <math>x</math> with integer coefficients. Find all functions <math>\theta : {\mathbb Z}[x] \to {\mathbb Z}</math> such that for any polynomials <math <math>\theta(p+1) = \theta(p)+1</math>, and5 KB (930 words) - 08:38, 22 February 2023
- ...> and <math>\omega</math>, respectively, such that line <math>PS</math> is tangent to <math>\omega</math> (at <math>S</math>). Determine the locus of <math>O< ...> of the plane good if <math>S</math> is unchanged upon rotation by <math>\theta</math> around any point of <math>S</math> (in both clockwise and counterclo5 KB (873 words) - 08:45, 24 February 2023
- ...gging as shown in the figure below <math>(APQRS)</math>. What angle <math>\theta</math><math>\angle PAB=\angle QPC=\angle RQB=\cdots</math> will produce in label("$\theta$", (-41.5,-10.5));15 KB (2,168 words) - 05:11, 4 February 2024
- ...Gamma_2</math> again in <math>C</math>, different from <math>B</math>; the tangent to <math>\Gamma_2</math> at <math>B</math> intersects <math>\Gamma_1</math> ...heta+\sin \theta)=P(\cos \theta-\sin \theta)</math>, for every real <math>\theta</math>. Prove that <math>P(x)</math> can be expressed in the form<cmath>P(x3 KB (568 words) - 05:43, 25 September 2023
- ...<math>a^2+b^2=16</math>. We want to find the line <math>81a-108b=k</math> tangent to circle <math>a^2+b^2=16</math>. ...theta</math> for obviously positive <math>\cos\theta</math> and <math>\sin\theta</math>.12 KB (1,842 words) - 19:26, 23 February 2024
- Note: I've omitted <math>\theta</math> because it's unnecessary and might clog things up a little. label("$\theta$",shift(dir(degrees(d)/2)/5)*O,dir(degrees(d)/2));16 KB (2,796 words) - 13:12, 21 January 2024
- ...ath> with radius <math>r = 6</math>). Next, from <math>A</math>, construct tangent lines to the incircle meeting the circumcirlce at point <math>B</math> and \theta \coloneqq \angle BAI = \angle IAP \qquad \text{and} \qquad \phi \coloneqq \17 KB (2,854 words) - 15:18, 23 April 2024
- <math>\theta = \odot ACB', \theta' = \odot BCA'.</math> ...e midpoints of the arcs <math>\overset{\Large\frown}{CB'}</math> of <math>\theta,D = AL \cap \omega.</math>23 KB (4,003 words) - 16:17, 21 April 2024