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));
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19 KB (3,268 words) - 20:55, 4 February 2025
...y pass through a point <math>P</math> lying on <math>\Omega</math> and are tangent to one line.
Prove that there exists a line tangent to two of these circles and passing through the vertex of <math>\triangle A
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122 KB (21,381 words) - 13:47, 3 August 2025
Let <math>\theta = \frac{\pi}{16}</math>. Then,
y = e^{8i\theta} = e^{\frac{\pi}{2} i} = (\cos \theta + i\sin \theta)^8 = 0 + i.
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17 KB (2,502 words) - 23:17, 7 November 2025
...cle <math>x^2+y^2=16</math>, which is perpendicular to the line connecting tangent point to circle's center <math>(0,0)</math>.
...theta</math> for obviously positive <math>\cos\theta</math> and <math>\sin\theta</math>.
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17 KB (2,684 words) - 11:05, 26 July 2025
Using the half-angle formula for tangent,
Denote by <math>\theta</math> the argument of point <math>P</math> on the circle.
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22 KB (3,480 words) - 18:33, 1 August 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 diagonals
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10 KB (1,670 words) - 22:31, 14 July 2025
...rt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math>
...2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}\cos^3{\theta}=\frac{1}{8}</math>. This means that <math>y=\frac{3\sqrt{3}}{8}\implies OC
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22 KB (3,598 words) - 14:44, 20 January 2025
...</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);
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7 KB (1,087 words) - 17:02, 25 August 2025
...ath> lies on line <math>\ell</math>. A circle of radius <math>12</math> is tangent to line <math>\
l</math> and is externally tangent to the triangle. The area of the region exterior to the triangle and the ci
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26 KB (3,978 words) - 00:32, 11 November 2025
...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 t
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25 KB (4,662 words) - 13:32, 26 March 2025
...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 bi
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11 KB (1,873 words) - 15:34, 23 August 2025
...<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 <math
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31 KB (5,093 words) - 08:55, 17 December 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>.
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15 KB (2,593 words) - 12:37, 29 January 2021
...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>
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6 KB (1,013 words) - 21:09, 21 November 2023
...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 w
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7 KB (1,184 words) - 12:25, 22 December 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>
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50 KB (8,748 words) - 16:42, 20 October 2025
...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 + \ta
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11 KB (1,755 words) - 17:58, 3 June 2025
...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>
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4 KB (662 words) - 23:51, 2 October 2023
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>.
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2 KB (301 words) - 17:50, 1 April 2018
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 the
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7 KB (1,214 words) - 17:49, 29 January 2018
