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- ...that <math>\triangle APO</math> is an [[isosceles triangle]]: <math>\angle AOP = 15^{\circ}</math> and <math>\angle OAP = OAB - 60^{\circ} = \frac{180^{\c2 KB (343 words) - 14:39, 14 June 2023
- Thus, <math>\angle{AOP}+\angle{POB}+\angle{COQ}+\angle{QOD}=180</math>, or <math>(\arctan(\tfrac{12 KB (399 words) - 16:37, 2 January 2024
- ..., and <math>P</math> is a point on <math>l</math> such that triangle <math>AOP</math> is isosceles. Let <math>p</math> denote the value of <math>AP \cdot6 KB (1,100 words) - 21:35, 9 January 2016
- ..., and <math>P</math> is a point on <math>l</math> such that triangle <math>AOP</math> is isosceles. Let <math>p</math> denote the value of <math>AP \cdot540 bytes (96 words) - 23:28, 22 December 2023
- ...angent. Let <math>O</math> be the centre of the hemisphere. Triangle <math>AOP</math> is a right triangle, since <math>P</math> is ninety degrees. Apply t2 KB (376 words) - 22:14, 5 January 2024
- ...^{\circ}</math>. Then <math>\angle O_2PA=75^{\circ}-\theta \implies \angle AOP = 180^{\circ}-2\angle O_2PA = 2\theta+30^{\circ} \implies \angle ABP=\theta13 KB (2,075 words) - 16:50, 9 November 2024
- ...AS similarity. Thus, <math>\angle AOP = \angle AQH,</math> so <math>\angle AOP</math> is a right angle. Since <math>\angle AOP</math> and <math>\angle AOQ</math> are both right angles, we get <math>\ang10 KB (1,733 words) - 18:15, 14 June 2020
- ...\overline {OP} = h</math>. Using the Pythagorean Theorem on triangle <math>AOP</math>, <math>BOP</math>, or <math>COP</math> (all three are congruent by S Then continue as before to use the Pythagorean Theorem on <math>\triangle AOP</math>, find <math>h</math>, and find the volume of the pyramid.7 KB (1,177 words) - 14:55, 5 January 2024
- In addition, <math>\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A</math>. Combining these two equati6 KB (968 words) - 11:05, 7 June 2024
- Let <math>\alpha=\angle AOP,\;\beta=\angle BPD,\;\theta=\angle APE</math>6 KB (1,039 words) - 12:00, 13 March 2024
- Because of how reflections work, we have that <math>\angle AOP' = \angle POA</math> and <math>\angle P'OB = \angle BOP''</math>; adding th9 KB (1,435 words) - 14:09, 29 September 2024
- ...and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. .... We can use the [[Law of Cosines]] on isosceles triangles <math>\triangle AOP, \, \triangle COP, \, \triangle BOP, \, \triangle DOP</math> to get22 KB (3,480 words) - 18:33, 1 August 2024
- ...</math> and <math>H</math> point <math>P</math>. We know that <math>\angle{AOP} = 20</math> and that <math>\angle{A} = 90</math>. Subtracting <math>20</ma4 KB (633 words) - 11:22, 20 October 2024