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{{About|the law of sines in trigonometry|the law of sines in physics|Snell's law}}
Most of these procedures either could not be repeated or the chances of side effects increase dramatically when mixing treatment options. Drugs containing Nitroglycerine, isosorbide mononitrate, isosorbide dinitrate are recommended as bothersome for making use of with tadalafil. For a lot of women, including young girls makeup is essential. To learn more about the  on your body and how our  product can help you, please visit our site to claim your free report: The Algae, Adult Stem Cell and Omega 3 Connection. Two) Consuming A satisfactory amount of Water - Tend not to disregard the intake of drinking water when targeting a healthy life. <br><br>This article has been flagged as spam, if you think this is an error please contact us. The hypnotist would probably create some recommendations that concentrate on this anxiety to discover the invisible problem. The increasing complexity faced by the organization means that few good quality decisions can be made in one functional area - now, decisions can only be made and implemented by everyone reaching across functions for the information, assistance and support they need. It is also intensified by venous leakage, a problem that occurs when the penile veins fails to tighten properly during an erection. Have you unwittingly let yourself take part in the great internet pornography experiment. <br><br>Men who are impotent from diabetes must seek treatment every month and each month, the attending doctor must draw a blood test called HBA1C, or another blood test called fructosamine. Respiration techniques are also useful if used regularly. In fact, we would love to get you started today with Top Ten Tips to Naturally Cure Male Impotence. If these key treatments do not assist you, your specialist may recommend you impotence drugs such as Viagra, Cialis and Levitra. The advantage that herbal impotence cures have over different cures is that they are much cheaper and are out there everywhere. <br><br>On getting the impulses, neurotransmitters are released and this relaxes the penile muscles and it fills up with blood. What about the health implications in this dangerous experiment. Robson became impotent as a result of radiotherapy to kill prostate cancer cells. Our online drugstore is offering all the medicines that are generic in nature. Drinking plenty of water is also extremely beneficial. <br><br>The best way to avoid this is to always use photo release forms when shooting. Natural yet effective love-making natural treatments for erectile dysfunction are plentiful online however , be sure to buy a having sex capsule which has Tongkat Ali (Eurycoma Longifolia or possibly Pasak Bumi as it is known throughout Philippines). 2) Overweight and a couch-potato lifestyle can be bad too. Of course, if your personal or family history suggests that no wine is best for you, then forego the vino and opt for red grape juice, which contains those same antioxidants. Patients with dysfunction of male organ are often advised to avoid smoking and intake of alcohol.<br><br>If you beloved this posting and you would like to receive much more information pertaining to how to overcome erectile dysfunction ([http://www.eiaculazione-precoce.info/sitemap/ eiaculazione-precoce.info]) kindly go to our own web-page.
[[File:Acute Triangle.svg|thumb|right|A triangle labelled with the components of the law of sines. Big A, B and C are the angles, and little a, b, c are the sides opposite them. (a opposite A, etc.)]]
{{Trigonometry}}
In [[trigonometry]], the '''law of sines''', '''sine law''', '''sine formula''', or '''sine rule''' is an [[equation]] relating the [[lengths]] of the sides of an arbitrary [[triangle]] to the [[sine]]s of its angles. According to the law,
 
:<math> \frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \,=\, D \!</math>
 
where ''a'', ''b'', and ''c'' are the lengths of the sides of a triangle, and ''A'', ''B'', and ''C'' are the opposite angles (see the figure to the right), and ''D'' is the [[diameter]] of the triangle's [[circumcircle]]. When the last part of the equation is not used, sometimes the law is stated using the [[Multiplicative inverse|reciprocal]]:
 
: <math> \frac{\sin A}{a} \,=\, \frac{\sin B}{b} \,=\, \frac{\sin C}{c} \!</math>
 
The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as [[triangulation]]. However calculating this may result in [[numerical error]] if an angle is close to 90 degrees. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the formula gives two possible values for the enclosed angle, leading to an ''ambiguous case''.
 
The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in a general triangle, with the other being the [[law of cosines]].
 
==Proof==
There are three cases to consider in proving the law of sines. The first is when all angles of the triangle are [[acute]]. The second is when one angle is a [[right angle]]. The third is when one angle is [[obtuse]].
 
===For acute triangles===
We make a triangle with the sides ''a'', ''b'', and ''c'', and angles ''A'', ''B'', and ''C''.  Then we draw the [[Altitude (triangle)|altitude]] from vertex ''B'' to side ''b''; by definition it divides the original triangle into two right angle triangles: ''ABR'' and ''R'BC''. Mark this line ''h<sub>1</sub>''.
 
[[File:Acute Triangle B.svg|thumb|Triangle ABC with altitude from B drawn]]
 
Using the definition of <math>\textstyle \sin \alpha = \frac{\text{opposite}} {\text{hypotenuse}}</math> we see that for angle ''A'' on the right angle triangle ''ABR'' and ''C'' on ''R'BC'' we have:
:<math>\sin A = \frac{h_1}{c}\text{; } \sin C = \frac{h_1}{a}</math>
 
Solving for ''h<sub>1</sub>''
 
:<math>h_1 = c \sin A\text{; }  h_1 = a \sin C \,</math>
 
Equating ''h<sub>1</sub>'' in both expressions:
 
:<math>h_1 = c \sin A  = a \sin C \,</math>
 
Therefore:
:<math>\frac{a}{\sin A} = \frac{c}{\sin C}.</math>
 
 
Doing the same thing from angle ''A'' to side ''a'' we call the altitude ''h<sub>2</sub>'' and the two right angle triangles ''ABR'' and ''AR'C'':
[[File:Acute Triangle A.svg|thumb|Triangle ABC with altitude from A drawn]]
:<math>\sin B = \frac{h_2}{c}\text{; } \sin C = \frac{h_2}{b}</math>
 
Solving for ''h<sub>2</sub>''
 
:<math>h_2 = c \sin B\text{; }  h_2 = b \sin C \,</math>
 
Therefore:
:<math>\frac{b}{\sin B} = \frac{c}{\sin C}</math>
 
 
Equating the <math>\textstyle \frac{c}{\sin C}</math> terms in both expressions above we have:
 
:<math>\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}</math>
 
===For right angle triangles===
We make a triangle with the sides ''a'', ''b'', and ''c'', and angles ''A'', ''B'', and ''C'' where ''C'' is a right angle.
 
[[File:Right Angle Triangle.svg|thumb|Triangle ''ABC'' with right angle ''C''.]]
 
Since we already have a right angle triangle we can use the definition of sine:
 
:<math>\sin A = \frac{a}{c} \text{; }  \sin B = \frac{b}{c}</math>
 
Solving for ''c'':
 
:<math>c = \frac{a}{\sin A} \text{; } c = \frac{b}{\sin B}</math>
 
Therefore:
 
:<math>\frac{a}{\sin A} = \frac{b}{\sin B}</math>
 
 
For the remaining angle ''C'' we need to remember that it is a right angle and <math>\textstyle \sin C = 1</math> in this case. Therefore we can rewrite ''c'' = ''c'' / 1 as:
 
:<math>c = \frac{c}{ \sin C} </math>
 
Equating ''c'' in both the equations above we again have:
 
:<math>\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}</math>
 
===For obtuse triangles===
We make a triangle with the sides ''a'', ''b'', and ''c'', and angles ''A'', ''B'', and ''C'' where ''A'' is an obtuse angle.In this case if we draw an altitude from any angle other than ''A'' the point where this line will touch the base of the triangle ''ABC'' will lie outside any of the lines ''a'', ''b'', or ''c''. We draw the altitude from angle ''B'', calling it ''h<sub>1</sub>'' and create the two extended right triangles ''RBA''' and ''RBC''.
 
[[File:Obtuse triangle altitude from B.svg|thumb|Obtuse triangle ''ABC'' with altitude drawn from B.]]
 
From the definition of sine we again have:
 
:<math>\sin A' = \frac{h_1}{c}\text{; } \sin C = \frac{h_1}{a}</math>
 
We use [[List of trigonometric identities#Symmetry|identity]] <math>\textstyle \sin \pi - \theta = \sin \theta</math> to express <math>\textstyle \sin A'</math> in terms of <math>\textstyle \sin A</math>. By definition we have:
 
:<math>A + A' = \pi</math>
 
:<math>A = \pi - A'</math>
 
:<math>\sin A = \sin (\pi - A') = \sin A'</math>
 
Therefore:
:<math>\sin A = \frac{h_1}{c}\text{; } \sin C = \frac{h_1}{a}</math>
and
:<math>\frac{a}{\sin A} = \frac{c}{\sin C}</math>
 
 
We now draw an altitude from ''A'' calling it ''h<sub>2</sub>'' and forming two right triangles ''ABR'' and ''AR'C''.
 
[[File:Obtuse triangle altitude from A.svg|thumb|Obtuse triangle ''ABC'' with altitude from A]]
 
From this we straightforwardly get: 
:<math>\sin B = \frac{h_2}{c}\text{; } \sin C = \frac{h_2}{b}</math>
and
:<math>\frac{b}{\sin B} = \frac{c}{\sin C}</math>
 
 
Equating the <math>\textstyle\frac{\sin C}{c}</math> in both equations above we again get:
 
:<math>\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}</math>
 
 
 
Proving the theorem in all cases.
 
==The ambiguous case==
When using the law of sines to solve triangles, there exists an ambiguous case where two separate triangles can be constructed (i.e., there are two different possible solutions to the triangle). In the case shown below they are triangle ''ABC'' and ''AB'C'''.
 
[[File:PictureAmbitext.svg]]
 
Given a general triangle the following conditions would need to be fulfilled for the case to be ambiguous:
 
* The only information known about the triangle is the angle ''A'' and the sides ''a'' and ''c''
* The angle ''A'' is [[Angle#Types of angles|acute]] (i.e., A < 90°).
* The side ''a'' is shorter than the side ''c'' (i.e., ''a'' < ''c'').
* The side ''a'' is longer than the altitude ''h'' from angle ''B'', where ''h'' = ''c'' sin ''A'' (i.e., ''a'' > ''h'').
 
Given all of the above premises are true, then either of the angles ''C'' or ''C' '' may produce a valid triangle; meaning, both of the following are true:
 
: <math> C = \arcsin {c \sin A \over a} \text{ or } C' = \pi - \arcsin {c \sin A \over a}</math>
 
From there we can find the corresponding ''B'' and ''b'' or ''B' ''and ''b' '' if required, where ''b'' is the side bounded by angles ''A'' and ''C'' and ''b' ''bounded by ''A'' and ''C' ''.
 
Without further information it is impossible to decide which is the triangle being asked for.
 
==Examples==
The following are examples of how to solve a problem using the law of sines:
 
Given: side ''a''&nbsp;=&nbsp;20, side ''c''&nbsp;=&nbsp;24, and angle ''C''&nbsp;=&nbsp;40°
 
Using the law of sines, we conclude that
 
:<math>\frac{\sin A}{20} = \frac{\sin 40^\circ}{24}.</math>
 
:<math> A = \arcsin\left( \frac{20\sin 40^\circ}{24} \right) \approx 32.39^\circ. </math>
 
Or another example of how to solve a problem using the law of sines:
 
If two sides of the triangle are equal to ''x'' and the length of the third side, the [[chord (geometry)|chord]], is given as 100&nbsp;feet and the angle ''C'' opposite the chord is given in degrees, then
 
: <math>\angle A = \angle B = \frac{180^\circ-\angle C}{2}= 90-\frac{\angle C}{2}\!</math>
 
and
 
:<math>{x \over \sin A}={\mbox{chord} \over \sin C}\text{ or }{x \over \sin B}={\mbox{chord} \over \sin C}\,\!</math>
 
<!-- extra blank line for legibility -->
 
:<math>{\mbox{chord} \,\sin A \over \sin C} = x\text{ or }{\mbox{chord} \,\sin B \over \sin C} = x.\!</math>
 
==Relation to the circumcircle==
In the identity
 
:<math> \frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C},\!</math>
 
the common value of the three fractions is actually the [[diameter]] of the triangle's [[circumcircle]].<ref>Coxeter, H. S. M. and Greitzer, S. L. ''Geometry Revisited''. Washington, DC: Math. Assoc. Amer., pp. 1–3, 1967</ref> It can be shown that this quantity is equal to
 
:<math>\begin{align}
\frac{abc} {2S} & {} = \frac{abc} {2\sqrt{s(s-a)(s-b)(s-c)}} \\[6pt]
& {} = \frac {2abc} {\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) }},
\end{align}</math>
 
where ''S'' is the area of the triangle and ''s'' is the [[semiperimeter]]
 
:<math>s = \frac{a+b+c} {2}.</math>
 
The second equality above is essentially [[Heron's formula]].
 
==Spherical case==
In the spherical case, the formula is:
:<math>\frac{\sin A}{\sin \alpha} = \frac{\sin B}{\sin \beta} = \frac{\sin C}{\sin \gamma}.</math>
 
Here, ''α'', ''β'', and ''γ'' are the angles at the center of the sphere subtended by the three arcs of the spherical surface triangle ''a,'' ''b,'' and ''c,'' respectively.  ''A'', ''B'', and ''C'' are the surface angles opposite their respective arcs.
 
It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since
 
:<math>\lim_{\alpha \rightarrow 0} \frac{\sin \alpha}{\alpha} = 1</math>
 
and the same for <math>{\sin \beta}</math> and <math>{\sin \gamma}</math>.
 
: See also [[Spherical law of cosines]] and [[Half-side formula]].
 
==Hyperbolic case==
In [[hyperbolic geometry]] when the curvature is &minus;1, the law of sines becomes
:<math>\frac{\sin A}{\sinh a} = \frac{\sin B}{\sinh b} = \frac{\sin C}{\sinh c} \,.</math>
 
In the special case when ''B'' is a right angle, one gets
:<math>\sin C = \frac{\sinh c}{\sinh b} \,</math>
 
which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse.
 
:See also [[hyperbolic triangle]].
 
==Unified formulation==
Define a generalized sine function, depending also on a real parameter <math>K</math>:
 
:<math>\sin_K x = x - \frac{K x^3}{3!} + \frac{K^2 x^5}{5!} - \frac{K^3 x^7}{7!} + \cdots.</math>
 
The law of sines in constant curvature <math>K</math> reads as<ref>{{cite web|last=Russell|first=Robert A.|title=Generalized law of sines|url=http://140.177.205.23/GeneralizedLawofSines.html|work=Wolfram Mathworld|accessdate=25 September 2011}}</ref>
 
:<math>\frac{\sin A}{\sin_K a} = \frac{\sin B}{\sin_K b} = \frac{\sin C}{\sin_K c} \,.</math>
 
By substituting <math>K=0</math>, <math>K=1</math>, and <math>K=-1</math>, one obtains respectively the euclidian, spherical, and hyperbolic cases of the law of sines described above.
 
Let <math>p_K(r)</math> indicate the circumference of a circle of radius <math>r</math> in a space of constant curvature <math>K</math>. Then <math>p_K(r)=2\pi \sin_K r</math>. Therefore the law of sines can also be expressed as:
 
:<math>\frac{\sin A}{p_K(a)} = \frac{\sin B}{p_K(b)} = \frac{\sin C}{p_K(c)} \,.</math>
 
This formulation was discovered by [[János Bolyai]].<ref>{{cite book|last=Katok|first=Svetlana|authorlink=Svetlana Katok|title=Fuchsian groups|year=1992|publisher=University of Chicago Press|location=Chicago|isbn=0-226-42583-5|page=22}}</ref>
 
==History==
According to [[Ubiratàn D'Ambrosio]] and [[Helaine Selin]], the spherical law of sines was discovered in the 10th century.  It is variously attributed to [[Abu-Mahmud al-Khujandi|al-Khujandi]], [[Abul Wafa Bozjani]], [[Nasir al-Din al-Tusi]] and [[Abu Nasr Mansur]].<ref name=Sesiano>Sesiano just lists al-Wafa as a contributor. Sesiano, Jacques (2000) "Islamic mathematics" pp. 137&#151; , page 157, in {{citation|title=Mathematics Across Cultures: The History of Non-western Mathematics|first1=Helaine|last1=Selin|first2=Ubiratan|last2=D'Ambrosio|year=2000|publisher=[[Springer Science+Business Media|Springer]]|isbn=1-4020-0260-2}}</ref>
 
[[Al-Jayyani]]'s ''The book of unknown arcs of a sphere'' in the 11th century introduced the general law of sines.<ref name="MacTutor Al-Jayyani">{{MacTutor|id=Al-Jayyani|title=Abu Abd Allah Muhammad ibn Muadh Al-Jayyani}}</ref> The plane law of sines was later described in the 13th century by [[Nasīr al-Dīn al-Tūsī]]. In his ''On the Sector Figure'', he stated the law of sines for plane and spherical triangles, and provided proofs for this law.<ref>{{cite book | first=J. Lennart | last=Berggren | title=The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook | chapter=Mathematics in Medieval Islam | publisher=Princeton University Press | year=2007 | isbn=978-0-691-11485-9 | page=518 }}</ref>
 
According to [[Glen Van Brummelen]], "The Law of Sines is really [[Regiomontanus]]'s foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles."<ref>Glen Van Brummelen (2009). "''[http://books.google.cz/books?id=bHD8IBaYN-oC&pg=&dq&hl=en#v=onepage&q=&f=false The mathematics of the heavens and the earth: the early history of trigonometry]''". Princeton University Press. p.259. ISBN 0-691-12973-8</ref> Regiomontanus was a 15th-century German mathematician.
 
==An equation with sines for tetrahedra==
[[Image:tetra.svg|thumb|right|A tetrahedron structure with vertices ''O'', ''A'', ''B'', ''C''. The product of the sines of ∠OAB, ∠OBC, ∠OCA appears on one side of the identity, and the product of the sines of ∠OAC, ∠OCB, ∠OBA on the other.]]
 
An equation involving sine functions and tetrahedra is as follows. For a [[tetrahedron]] with vertices ''O'', ''A'', ''B'', ''C'', it is true that
 
: <math>
\begin{align}
& {} \quad \sin\angle OAB\cdot\sin\angle OBC\cdot\sin\angle OCA \\
& = \sin\angle OAC\cdot\sin\angle OCB\cdot\sin\angle OBA.
\end{align}
</math>
 
One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface.
 
Putting any of the four vertices in the role of ''O'' yields four such identities, but in a sense at most three of them are independent: If the "clockwise" sides of three of them are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides, the result is the fourth identity.  One reason to be interested in this "independence" relation is this: It is widely known that three angles are the angles of some triangle if and only if their sum is a half-circle.  What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron?  Clearly the sum of the angles of any side of the tetrahedron must be a half-circle.  Since there are four such triangles, there are four such constraints on sums of angles, and the number of [[Degrees of freedom (statistics)|degrees of freedom]] is thereby reduced from 12 to 8.  The four relations given by this sines law further reduce the number of degrees of freedom, ''not'' from 8 down to 4, but only from 8 down to 5, since the fourth constraint is not independent of the first three.  Thus the space of all shapes of tetrahedra is 5-dimensional.
 
==See also==
* [[Gersonides]]
* [[Half-side formula]]&nbsp;– for solving [[spherical triangles]]
* [[Law of tangents]]
* [[Mollweide's formula]]&nbsp;– for checking solutions of triangles
*[[Solution of triangles]]
* [[Surveying]]
 
==References==
{{Reflist}}
 
==External links==
* {{springer|title=Sine theorem|id=p/s085520}}
* [http://www.cut-the-knot.org/proofs/sine_cosine.shtml#law The Law of Sines] at [[cut-the-knot]]
* [http://www.du.edu/~jcalvert/railway/degcurv.htm Degree of Curvature]
* [http://www.efnet-math.org/Meta/sine1.htm Finding the Sine of 1 Degree]
 
{{DEFAULTSORT:Law Of Sines}}
[[Category:Trigonometry]]
[[Category:Angle]]
[[Category:Triangle geometry]]
[[Category:Articles containing proofs]]
[[Category:Theorems in plane geometry]]
 
{{Link FA|km}}

Latest revision as of 18:36, 1 October 2014

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