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From Wikipedia
In geometry, tangent circles (also known as kissing circles) are circles that intersect in a single point. There are two types of tangency: internal and external. Many problems and constructions in geometry are related to tangent circles; such problems often have reallife applications such as trilateration and maximizing the use of materials.
External/internal tangency; chains of tangent circles
One given circle
Two given circles
Steiner chains
Pappus chains
Three given circles: Apollonius' problem
Apollonius' problem is to construct circles that are tangent to three given circles.
Apollonian gasket
If a circle is iteratively inscribed into the interstital curved triangles between three mutually tangent circles, an Apollonian gasket results, one of the earliest fractals described in print.
Malfatti's problem
Malfatti's problem is to carve the largest three cylinders from a triangular block of marble.
Six circles theorem
A chain of six circles can be drawn such that each circle is tangent to two sides of a given triangle and also to the preceding circle in the chain. The chain closes; the sixth circle is always tangent to the first circle.
Seven circles theorem
A chain of six tangent circles is given, each of which is tangent to a seventh given circle. The tangent points of the chain circles with the seventh circle are connected pairwise between opposite circles in the chain, i.e., between circles 1 and 4, 2 and 5 and 3 and 6. These lies are concurrent, i.e., they intersect in the same point.
Generalizations
Problems involving tangent circles are often generalized to spheres. For example, the Fermat problem of finding sphere(s) tangent to four given spheres is a generalization of Apollonius' problem, whereas Soddy's hexlet is a generalization of a Steiner chain.
In trigonometry, the law of sines (also known as the sine law, sine formula, or sine rule) is an equation relating the lengths of the sides of an arbitrary triangle to the sines of its angles. According to the law,
 \frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C},
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). Sometimes the law is stated using the reciprocal of this equation:
 \frac{\sin A}{a} \,=\, \frac{\sin B}{b} \,=\, \frac{\sin C}{c}.
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. It can also be used when two sides and one of the nonenclosed 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, the other being the law of cosines.
Examples
The following are examples of how to solve a problem using the law of sines:
Given: side a = 20, side c = 24, and angle C = 40Â°
Using the law of sines, we conclude that
 \frac{\sin A}{20} = \frac{\sin 40^\circ}{24}.
 A = \arcsin\left( \frac{20\sin 40^\circ}{24} \right) \cong 32.39^\circ.
Or another example of how to solve a problem using the law of sines:
If two sides of the triangle are equal to R and the length of the third side, the chord, is given as 100 feet and the angle C opposite the chord is given in degrees, then
 \angle A = \angle B = \frac{180C}{2}= 90\frac{C}{2}
and
 {R \over \sin A}={\mbox{chord} \over \sin C}\text{ or }{R \over \sin B}={\mbox{chord} \over \sin C}\,
 {\mbox{chord} \,\sin A \over \sin C} = R\text{ or }{\mbox{chord} \,\sin B \over \sin C} = R.
Numeric problems
Like the law of cosines, although the law of sines is mathematically true, it has problems for numeric use. Much precision may be lost if an arcsine is computed when the sine of an angle is close to one.
Some applications
 The sine law can be used to prove the angle sum identity for sine when Î± and Î² are each between 0 and 90 degrees.
 To prove this, make an arbitrary triangle with sides a, b, and c with corresponding arbitrary angles A, B and C. Draw a perpendicular to c from angle C. This will split the angle C into two different angles, Î± and Î², that are less than 90 degrees, where we choose to have Î± to be on the same side as A and Î² be on the same side as B. Use the sine law identity that relates side c and side a. Solve this equation for the sine of C. Notice that the perpendicular makes two right angles triangles, also note that sin(A) = cos(Î±), sin(B) = cos(Î²) and that c = a sin(Î²) + b sin(Î±). After making these substitutions you should have sin(C) =sin(Î± + Î²) = sin(Î²)cos(Î±) + (b/a)sin(Î±)cos(Î±). Now apply the sine law identity that relates sides b and a and make the substitutions noted before. Now substitute this expression for (b/a) into the original equation for sin(Î± + Î²) and you will have the angle sum identity for Î± and Î² in terms of sine.
 The only thing that was used in the proof that was not a definition was the sine law. Thus the sine law is equivalent to the angle sum identity when the angles sum is between 0 and 180 degrees and when each individual angle is between 0 and 90 degrees.
 The sine law along with the prosthaphaeresis and shift identities can be used to prove the law of tangents and Mollweide's formulas (Dresdin 2009, Plane Trigonometry pg. 76–78 ).
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).
Given a general triangle ABC, 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 b, where the angle A is not the included angle of the two sides (in the above image, the angle C is the included angle).
 The angle A is acute (i.e., A < 90Â°).
 The side a is shorter than the side b (i.e., a< b).
 The side a is longer than the altitude of a right angled triangle with angle A and hypotenuse b (i.e., a> b sin A).
Given all of the above premises are true, the angle B may be acute or obtuse; meaning, one of the following is true:
 B = \arcsin {b \sin A \over a}
OR
 B= 180^\circ  \arcsin {b \sin A \over a}
Relation to the circumcircle
In the identity
 \frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C},
the common value of the three fractions is actually the diameter of the triangle's circumcircle. It can be shown that this quantity is equal to
 \begin{align}
\frac{abc} {2S} & {} = \frac{abc} {2\sqrt{s(sa)(sb)(sc)}} \\[6pt] & {} = \frac {2abc} {\sqrt{(a^2+b^2+c^2)^22(a^4+b^4+c^4) }}, \end{align}
where S is the area of the triangle and s is the semiperimeter
 s = \frac{a+b+c} {2}.
The second equality above is essentially Heron's formula.
Spherical case
In the spherical case, the formula is:
 \frac{\sin A}{\sin \alpha} = \frac{\sin B}{\sin \beta} = \frac{\sin C}{\sin \gamma}.
Here, Î±, Î², and Î³ 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.