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# example of law of tangent

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Tangent circles

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 real-life applications such as trilateration and maximizing the use of materials.

## Three given circles: Apollonius' problem

Apollonius' problem is to construct circles that are tangent to three given circles.

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.

Law of sines

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&mdash;a technique known as triangulation. 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, 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&nbsp;=&nbsp;20, side c&nbsp;=&nbsp;24, and angle C&nbsp;=&nbsp;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&nbsp;feet and the angle C opposite the chord is given in degrees, then

\angle A = \angle B = \frac{180-C}{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&nbsp;C. Draw a perpendicular to c from angle&nbsp;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&nbsp;a. Solve this equation for the sine of C. Notice that the perpendicular makes two right angles triangles, also note that sin(A) =&nbsp;cos(Î±), sin(B) =&nbsp;cos(Î²) and that c&nbsp;=&nbsp;a&nbsp;sin(Î²)&nbsp;+&nbsp;b&nbsp;sin(Î±). After making these substitutions you should have sin(C) =sin(Î±&nbsp;+&nbsp;Î²) =&nbsp;sin(Î²)cos(Î±)&nbsp;+&nbsp;(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±&nbsp;+&nbsp;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 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(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}

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.

Question:examples of 3 laws of motion

Answers:Newton first law of motion:every body continues to remain in its state of rest or uniform motion along a straight line unless it is compelled by external force to change that state. First law of newton is also called the law of inertia. ex : A person sitting in a vehicle at rest has his whole body at rest.when the vehicle suddenly starts moving forward,the lower part of in contact with the vehicle moves forward.But the upper part of the body continues too remain at rest due to inertia.As a result,the person has a tendency to fall back. Newton second law of motion:The acceleration given to a body is directly proportional to the force and inversely proportional to the mass and it takes place in the direction of force. ex : A force of 0.12N acts for 3 seconds on a body of mass 0.4 kg at rest.Find the velocity gained by the body. solution : F=0.12 N, m =0.4 kg and t = 3 s. using the equation F = ma, 0.12 = 0.41 therefore a = 0.12/0.4 = 0.3 meter per second square Initial velocity of the body U = 0 so final velocity of the body is given by v = u + at 0 + 0.3 * 3 = 0.9 meter per second square Third law of newton states that to every action there is any equal and opposite reaction. ex:When a body reaming at rest on a table.The body exerts a downward force on the table equal to its weight (action).The table in turn,exerts on the body an equal force in the opposite direction (reaction).

Question:i need 3 pictures for each but mostly i need 3 written examples for each Laws of Motion law of universal gravitation Newtons 1st law of motion Newtons 2nd law of motion Newtons 3rd law of motion PLEASE HELP DUE Tomorrow ! and i cant think of examples i didnt pay attention in class because i was late on notes i started this school 3 days ago please help!!

Answers:Universal gravitation - earth's pull on moon, or moon's pull on earth, causing the tides 1st law - object at rest (in motion) tend to stay at rest (in motion) unless acted upon by external forces - billiards table, satelites in space that go on forever (or the toolbag that the astronauts dropped recently) 2nd law - reduces down to force=mass*acceleration - anything that moves is an example; car accelerating, golf club hitting ball, rocket ship blastoff 3rd law - every action has an opposite reaction - billiards ball reactions are a good example here, recoil of a gun/cannon

Question:Could someone give some real life examples of Newtons three laws of motion.? Thanks in advance

Answers:Inertia...car skids out of control on a curve and ends up in a corn field. F = ma...car of mass m accelerates down the freeway when the driver pushes the pedal to the floor. Balanced forces...you sitting in your chair as you read this answer.

Question:1 2 3 4

Answers:Try drawing it! For example if each circle is an O then it is going to look like this: OO with a line between them I can't think of any way it could look like 2 or 3 or 4.