Difference between revisions of "Law of Sines"
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− | + | The '''Law of Sines''' is a useful identity in a [[triangle]], which, along with the [[law of cosines]] and the [[law of tangents]] can be used to determine sides and angles. The law of sines can also be used to determine the [[circumradius]], another useful function. | |
+ | |||
+ | == Statement == | ||
+ | In triangle <math>\triangle ABC</math>, where <math>a</math> is the side opposite to <math>A</math>, <math>b</math> opposite to <math>B</math>, <math>c</math> opposite to <math>C</math>, and where <math>R</math> is the circumradius: | ||
+ | |||
+ | <cmath>\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R</cmath> | ||
==Proof == | ==Proof == | ||
=== Method 1 === | === Method 1 === | ||
− | + | <asy> | |
+ | real r = 1; | ||
+ | pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; | ||
+ | O = circumcenter(A,B,C); | ||
+ | D = (B+C)/2; | ||
+ | |||
+ | draw(C--A--B--C--O--B); | ||
+ | draw(O--D); | ||
+ | draw(circumcircle(A,B,C)); | ||
− | + | label("$A$",A,(-1,1));label("$B$",B,(-1,-1));label("$C$",C,(1,-1));label("$D$",D,(0,-1));label("$O$",O,(0,1)); | |
+ | label("$a/2$",(B+D)/2,(0,-1));label("$R$",(3*O+2*B)/5,(-1,1));label("$\theta$",O,(-0.8,-1.2));label("$\theta$",A,(0,-1.5)); | ||
− | The same holds for b and c thus establishing the identity. | + | dot(A^^B^^C^^D^^O); |
+ | </asy> | ||
+ | |||
+ | In the diagram above, point <math> O </math> is the [[circumcenter]] of <math> \triangle ABC </math>. Point <math>D</math> is on <math>BC</math> such that <math> OD </math> is [[perpendicular]] to <math> BC </math>. Since <math> \triangle ODB \cong \triangle ODC </math>, <math> BD = CD = \frac a2 </math> and <math> \angle BOD = \angle COD </math>. But <math> 2\angle BAC = \angle BOC </math> making <math> \angle BOD = \angle COD = \theta </math>. We can use simple trigonometry in [[right triangle]] <math> \triangle BOD </math> to find that | ||
+ | |||
+ | <center><math> \sin \theta = \frac{\frac a2}R \iff \frac a{\sin\theta} = 2R. </math> </center> | ||
+ | |||
+ | The same holds for <math>b</math> and <math>c</math>, thus establishing the identity. | ||
− | |||
− | |||
− | |||
=== Method 2 === | === Method 2 === | ||
This method only works to prove the regular (and not extended) Law of Sines. | This method only works to prove the regular (and not extended) Law of Sines. | ||
− | The formula for the area of a triangle is | + | The formula for the area of a triangle is <math>[ABC] = \frac{1}{2}ab\sin C </math>. |
− | <math>[ABC] = \frac{1}{2}ab\sin C </math> | ||
Since it doesn't matter which sides are chosen as <math>a</math>, <math>b</math>, and <math>c</math>, the following equality holds: | Since it doesn't matter which sides are chosen as <math>a</math>, <math>b</math>, and <math>c</math>, the following equality holds: | ||
− | < | + | <cmath>\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C </cmath> |
− | Multiplying the equation by <math>\frac{2}{abc} </math> | + | Assuming the triangle in question is nondegenerate, <math>abc \ne 0</math>. Multiplying the equation by <math>\frac{2}{abc} </math> yields: |
− | < | + | <cmath>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} </cmath> |
− | ==See | + | ==Problems== |
+ | ===Introductory=== | ||
+ | *If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle? | ||
+ | |||
+ | <center><math> | ||
+ | \mathrm{(A) \ 2} | ||
+ | \qquad \mathrm{(B) \ } 8/\sqrt{15} | ||
+ | \qquad \mathrm{(C) \ } 5/2 | ||
+ | \qquad \mathrm{(D) \ } \sqrt{6} | ||
+ | \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2 | ||
+ | </math></center> | ||
+ | |||
+ | ([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]]) | ||
+ | |||
+ | ===Intermediate=== | ||
+ | *[[Triangle]] <math>ABC</math> has sides <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math> of [[length]] 43, 13, and 48, respectively. Let <math>\omega</math> be the [[circle]] [[circumscribe]]d around <math>\triangle ABC</math> and let <math>D</math> be the [[intersection]] of <math>\omega</math> and the [[perpendicular bisector]] of <math>\overline{AC}</math> that is not on the same side of <math>\overline{AC}</math> as <math>B</math>. The length of <math>\overline{AD}</math> can be expressed as <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are [[positive integer]]s and <math>n</math> is not [[divisibility | divisible]] by the [[square]] of any [[prime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>. | ||
+ | |||
+ | ([[Mock AIME 4 2006-2007 Problems/Problem 15|Source]]) | ||
+ | ===Olympiad=== | ||
+ | Let <math>ABCD </math> be a convex quadrilateral with <math>AB=BC=CD </math>, <math>AC \neq BD </math>, and let <math>E </math> be the intersection point of its diagonals. Prove that <math>AE=DE </math> if and only if <math> \angle BAD+\angle ADC = 120^{\circ} </math>. | ||
+ | |||
+ | ([[2007 BMO Problems/Problem 1|Source]]) | ||
+ | |||
+ | ==See Also== | ||
* [[Trigonometry]] | * [[Trigonometry]] | ||
* [[Trigonometric identities]] | * [[Trigonometric identities]] | ||
* [[Geometry]] | * [[Geometry]] | ||
* [[Law of Cosines]] | * [[Law of Cosines]] | ||
+ | |||
+ | [[Category:Theorems]] | ||
+ | [[Category:Trigonometry]] |
Latest revision as of 20:52, 4 February 2025
The Law of Sines is a useful identity in a triangle, which, along with the law of cosines and the law of tangents can be used to determine sides and angles. The law of sines can also be used to determine the circumradius, another useful function.
Contents
[hide]Statement
In triangle , where
is the side opposite to
,
opposite to
,
opposite to
, and where
is the circumradius:
Proof
Method 1
In the diagram above, point is the circumcenter of
. Point
is on
such that
is perpendicular to
. Since
,
and
. But
making
. We can use simple trigonometry in right triangle
to find that
![$\sin \theta = \frac{\frac a2}R \iff \frac a{\sin\theta} = 2R.$](http://latex.artofproblemsolving.com/4/7/a/47a05c63201d5a7fcbf77d22abb8c2aed4583246.png)
The same holds for and
, thus establishing the identity.
Method 2
This method only works to prove the regular (and not extended) Law of Sines.
The formula for the area of a triangle is .
Since it doesn't matter which sides are chosen as ,
, and
, the following equality holds:
Assuming the triangle in question is nondegenerate, . Multiplying the equation by
yields:
Problems
Introductory
- If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
![$\mathrm{(A) \ 2} \qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2$](http://latex.artofproblemsolving.com/b/4/c/b4ca0aa20318b824e736a8e98ee1b6b667f5e2d7.png)
(Source)
Intermediate
- Triangle
has sides
,
, and
of length 43, 13, and 48, respectively. Let
be the circle circumscribed around
and let
be the intersection of
and the perpendicular bisector of
that is not on the same side of
as
. The length of
can be expressed as
, where
and
are positive integers and
is not divisible by the square of any prime. Find the greatest integer less than or equal to
.
(Source)
Olympiad
Let be a convex quadrilateral with
,
, and let
be the intersection point of its diagonals. Prove that
if and only if
.
(Source)