Difference between revisions of "Law of Sines"
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Revision as of 19:42, 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
Theorem
In triangle 
, where 
is the side opposite to 
, 
opposite to 
, 
opposite to 
, and where 
is the circumradius:
![\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\]](http://latex.artofproblemsolving.com/2/6/0/260d7ef25c7fad85f7793a9b3dfa5baff63f1cd2.png)
Proof
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)); dot(A^^B^^C^^D^^O); [/asy]](http://latex.artofproblemsolving.com/5/1/e/51ebac9684b16f3229628f5bf9ce3ef88ff85435.png)
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
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 ![$[ABC] = \frac{1}{2}ab\sin C$](http://latex.artofproblemsolving.com/3/b/0/3b0352460a419194158d53dd89624c2143b20d05.png)
.
Since it doesn't matter which sides are chosen as , , and , the following equality holds:
![\[\frac{1}{2}bc\sin A = \frac{1}{2}ac\sin B = \frac{1}{2}ab\sin C\]](http://latex.artofproblemsolving.com/b/3/c/b3c723d9efc56c7f0d769b4aa916faa52c0b7045.png)
Assuming the triangle in question is nondegenerate, 
. Multiplying the equation by 
yields:
![\[\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\]](http://latex.artofproblemsolving.com/e/3/0/e30afe510a434364b7b98869402ff35fc9691404.png)
Problems
Introductory
- If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
(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 
.
has sides 
, 
, and 
of 
be the 
that is not on the same side of 
can be expressed as 
, where 
and 
are 
.(Source)
Olympiad
Let 
be a convex quadrilateral with 
, 
, and let 
be the intersection point of its diagonals. Prove that 
if and only if 
.
(Source)
