Law of Sines
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]Theorem
In triangle , where
is the side opposite
,
opposite
, and
opposite
:
Proof
Method 1
In the diagram below, circle circumscribes triangle
.
is perpendicular to
. Since
,
and
. But
making
. Therefore, we can use simple trig in right triangle
to find that
![$\sin \theta = \frac{\frac a2}R \Leftrightarrow \frac a{\sin\theta} = 2R.$](http://latex.artofproblemsolving.com/0/3/f/03f858d898d88e6bb7665ea3514b180f2e8ebcf5.png)
The same holds for and
, thus establishing the identity.
This picture could be replaced by an asymptote drawing. It would be appreciated if you do this.
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:
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/a/d/3/ad39e455fe156ef0c687ce7c7d910ebd4f9c9f27.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)
Advanced
Let be a convex quadrilateral with
, $AC \neqBD$ (Error compiling LaTeX. Unknown error_msg), and let
be the intersection point of its diagonals. Prove that
if and only if
.
(Source)