Law of Tangents

Revision as of 12:10, 2 February 2008 by Temperal (talk | contribs) (hm)

The Law of Tangents is a useful trigonometric identity that, along with the law of sines and law of cosines, can be used to determine angles in a triangle. Note that the law of tangents usually cannot determine sides, since only angles are involved in its statement.

Theorem

The law of tangents states that in triangle $\triangle ABC$, if $A$ and $B$ are angles of the triangle opposite sides $a$ and $b$ respectively, then $\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2}$.

Proof

First we can write the RHS in terms of sines and cosines: \[\frac{a-b}{a+b}=\frac{\sin (A-B)/2 \cos (A+B)/2}{\sin (A+B)/2\cos (A-B)/2}\] We can use various sum-of-angle trigonometric identities to get: \[\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A +\sin B}\] By the law of sines, we have \[\frac{a-b}{a+b}=\frac{a/2R-b/2R}{a/2R+b/2R}\] where $R$ is the circumradius of the triangle. Applying the law of sines again, \[\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2}\] as desired.

Problems

Introductory

This problem has not been edited in. If you know this problem, please help us out by adding it.

Intermediate

In $\triangle ABC$, LET $d$ BE A POINT IN $bc$ SUCH THAT $ad$ bisects $\angle A$. Given that $AD=6,BD=4$, and $DC=3$, find $AB$.

Olympiad

Show that $[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$.

(AoPS Vol. 2)

See Also