Difference between revisions of "Law of Tangents"

Revision as of 13:11, 2 February 2008

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 $AB$ .

(Mu Alpha Theta 1991)

Olympiad

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

(AoPS Vol. 2)

@@ Line 20: / Line 20: @@
 {{problem}}
 ===Intermediate===
-In <math>\triangle ABC</math>, LET <math>d</math> BE A POINT IN <math>bc</math> SUCH THAT <math>ad</math> bisects <math>\angle A</math>. Given that <math>AD=6,BD=4</math>, and <math>DC=3</math>, find <math>AB</math>.
+In <math>\triangle ABC</math>, let <math>D</math> be a point in <math>BC</math> such that <math>AD</math> bisects <math>\angle A</math>. Given that <math>AD=6,BD=4</math>, and <math>DC=3</math>, find <math>AB</math>.
 <div align="right">([[Mu Alpha Theta]] 1991)</div>
 ===Olympiad===

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