# Difference between revisions of "Law of Tangents"

(→Proof) |
|||

Line 12: | Line 12: | ||

By the angle addition identities, | By the angle addition identities, | ||

<cmath> \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2} </cmath> | <cmath> \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2} </cmath> | ||

− | as desired. <math>\ | + | as desired. <math>\square</math> |

==Problems== | ==Problems== |

## Revision as of 00:39, 15 December 2009

The **Law of Tangents** is a rather obscure trigonometric identity that is sometimes used in place of its better-known counterparts, the law of sines and law of cosines, to calculate angles or sides in a triangle.

## Statement

If and are angles in a triangle opposite sides and respectively, then

## Proof

Let and denote , , respectively. By the Law of Sines, By the angle addition identities, 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 , let be a point in such that bisects . Given that , and , find .

(Mu Alpha Theta 1991)

### Olympiad

Show that .

(AoPS Vol. 2)