Difference between revisions of "Law of Tangents"
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<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)