Difference between revisions of "Law of Tangents"
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<cmath> \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . </cmath> | <cmath> \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . </cmath> | ||
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) | + | <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 \frac{1}{2} (A-B)}{\tan \frac{1}{2} (A+B)} </cmath> |
as desired. <math>\square</math> | as desired. <math>\square</math> | ||
Revision as of 17:48, 2 March 2017
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. 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)