Law of Tangents
Let a triangle have sides of length 
, 
, and 
and let the angles opposite these sides be denoted 
, 
, and 
. The law of tangents
states
![(a-b)/(a+b)=(tan[1/2(A-B)])/(tan[1/2(A+B)]).](/images/equations/LawofTangents/NumberedEquation1.gif) |
An analogous result for oblique spherical triangles states that
![(tan[1/2(a-b)])/(tan[1/2(a+b)])=(tan[1/2(A-B)])/(tan[1/2(A+B)]).](/images/equations/LawofTangents/NumberedEquation2.gif) |
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Law of Tangents
Let a triangle have sides of length
An analogous result for oblique spherical triangles states that
Wolfram Web Resources
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blancmange function, n=8