Law of sines, cosines and tangents
Law of sines
The area of a triangle ABC is given by the formulas:
=>
dividing by $a\times b\times c$, we get the sine formula:
$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$
Let R be the radius of a circle with center O through points A,B and C(for every 3 points that do not lie on a straight line there is exactly 1 circle through these points) of a triangle ABC.
Let B' be the second intersection point of BO and the circle. The
angle B' in the triangle BB'C is equal to A, and the triangle BB'C is a right triangle
=> a = 2Rsin(B') = 2Rsin(A) therefore:
$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}=2R$
Law of cosines
Let a(the length of BC), b(the length of CA), c(the length of AB) be the lengths of the sides of a triangle ABC.
b2 = c2 + a2 - 2ca cos(∠B)
c2 = a2 + b2 - 2ab cos(∠C)
Law of tangents
$\frac{a-b}{a+b}=\frac{ \tan\left[\frac{1}{2}(A - B)\right] }{ \tan\left[\frac{1}{2}(A + B)\right] }$
$\frac{b-c}{b+c}=\frac{ \tan\left[\frac{1}{2}(B - C)\right] }{ \tan\left[\frac{1}{2}(B + C)\right] }$
$\frac{a-c}{a+c}=\frac{ \tan\left[\frac{1}{2}(A - C)\right] }{ \tan\left[\frac{1}{2}(A + C)\right] }$