Law of sines, cosines and tangents

Law of sines

Triangle ABC

The area of a triangle ABC is given by the formulas:

$A = \frac{a\times c \times \sin(B)}{2} = \frac{b \times c \times \sin(A)}{2} = \frac{a \times b \times \sin(C)}{2}$


$a\times c \times \sin(B) = b\times c \times \sin(A) = a \times b \times \sin(C)$

dividing by $a\times b\times c$, we get the sine formula:


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.

circle, triangle

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:


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.

Triangle ABC
a2 = b2 + c2 - 2bc cos(∠A)
b2 = c2 + a2 - 2ca cos(∠B)
c2 = a2 + b2 - 2ab cos(∠C)

Law of tangents

$\frac{a-b}{a+b}=\frac{ \tan\left[\frac{1}{2}(A - B)\right] }{ \tan\left[\frac{1}{2}(A + B)\right] }$

$\frac{b-c}{b+c}=\frac{ \tan\left[\frac{1}{2}(B - C)\right] }{ \tan\left[\frac{1}{2}(B + C)\right] }$

$\frac{a-c}{a+c}=\frac{ \tan\left[\frac{1}{2}(A - C)\right] }{ \tan\left[\frac{1}{2}(A + C)\right] }$


Sine rule problems
Cosine rule problems

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