Sine Rule
In a triangle ABC the area is:
S = a.c.sin(B)/2 = b.c.sin(A)/2 = a.b.sin(C)/2
=>
a.c.sin(B) = b.c.sin(A) = a.b.sin(C)
dividing through by a.b.c, we get the sine formula
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Say R is the radius of the circle with center O through the points A,B and C(for every 3 points that do not lie in a line there is 1 circle that the points belong to) of triangle ABC.
Let B' be the second intersection point of BO and the circle. The
angle B' in triangle BB'C is equal to A, and the triangle BB'C is a right triangle
=> a = 2Rsin(B') = 2Rsin(A) then we have:
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= | 2R |
Cosine(Cos) Rule
Let's a(the length of BC), b(the length of CA), c(the length of AB) are the lengths of the sides of a triangle ABC. Cosine formulas are:
a2 = b2 + c2 - 2bc cos(∠A)
b2 = c2 + a2 - 2ca cos(∠B)
c2 = a2 + b2 - 2ab cos(∠C)
where cos(∠A) is cos(∠CAB), cos(∠B) is cos(∠ABC) and cos(∠C) is cos(∠BCA)
b2 = c2 + a2 - 2ca cos(∠B)
c2 = a2 + b2 - 2ab cos(∠C)
Problems
Sine rule problems
Cosine rule problems
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