A polygon with three vertices. As demonstrated in the first figure, the angles of a triangle ABC add up to 180°. By considering separate areas, illustrated in the second figure, the area of the triangle equals ‘half base times height’.

If now A, B, and C denote the angles of the triangle, and a, b, and c the lengths of the sides opposite them, the following results hold:

Proving the angle sum formula

Determining the area of a triangle

1. (i) The area of the triangle equals $\frac{1}{2}bc\text{sin}A$.

2. (ii) The sine rule:

   $$\frac{a}{\text{sin}A}=\frac{b}{\text{sin}B}=\frac{c}{\text{sin}C}=2R,$$

   where R is the radius of the circumcircle.

3. (iii) The cosine rule:

   $$a^2=b^2+c^2-2bc\text{cos}A.$$

4. (iv) The tangent rule:

   $$\text{tan}\frac{B-C}{2}=\frac{b-c}{b+c}\text{cot}\frac{A}{2}.$$

5. (v) Hero’s formula: Let $s=\frac{1}{2}(a+b+c)$. Then the area of the triangle equals

   $$\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.$$

See acute, obtuse, oblique, equilateral, isosceles triangle, right-angled triangle, scalene triangle, solution of triangles, spherical triangle.