The Laws of Sines and Cosines are commonly used to solve Oblique triangles and thus form the basis for important trigonometric applications. The congruence criteria state the required measures of a triangle to determine the size and shape of that triangle. Laws of Sines and Cosines are trigonometric counterparts of the Geometric criteria of Congruence AAS, ASA, SAS and SSS.Let us learn the formulas given by the laws of Sines and Cosines, and how these laws are used in solving Oblique Triangles and also solve few word problems based on these.

According to congruence criteria SAS and SSS, the size and shape of a triangle is determined by the measures of two sides and the included angle and three sides respectively. Law of Cosines establishes this fact trigonometrically and is applied in solving Oblique triangle under these situations.
The sides in ΔABC are represented by lower case letters a, b, c corresponding to angles opposite to them A, B and C.

Triangle Notation

The version of Law of Cosines relating a side to the angle opposite to it and the other two sides is as follows:
a2 = b2 + c2 - 2bc Cos A
b2 = c2 + a2 - 2ca Cos B
c2 = a2 + b2 - 2ab Cos C

The version of law of cosines expressing the cosine of an angle in terms of the sides is
Cos A = $\frac{b^{2}+c^{2}-a^{2}}{2bc}$

Cos B = $\frac{c^{2}+a^{2}-b^{2}}{2ca}$

Cos C = $\frac{a^{2}+b^{2}-c^{2}}{2ab}$

There is an element of symmetry in the three formulas given for each group. By remembering one formula in each of two above groups, the other formulas can be got successively by replacing a, b and c by b, c and a.

Let us see how the law of cosines is used to solve triangles:

Case (1): Two sides and the included angle are known.
Solve ΔABC fully given that b = 12, c = 8 and m ∠A = 25º rounding to the tenth of the measure.
We can first find the third side a, using law of cosines.
a2 = b2 + c2 - 2bc Cos A
= 122 + 82 - 2(12)(8) Cos 25
= 144 + 64 - 192 Cos 25 = 208 - 192 Cos 25 ≈ 33.99

a = √33.99 = 5.8 (rounded to the tenth)
Now to find the measure of angle B applying cosine rule,
Cos B = $\frac{c^{2}+a^{2}-b^{2}}{2ca}$

= $\frac{8^{2}+(5.8)^{2}-12^{2}}{2\times 8\times 5.8}$

≈ - 0.4995

m ∠B = Cos-1(-0.4995) ≈ 120º.
m ∠C = 180 - (m ∠A + m ∠B)
= 180 - ( 25 + 120) = 35º.

Case (2): Three sides of the triangle are known:
Find the angles of ΔABC given that a = 15, b = 12 and c = 8.
Two angles of the triangle can be found using Law of Cosines twice.
Cos A = $\frac{b^{2}+c^{2}-a^{2}}{2bc}$

= $\frac{12^{2}+8^{2}-15^{2}}{2\times 12\times 8}$

≈ -0.0885
m ∠A = Cos(-0.0885) = 95.1º

Cos B = $\frac{c^{2}+a^{2}-b^{2}}{2ca}$

= $\frac{8^{2}+15^{2}-12^{2}}{2\times 8\times 15}$

≈ 0.6042
m ∠B = Cos-1(0.6042) = 52.8º
m ∠C = 180 - (m ∠A + m ∠B)
= 180 - (95.1 + 52.8) = 32.1º