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
Law of Sines is the trigonometric counterpart of the geometrical criteria of congruence AAS and ASA,
The law of sines states that in any triangle the ratio of the Sine of each angle to its opposite side is a constant.
Thus in ΔABC,

$\frac{SinA}{a}$ = $\frac{SinB}{b}$ = $\frac{SinC}{c}$

or equivalently taking the reciprocal rations,

$\frac{a}{SinA}$ = $\frac{b}{SinB}$ = $\frac{c}{SinC}$

Law of sines is used to solve triangles when
  • Two angles and any side is given, or
  • Two sides and an angle opposite to any one of them is given.

The second case SSA does not correspond to a geometrical criterion of congruence. This may lead to two solutions (ambiguous case), one with an acute angle and the other with an obtuse angle which is the supplement of the acute angle solved.

Let us solve example problems for each situation
Case (1): Two angles and a side are known:

In ΔABC shown here, find the exact length a.
Two angles A and B are given and b opposite
to angle B is given. Hence using Sine rule
$\frac{a}{SinA}$ = $\frac{b}{SinB}$
$\frac{a}{Sin60}$ = $\frac{12}{Sin45}$
Solving for a
a = Sin 60 x $\frac{12}{Sin45}$
Now we use the exact values Sin 45 and Sin 60
from a table or from unit circle,
a = $\frac{\sqrt{3}}{2}$ x $\frac{12}{\frac{\sqrt{2}}{2}}$
= $\frac{\sqrt{3}}{2}$ x 12√2 = 6√6
Sine Rule

Case (2): Two sides and an angle opposite to one of them is known:
Solve for ΔPQR, given that p = 16, q = 8 and ∠P = 48º
For this case the angle opposite to the second side given is always found first.
Applying sine rule for the situation,

$\frac{SinP}{p}$ = $\frac{SinQ}{q}$ = $\frac{SinR}{r}$

$\frac{Sin48}{16}$ = $\frac{SinQ}{8}$

Sin Q = $\frac{8}{16}$ x Sin 48 ≈ 0.3716
∠Q = Sin-1 (0.3716) = 21.8º Acute angle solved
∠Q' = 180 - 21.8 = 158.2º Corresponding Obtuse angle.
But ∠P + ∠Q' = 48 + 158.2 > 180. Thus the third angle is not possible.
Hence we have only one solution for the given conditions.

∠R = 180 - (∠P + ∠Q) = 180 - (48 + 21.8) = 110.2º
Applying sine rule again to find r,

$\frac{p}{SinP}$ = $\frac{r}{SinR}$

r = Sin R x $\frac{p}{SinP}$

= Sin(110.2) x $\frac{16}{sin48}$

≈ 20.2

Solved Examples

Question 1: In a children's amusement park, the administration wanted to build a walker's bridge along a Pond in the midst of the park from point A to B. For this purpose an Engineer identified the approximate straight edge of the Pond near B and marked a point C at a distance 12 m from B. He was able to measure the angles ACB and ABC as 102º and 68º to estimate the length of the bride to be built. Find the approximate length of the bridge rounded to a meter.

Law of Sine and Cosines Word Problem
Two angles and one side are given. Hence the problem is to be solved using Law of Sines.
     As the measure of side BC is given as 12 m we need to know the measure of the angle opposite to it that is ∠A.
     ∠A = 180 - (∠B + ∠C) = 180 - (68 + 102) = 10º.
    The Length of the bridge is represented by AB which is the side opposite to ∠C.
     Using sine rule,

     $\frac{a}{SinA}$ = $\frac{c}{SinC}$

     $\frac{12}{Sin10}$ = $\frac{c}{Sin102}$

     c = Sin 102 x $\frac{12}{Sin10}$                Proportion solved for 'c'.

        ≈ 68 m                                                     rounded to the nearest meter.

     Hence the estimated length of the proposed bridge = 68 m.

Question 2: To solve the triangle given below Amy and Molly described the method to follow with reasoning.

     Law of Sines and Cosines Word Problems

     Molly wrote: "Begin by using Law of Cosines as you are given two sides and the included angle."
     Amy wrote:  "Begin by using Law of Sines as you are given two sided and an angle opposite to one of them."
     Explain who is right and solve the problem using that method.
Amy is correct. Molly has noted the angle A wrongly as the included angle. It is indeed the angle opposite to side BC
     which is given as 22.
     Applying law of Sines,
     $\frac{SinA}{a}$ = $\frac{SinB}{b}$ = $\frac{SinC}{c}$

     $\frac{Sin35}{22}$ = $\frac{SinB}{38}$

     Sin B = 38 x $\frac{Sin35}{22}$ = 0.9907

     Two solutions are possible.  ∠B = Sin-1 (0.9907) ≈ 82º.
      The obtuse angle for the other solution ∠B' = 180 - 82 = 98º.

     The corresponding third angles are hence,
     ∠C = 180 - (∠A + ∠B) = 180 - (35 + 82) = 63º
     ∠C' = 180 -(∠A + ∠B') = 180 - (35 + 98) = 47º

     The corresponding third sides c and c' are calculated using the sine rule as follows:

      $\frac{c}{Sin63}$ = $\frac{22}{Sin35}$

       c = Sin 63 x $\frac{22}{Sin35}$  = 34

     $\frac{c'}{Sin47}$ = $\frac{22}{Sin35}$

      c' = Sin 47 x $\frac{22}{Sin35}$  = 28.

    Hence the two solutions are A = 35º, B = 82º, C = 63º, a = 22, b = 38 and c = 34
                                                A = 35º, B = 98º, C = 47º, a = 22,  b =38 and c = 28.
Note: Do not go by the appearance of the angle in the picture. If two solutions are possible for the given info, find both the solutions.

Question 3: A hypothetical flight route between two cities A and B is of distance 125 miles. Due to bad weather, the pilot first flew from city A to city C which was at a distance of 55 miles and then turned the flight and flew to city B. The distance between the cities B and C is 140 miles. Find the angles between the directions between the three cities.

Law of Sines and Cosines Problems
The three sides of the triangle are known and the triangle is to be solved for the angles. Hence this is an application of Law of Cosines.

    Cos A = $\frac{b^{2}+c^{2}-a^{2}}{2bc}$

              = $\frac{55^{2}+125^{2}-140^{2}}{2\times 55\times 125}$

              ≈ -0.0691

           A = Cos-1 (-0.0691) ≈ 94º

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

              = $\frac{125^{2}+140^{2}-55^{2}}{2\times 125\times 140}$

              = 0.92

          B = Cos-1 (0.92) = 23º

          C = 180 - (A + B) = 180 - (94 + 23) = 63º.

Deviation between directions AB and AC = 94º.
Deviation between directions BA and BC = 23º
Deviation between directions CA and CB = 63º.