Law of Sines

The Law of Sines is the relationship between the sides and angles of non-right (oblique) triangles.  Simply, it states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle.

In ΔABC is an oblique triangle with sides a,b and c , then a sinA = b sinB = c sinC .

To use the Law of Sines you need to know either two angles and one side of the triangle (AAS or ASA) or two sides and an angle opposite one of them (SSA).  Notice that for the first two cases we use the same parts that we used to prove congruence of triangles in geometry but in the last case we could not prove congruent triangles given these parts.  This is because the remaining pieces could have been different sizes.  This is called the ambiguous case and we will discuss it a little later.

Example 1: Given two angles and a non-included side (AAS).

Given ΔABC with mA=30° , mB=20° and a=45 m.  Find the remaining angle and sides.

       

       The third angle of the triangle is

              mC=180°mAmB=180°30°20°=130°

        By the Law of Sines,

               45 sin30° = b sin20° = c sin130°

       By the Properties of Proportions

               b= 45sin20° sin30° 30.78 m and c= 45sin130° sin30° 68.94 m

Example 2: Given two angles and an included side (ASA).

Given mA=42° , mB=75° and c=22 cm.  Find the remaining angle and sides.

       

       The third angle of the triangle is:

               mC=180°mAmB=180°42°75°=63°

       By the Law of Sines,

               a sin42° = b sin75° = 22 sin63°

       By the Properties of Proportions

               a= 22sin42° sin63° 16.52 cm and b= 22sin75° sin63° 23.85 cm

The Ambiguous Case

If two sides and an angle opposite one of them is given, three possibilities can occur.

            (1) No such triangle exists.

            (2) Two different triangles exist.

            (3) Exactly one triangle exists.

Consider a triangle in which you are given a,b and A .  (The altitude h from vertex B to side AC ¯ , by the definition of sines is equal to bsinA .)

            (1) No such triangle exists if A is acute and a<h or A is obtuse and ab .

                       

            (2) Two different triangles exist if A is acute and h<a<b .

                       

            (3) In every other case, exactly one triangle exists.

                       

Example 1: No Solution Exists

Given a=15,b=25 and mA=80° .  Find the other angles and side. 

               h=bsinA=25sin80°24.6

       

       Notice that a<h . So it appears that there is no solution.  Verify this using the Law of Sines.

               a sinA = b sinB 15 sin80° = 25 sinB sinB= 25sin80° 15 1.641>1

       This contrasts the fact that the 1sinB1 .  Therefore, no triangle exists.

Example 2: Two Solutions Exist

Given a=6,b=7 and mA=30° .  Find the other angles and side. 

               h=bsinA=7sin30°=3.5

              h<a<b therefore, there are two triangles possible.

       

       By the Law of Sines, a sinA = b sinB

               sinB= bsinA a = 7sin30° 6 0.5833

       There are two angles between 0° and 180° whose sine is approximately 0.5833, are 35.69° and 144.31° .

                   If  B35.69° C180°30°35.69°=114.31° c= asinC sinA 6sin114.31° sin30° 10.94 If  B144.31° C180°30°144.31°=5.69° c 6sin5.69° sin30° 1.19

Example 3: One Solution Exists

Given a=22,b=12 and mA=40° .  Find the other angles and side.

                 a>b

       

       By the Law of Sines, a sinA = b sinB

               sinB= bsinA a = 12sin40° 22 0.3506 B20.52°

        B is acute.

               mC=180°mAmB=180°40°20.52°=29.79°

       By the Law of Sines,

               c sin119.48° = 22 sin40° c= 22sin119.48° sin40° 29.79

If we are given two sides and an included angle of a triangle or if we are given 3 sides of a triangle, we cannot use the Law of Sines because we cannot set up any proportions where enough information is known.  In these two cases we must use the Law of Cosines.