Law of Sines

• Given the triangle below …

… the law of sines is given by …

sin sin sin

a b c

Law of Sines• Note that in each ratio, the sine of the angle is

written over the length of the side opposite that angle.

sin sin sin

a b c

• Note also that the triangle is not a right triangle, so the pythagorean theorem cannot be used.

Example 1:Solve the triangle with the given measures:

30 , 110 , 10 b cm

180

• Since the three angles of a triangle add up to 180 degrees …

30 110 180

40

30 , 110 , 10 b cm

sin sin

a b

• Now use the law of sines. Since side b is given, one ratio will include side b and angle β (the angle opposite side b).

• The other ratio is our choice since we know the value of both angles.

sin 30 sin110

10

a

sin 30 sin110

10

a

sin110 10sin 30a

10sin 30

sin110a

5.3a cm

• Note: make sure the calculator is set to degree mode.

30 , 110 , 10 b cm

sin sin

c b

• Now find the last side c. Use side b for the other ratio since it is given. Using the rounded value of a would lead to further rounding error.

sin 40 sin110

10

c

5.3a40

• All missing measures have been found and the triangle is solved.

sin 40 sin110

10

c

sin110 10sin 40c

10sin 40

sin110c

6.8c cm

Example 2:Solve the triangle with the given measures:

52 , 6 , 5 a in b in

• Since side b and angle β are both given, use them for one ratio.

• Since side a is given, use it for the other ratio.

sin sin

a b

52 , 6 , 5 a in b in

sin sin 52

6 5

5sin 6sin 52

6sin52sin

5

1 6sin52sin5

71

6sin52sin

5

• Note: make sure the calculator is set to degree mode.

• We know two angles now, so find the third angle.

52 , 6 , 5 a in b in

71

180

71 52 180

57

• Find the remaining side.

52 , 6 , 5 a in b in

71 57

sin sin

c b

sin 57 sin 52

5

c

sin 57 sin 52

5

c

sin 52 5sin 57 c

5sin57

sin52c

5.3c in

• All missing measures have been found and the triangle is solved.

• Before we move on, consider one of the calculations that was made in this problem.

1 6sin52sin5

71

• The calculator gave us a value of …

1sin 0.9456

• Consider a unit circle with the given information.

• There is another possible value for the angle. Using a reference angle of 71 degrees …

… we find another angle that will also solve the equation.

180 71 109

• Note that 109 degrees is possible for an angle in a triangle.

• Recall the sides and angles that have been found up to this point

52 , 6 , 5 a in b in• Given:

• Determined:

71

57

5.3c in

• Now consider a second possible triangle.

• Use the second value found for α and re-solve the triangle.

52 , 6 , 5 a in b in

109

180

109 52 180

19

52 , 6 , 5 a in b in

109 19

sin sin

c b

sin19 sin 52

5

c

sin 52 5sin19 c

sin 52 5sin19 c

5sin19

sin52c

2.1c in

• The given information led to two different triangles.

71

57

5.3c in

109

19

2.1c cm