Download Geometry Section 9.7 The Law of Sines

Geometry Section 9.7
The Law of Sines
What you will learn:
1. Find areas of triangles
2. Use the Law of Sines to solve triangles.
3. Use the Law of Cosines to solve triangles
Example: Find the area of each triangle. The
1 bh
formula for the area of a triangle is ________
2
h
h
sin 56 
8
8 sin 56  h
h  6.632
A  1 12  6.632  39.792
2
h
70
sin 70 
h
12
12 sin 70  h
h  11.276
A  1 15 11.276  84.57
2
In sections 9.4 - 9.6, we learned how to use
trigonometry to solve right triangles. The Law
of Sines allows us to find sides and/or angles
in any triangle.
Theorem 9.9: Law of Sines
sin A
a
sin B
b
sin C
c
Examples: Solve each triangle. Round to the
nearest 1000th as necessary.
BC sin 24  2 sin 56
100
2 sin 56
BC 
sin 24
 4.077
AB sin 24  2 sin 100
mC  180  56  24  100
sin 24 sin 56 sin 100


2
BC
AB
2 sin 100
AB 
sin 24
 4.842
Examples: Solve each triangle. Round to the
nearest 1000th as necessary.
sin 29 sin A sin C


2
2.34
AB
34.557
2 sin A  2.34 sin 29
mC  180  29  34.557  116.443
2.34 sin 29
sin A 
2
1  2.34 sin 29 
A  sin 

2


AB sin 29  2 sin 116.443
 34.557
2 sin 116.443
AB 
sin 29
 3.694
Note that you cannot, at least initially, apply the
Law of Sines to solve the triangle at the right. Why
not.
sin 40 sin A sin B


c
20
25
To use the Law of Sines you must know an angle AND the opposite side.
Theorem 9.10: Law of Cosines
For any triangle ABC with sides a, b and c
opposite angles A, B and C respectively,
a  b  c  2bc cos A
2
2
2
b  a  c  2ac cos B
2
2
2
c  a  b  2ab cos C
2
2
2
Example: Solve the triangle at the top right.
c  a  b  2ab cos C
2
2
2
c  20  25  2(20)(25) cos 40
2
2
2
c  258.9555569
c  16.092
2
sin 40 sin A sin B


16.092
20
25
sin 40 sin A

16.092
20
16.092 sin A  20 sin 40
20 sin 40
sin A 
16.092
 20 sin 40 
A  sin 
  53.024
 16.092 
1
B  180  40  53.024  86.976
Example: Solve the triangle at the right.
NOTE: When given the 3 sides of a triangle, you
should always solve for the largest angle first.
b  a  c  2ac cos B
2
2
2
27  12  20  2(12)(20) cos B
729  144  400  480 cos B
 144
 400
1  185 
B  cos 

  480 
185  480 cos B
2
2
2
185
cos B 
 480
B  112.670
Example: Solve the triangle at the right.
NOTE: When given the 3 sides of a triangle, you
should always solve for the largest angle first.
sin C sin A sin 122.670


20
12
27
sin A sin 122.670

12
27
27 sin A  12 sin 122.670
12 sin 122.670
sin A 
27
 12 sin 122.670 
A  sin 

27


1
A  21.971
C  180  122.670  21.971  35.359
HW: pp513 – 514 /
7-10, 13, 14, 20, 21, 28, 30, 32