Use the Law of Sines to solve triangles

1. 8.1 Law of Sines
Chapter 8 Applications of Trigonometry

2. Concepts and Objectives
 Law of Sines
 Use the law of sines to solve a triangle for a missing
side or angle.
 Resolve ambiguous cases of the law of sines.
 Find the area of a triangle using sine.

3. Beyond Right Angle Trigonometry
 When we first started talking about trigonometry, we
started off with right triangles and defined the trig
functions as ratios between the various sides.
 It turns out we can extend these ratios to all triangles,
such that if any three of the six side and angle measures
of a triangle are known (including at least one side), then
the other three can be found.

4. Law of Sines
In any triangle ABC, with sides a, b, and c,
(i.e. the lengths of the sides in a triangle are
proportional to the sines of the measures of the
angles opposite them).
 
sin sin sin
a b c
A B C
CB
A
c b
a

5. Law of Sines
To solve for an angle, we can also write this as
 
sin sin sinA B C
a b c
CB
A
c b
a
When using the law of sines, a good strategy is to select
an equation so that the unknown variable is in the
numerator and all other variables are known.

6. Law of Sines
 Example: Solve triangle BUG if B = 32.0°, U = 81.8°, and
b = 42.9 cm.
G
UB g
u 42.9 cm
32.0° 81.8°

7. Law of Sines
 Example: Solve triangle BUG if B = 32.0°, U = 81.8°, and
b = 42.9 cm.
G
UB g
u 42.9 cm
32.0° 81.8°

sin sin
b u
B U

 
42.9
sin32.0 sin81.8
u



42.9sin81.8
sin32.0
u
80.1 cm

8. Law of Sines
 Example: Solve triangle BUG if B = 32.0°, U = 81.8°, and
b = 42.9 cm.
G
UB g
80.1 cm 42.9 cm
32.0° 81.8°
To find G, recall that the angles
of a triangle add up to 180°.
Therefore, G = 66.2°.
Now that we have G, we can
find g:

 
42.9
sin32.0 sin66.2
g
74.1 cmg

9. Law of Sines
 If we are given the lengths of two sides and the angle
opposite one of them, then zero, one, or two such
triangles may exist. This situation is called the
ambiguous case of the law of sines.

10. Law of Sines
 Example: Solve triangle PIE if I = 55°, i = 8.94 m, and
p = 25.1 m.

11. Law of Sines
 Example: Solve triangle PIE if I = 55°, i = 8.94 m, and
p = 25.1 m.
Since the sine of an angle cannot be greater than 1, there
can be no such angle.

sin sinP I
p i


sin sin55
25.1 8.94
P

 
25.1sin55
sin 2.29986
8.94
P

12. Law of Sines
 Applying the law of sines:
 For any angle  of a triangle, 0 < sin   1. (If sin  = 1,
then  = 90° and the triangle is a right triangle.)
 sin  = sin180° –  (Supplementary angles have the
same sine value.)
 The smallest angle is opposite the shortest side, the
largest angle is opposite the longest side, and the
middle-valued angle is opposite the intermediate
side.

13. Law of Sines
 Example: Solve triangle ABC if A = 55.3°, a = 22.8 ft, and
b = 24.9 ft.

14. Law of Sines
 Example: Solve triangle ABC if A = 55.3°, a = 22.8 ft, and
b = 24.9 ft.
Use the law of sines to find angle B:


sin55.3 sin
22.8 24.9
B


24.9sin55.3
sin
22.8
B
sin .8978678B

15. Law of Sines
 Example: Solve triangle ABC if A = 55.3°, a = 22.8 ft, and
b = 24.9 ft.
sin B ≈ .8978678
There are two angles B between 0° and 180° that satisfy
this condition.
sin–1 .8978678 ≈ 63.9° (B1)
Supplementary angles have the same sine value, so
another possible value of B is
180° – 63.9° = 116.1° (B2)
Since A + B < 180°, both values will work.

16. Law of Sines
 Example: Solve triangle ABC if A = 55.3°, a = 22.8 ft, and
b = 24.9 ft.
B1 = 63.9° B2 = 116.1°
C1 = 60.8° C2 = 8.6°
22.8
sin60.8 sin55.3
c

 
22.8
sin8.6 sin55.3
c

 
c1 = 24.2 ft
22.8sin60.8
sin55.3
c



c2 = 4.1 ft
22.8sin8.6
sin55.3
c




17. Area Formulas
 If a triangle has sides of lengths a, b, and c, the area is
given by the following formula:
 If the included angle measures 90°, its sine is 1, and the
formula becomes the familiar
  A A A
1 1 1
sin or sin or sin
2 2 2
bc A ab C ac B
A=
1
2
bh

18. Area Formulas
 Example: Find the area of triangle ABC if B = 55°10,
a = 34.0 ft, and c = 42.0 ft.

19. Area Formulas
 Example: Find the area of triangle ABC if B = 55°10,
a = 34.0 ft, and c = 42.0 ft.
A
1
sin
2
ac B
    
   
 
1 10
34.0 42.0 sin 55
2 60
 2
586 ft
(You can also use the nSpire’s t function to
convert the minutes.)

20. Classwork
 College Algebra
 Page 742: 8-20 (4), page 606: 32-42 (even),
page 595: 48-64 (4)