The document discusses the Law of Sines theorem and how to use it to solve for unknown angles and side lengths in triangles. It provides examples of applying the Law of Sines to find the distance between a cell phone tower and a school, as well as to calculate unknown angles in different triangles where some combination of angles and side lengths are given. In one example, it is not possible to find the measure of the unknown angle because there is insufficient information provided.

1. Section 5-4
T h e L a w o f S i n e s

2. Theorem:
SAS Area Formula for a
triangle

3. Theorem:
SAS Area Formula for a
triangle
The area for any triangle is half the product of the lengths of
two sides and the sine of the angle between them

4. Theorem:
SAS Area Formula for a
triangle
The area for any triangle is half the product of the lengths of
two sides and the sine of the angle between them
This leads us to our next theorem...

5. Theorem:
Law of Sines

6. Theorem:
Law of Sines
In any triangle ABC:
sin A sin B sinC
= =
a b c

7. Theorem:
Law of Sines
In any triangle ABC:
sin A sin B sinC
= =
a b c
C
b
a A
B c

8. Example 1
From a specific point in a school classroom, students can see the
the top of a cell phone tower if they look 41° west of north.
Traveling 4.1 miles due west from the school, the same cell
phone tower is seen in the direction 26° west of north. How far
is the school from the top of the cell phone tower?

9. Example 1
From a specific point in a school classroom, students can see the
the top of a cell phone tower if they look 41° west of north.
Traveling 4.1 miles due west from the school, the same cell
phone tower is seen in the direction 26° west of north. How far
is the school from the top of the cell phone tower?
First, we need to set up our triangle.

10. Example 1
From a specific point in a school classroom, students can see the
the top of a cell phone tower if they look 41° west of north.
Traveling 4.1 miles due west from the school, the same cell
phone tower is seen in the direction 26° west of north. How far
is the school from the top of the cell phone tower?
First, we need to set up our triangle.
Tower
15°
x
116° 49°
School
4.1 miles

11. Example 1
From a specific point in a school classroom, students can see the
the top of a cell phone tower if they look 41° west of north.
Traveling 4.1 miles due west from the school, the same cell
phone tower is seen in the direction 26° west of north. How far
is the school from the top of the cell phone tower?
First, we need to set up our triangle.
Tower sin116° sin15°
=
15°
x x 4.1
116° 49°
School
4.1 miles

12. Example 1
From a specific point in a school classroom, students can see the
the top of a cell phone tower if they look 41° west of north.
Traveling 4.1 miles due west from the school, the same cell
phone tower is seen in the direction 26° west of north. How far
is the school from the top of the cell phone tower?
First, we need to set up our triangle.
Tower sin116° sin15°
=
15°
x x 4.1
4.1sin116°
116°
4.1 miles
49°
School x=
sin15°

13. Example 1
From a specific point in a school classroom, students can see the
the top of a cell phone tower if they look 41° west of north.
Traveling 4.1 miles due west from the school, the same cell
phone tower is seen in the direction 26° west of north. How far
is the school from the top of the cell phone tower?
First, we need to set up our triangle.
Tower sin116° sin15°
=
15°
x x 4.1
4.1sin116°
116°
4.1 miles
49°
School x=
sin15°
x ≈ 14.23796149

14. Example 1
From a specific point in a school classroom, students can see the
the top of a cell phone tower if they look 41° west of north.
Traveling 4.1 miles due west from the school, the same cell
phone tower is seen in the direction 26° west of north. How far
is the school from the top of the cell phone tower?
First, we need to set up our triangle.
Tower sin116° sin15°
=
15°
x x 4.1
4.1sin116°
116°
4.1 miles
49°
School x=
sin15°
x ≈ 14.23796149 miles

15. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.

16. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
X
38°
12
Y Z
10

17. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
sin 38° sin Z
X =
10 12
38°
12
Y Z
10

18. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
sin 38° sin Z
X =
10 12
38°
12 12sin 38°
= sin Z
10
Y Z
10

19. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
sin 38° sin Z
X =
10 12
38°
12 12sin 38°
= sin Z
10
Y Z
10 sin Z ≈ .7387937704

20. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
sin 38° sin Z
X =
10 12
38°
12 12sin 38°
= sin Z
10
Y Z
10 sin Z ≈ .7387937704
⎛ 12sin 38° ⎞
sin −1
( ) −1
sin Z ≈ sin ⎜
⎝ 10 ⎟ ⎠

21. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
sin 38° sin Z
X =
10 12
38°
12 12sin 38°
= sin Z
10
Y Z
10 sin Z ≈ .7387937704
⎛ 12sin 38° ⎞
sin −1
( ) −1
sin Z ≈ sin ⎜
⎝ 10 ⎟ ⎠
Z ≈ 47.62876444°

22. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
sin 38° sin Z
X =
10 12
38°
12 12sin 38°
= sin Z
10
Y Z
10 sin Z ≈ .7387937704
⎛ 12sin 38° ⎞
sin −1
( )
sin Z ≈ sin ⎜
⎝
−1
10 ⎟ ⎠
Z ≈ 47.62876444°
or

23. Example 2
In  XYZ , XY = 12, YZ = 10, and m∠X = 38°. Find m∠Z.
sin 38° sin Z
X =
10 12
38°
12 12sin 38°
= sin Z
10
Y Z
10 sin Z ≈ .7387937704
⎛ 12sin 38° ⎞
sin −1
( )
sin Z ≈ sin ⎜
⎝
−1
10 ⎟ ⎠
Z ≈ 47.62876444°
or
Z ≈ 132.3712356°

24. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.

25. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.
X
38°
12
Y Z
20

26. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.
X sin 38° sin Z
=
38° 20 12
12
Y Z
20

27. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.
X sin 38° sin Z
=
38° 20 12
12
12sin 38°
= sin Z
Y Z 20
20

28. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.
X sin 38° sin Z
=
38° 20 12
12
12sin 38°
= sin Z
Y Z 20
20
−1 ⎛ 12sin 38° ⎞
sin −1
( )
sin Z = sin ⎜
⎝ 20 ⎠ ⎟

29. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.
X sin 38° sin Z
=
38° 20 12
12
12sin 38°
= sin Z
Y Z 20
20
−1 ⎛ 12sin 38° ⎞
sin −1
( )
sin Z = sin ⎜
⎝ 20 ⎠ ⎟
Z ≈ 21.67842645°

30. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.
X sin 38° sin Z
=
38° 20 12
12
12sin 38°
= sin Z
Y Z 20
20
−1 ⎛ 12sin 38° ⎞
sin −1
( )
sin Z = sin ⎜
⎝ 20 ⎠ ⎟
Z ≈ 21.67842645°
or

31. Example 3
In  XYZ , XY = 12, YZ = 20, and m∠X = 38°. Find m∠Z.
X sin 38° sin Z
=
38° 20 12
12
12sin 38°
= sin Z
Y Z 20
20
−1 ⎛ 12sin 38° ⎞
sin −1
( )
sin Z = sin ⎜
⎝ 20 ⎠ ⎟
Z ≈ 21.67842645°
or
Z ≈ 152.3215736°

32. Example 4
In  XYZ , XY = 12, YZ = 5, and m∠X = 38°. Find m∠Z.

33. Example 4
In  XYZ , XY = 12, YZ = 5, and m∠X = 38°. Find m∠Z.
X
38°
12
Y Z
5

34. Example 4
In  XYZ , XY = 12, YZ = 5, and m∠X = 38°. Find m∠Z.
X
38° Wait a minute! I can’t get an answer?!
12
Y Z
5

35. Example 4
In  XYZ , XY = 12, YZ = 5, and m∠X = 38°. Find m∠Z.
X
38° Wait a minute! I can’t get an answer?!
12
Why not?
Y Z
5

36. Example 4
In  XYZ , XY = 12, YZ = 5, and m∠X = 38°. Find m∠Z.
X
38° Wait a minute! I can’t get an answer?!
12
Why not?
Y Z
5
12sin 38°
≈
5

37. Example 4
In  XYZ , XY = 12, YZ = 5, and m∠X = 38°. Find m∠Z.
X
38° Wait a minute! I can’t get an answer?!
12
Why not?
Y Z
5
12sin 38°
≈ 1.477587541
5

38. Example 4
In  XYZ , XY = 12, YZ = 5, and m∠X = 38°. Find m∠Z.
X
38° Wait a minute! I can’t get an answer?!
12
Why not?
Y Z
5
12sin 38°
≈ 1.477587541 That’s outside the domain.
5

39. If we have two angles, how many possibilities will we look for in
the answers?

40. If we have two angles, how many possibilities will we look for in
the answers?
Only one.

41. If we have two angles, how many possibilities will we look for in
the answers?
Only one.
If we have two sides, how many possibilities will we look for in
the answers?

42. If we have two angles, how many possibilities will we look for in
the answers?
Only one.
If we have two sides, how many possibilities will we look for in
the answers?
There are two possible answers.

43. Homework

44. Homework
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