Download Law of Sines

CH. 6 – ADDITIONAL TOPICS IN
TRIGONOMETRY
6.1 – Law of Sines
In this chapter, ΔABC has vertices A, B, and C, and
the sides opposite those vertices are a, b, and c,
respectively.
a
b
c
 Law of Sines:


sin A sin B sin C

This law can be used to find all angles and sides of an
oblique (not right) triangle
 You must know 3 angle or side measures to use this law,
and 2 of the knowns must be an opposite side/angle pair

C

Recall: Angles in a Δ sum to 180°!
a
b
A
c
B
C
a
b
c


sin A sin B sin C
102.3°
Ex: Solve the triangle.
27.4 ft
 To find A, subtract angles from 180°!


a
180 – 102.3 – 28.7 = 49°!
28.7°

To find c, use Law of Sines!
A
B
c
Hint: Use the given info instead of determined info when
possible in case you made a mistake!
 Hint: Don’t evaluate or round until you have the answer!
 Hint: We’re in degrees!

27.4
c

sin 28.7 sin102.3

27.4sin102.3
c
sin 28.7
c  55.75 ft
To find a, use Law of Sines!
27.4
a

sin 28.7 sin 49
27.4sin 49
a
sin 28.7
a  43.06 ft
C

Ex: Solve the triangle.


Can’t find anything but A right now!
To find A, use Law of Sines!
22
12

sin 42 sin A

Cross-multiply to get…
A
 12sin 42 
A  sin 1 

 22 
 To find C, subtract angles from 180°!
22sin A  12sin 42

12 in
22 in
42°
B
c
A  21.41
180 – 42 – 21.41 = 116.59°!
To find c, use Law of Sines!
22
c
22sin116.59

c
sin 42 sin116.59
sin 42


c  29.40 in
To get the best answer, try not to round until the end!
C
SPECIAL CASE #1

Ex: Solve the triangle.


Can’t find anything but A right now!
To find A, use Law of Sines!
11
31

sin 75 sin A

A
75°
c
Cross-multiply to get…
11sin A  31sin 75

31 in
11 in
 31sin 75 
A  sin 

11


1
ERR: DOMAIN!
Your calculator is telling you that there is no angle
measure for A that will create a triangle.
 Why? Because in reality, 11 inches is not long enough to
reach down to the 3rd side.


Final answer: NO SOLUTION!
B
C
SPECIAL CASE #2
Ex: Solve the triangle.
 To find A, use Law of Sines!
29
46

sin 31 sin A

Cross-multiply to get…
46 in
29 in

A
31°
c
 46sin 31 
A  sin 1 
A  54.78

 29 
 BUT WAIT! You just remembered that there is a 2nd
quadrant angle with the same sine, mainly 180-A =
125.22°!
 Therefore, there are 2 separate triangles that exist with the
given measurements! You must solve both triangles!
29sin A  46sin31
B
C
SPECIAL CASE #2 (CONT’D)
Triangle #1: A = 54.78°
29 in
 To find C, subtract the angles from 180!
46 in



180 – 31 – 54.78 = 94.21°!
To find c, use Law of Sines!
29
c

sin 31 sin 94.21
A
29sin 94.21
c
sin 31
31°
c
c  56.15 in
B
C
SPECIAL CASE #2 (CONT’D)
Triangle #2: A = 125.22°
29 in
 To find C, subtract the angles from 180!
46 in



180 – 31 – 125.22 = 23.78°!
To find c, use Law of Sines!
29
c

sin 31 sin 23.78

29sin 23.78
c
sin 31
c
c  22.71 in
When do we know if there will be 2 triangles formed?
1.
2.

A
31°
Only in the ASS condition
Only when the first angle found is greater than the given angle
Basically, consider a 2nd angle whenever you do sin-1!
B
AREA OF A TRIANGLE
1
1
1
 Area  ab sin C  ac sin B  bc sin A
2
2
2
C
a
b
A
c
B