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LESSON 37 (6.1) LAW OF SINES
You should learn to:
1. Use the Law of Sines to solve oblique triangles (AAS or ASA – non-ambiguous cases).
2. Use the Law of Sines to solve oblique triangles (SSA – ambiguous case).
Terms to know: oblique triangle, Law of Sines, ambiguous case
So far, you have solved only right triangles. You will now learn two methods for solving oblique triangles (triangles
with no right angle). To solve oblique triangles, you need to know the measure of at least one side and the measures of
two other parts of the triangle. There are four possible cases for what you are given:
You will use Law of Sines when you are given:
(Use the Law of Sines whenever you can!)
1. Two angles and any side (AAS or ASA).
2. Two sides and an angle opposite one of them (SSA).
You will use Law of Cosines when you are given:
3. Two sides and their included angle (SAS).
4. Three sides (SSS)
Case 2 is particularly tricky. It is known as the ambiguous case for the Law of Sines.
Law of Sines: If ABC is a triangle with sides a, b, and c, then:
a
b
c
sin A sin B sin C


or


sin A sin B sin C
a
b
c
Example 1: (AAS) For the triangle ABC, A  30 , B  45 , and a  32 feet . Find the remaining angle and
sides. (Express the angle to the nearest degree and the sides to the nearest tenth of a foot).
C  180  30  45
B
c
C  105
45
32
A
30
b
C
b
32

sin 45 sin 30
32sin 45
b
sin 30
b  45.3 ft.
c
32

sin105 sin 30
32sin105
c
sin 30
c  61.8 ft.
Example 2: (ASA) Because of prevailing winds, a tree grew so that it was leaning 6 east of vertical. At a
point 30 meters west of the base of the tree, the angle of elevation to the top of the tree is 23 .
Find the length of the tree to the nearest tenth of a meter.
B
B  180  96  23
B  61
a
96
A
6
23
30 m.
C
a
30

sin 23 sin 61
30sin 23
a
sin 61
a  13.4 m.
In Examples 1 and 2, you found that two angles and one side (Case 1) determine exactly one triangle. However, if two
sides and one opposite angle are given (Case 2), three possible situations can occur:
1. No triangle exists.
2. Exactly one triangle exists.
3. Two different triangles exist.
You must decide which situation applies to be able to successfully solve the triangle(s).
Here is a suggested method (recipe) for solving triangles of the type SSA.
Start to solve “the triangle(s)” using the Law of Sines, and see which of the following 3 cases applies.
1. If there is no triangle, your calculator will give you an error message, because you will be trying to do an
inverse sine for a ratio bigger than 1.
2. If there is one triangle, the angle that you find using the Law of Sines will be smaller than the given angle,
or the angle that you find will be exactly 90 degrees.
3. If there are two triangles, the angle that you find will be larger than the given angle (but not 90 degrees).
To find the second possible angle (for your second triangle), subtract the angle value you found on the
calculator from 180 degrees.
Example 3: Suppose A  60 , a  4, and b  14. Find the remaining sides and angles for any possible
triangles.
sin B sin 60

14
4
14sin 60
sin B 
4
 14sin 60 
B  sin 1 

4


C
14
A
4
60
B
c
(The side length that should be 4
is actually not long enough to reach
the third side if A must be 60 )
Not Possible
Example 4: Suppose A  31 , a  12 inches, and b  5 inches. Find the remaining sides and angles for any
possible triangles. Express angles to the nearest degree and sides to the nearest tenth of an inch.
C
12
5
A
31
B
c
sin B sin 31

5
12
5sin 31
sin B 
12
 5sin 31 
B  sin 1 

 12 
B  12
C  180  31  12
C  137
c
12

sin137 sin 31
12sin137
c
sin 31
c  15.9 in.
Example 5: Suppose A  58 , a  4.5, and b  5. Find the remaining sides and angles for any possible
triangles. Express angles to the nearest degree and sides to the nearest tenth.
C
5
4.5
58
A
B
c
sin B sin 58

5
4.5
5sin 58
sin B 
4.5
 5sin 58 
B  sin 1 

 4.5 
B  70
C  180  58  70
C  52
c
4.5

sin 52 sin 58
4.5sin 52
c
sin 58
c  4.2
OR there’s a second case; the length of side “a” could swing “in” instead of “out”.
C
B  180  70
B  110
5
4.5
A
58
c
c
4.5

sin12 sin 58
4.5sin12
c
sin 58
c  1.1
C  180  58  110
C  12
For triangles fitting the SSA type, you should draw accurate sketches showing which situation applies.
ASSIGNMENT 37 (6.1): Pages 414-416 (1, 2, 4, 7, 14, 17, 26 (Note: 20 50  20.833 ), 39, 43, 44)