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Trigonometric Functions:
Right Triangle Approach
Copyright © Cengage Learning. All rights reserved.
6.5 The Law of Sines
Copyright © Cengage Learning. All rights reserved.
The Law of Sines!
Be able to use the Law of Sines to find unknowns
in triangles!
Quick Review:
What does Soh-Cah-Toa stand for?
adj
opp
opp
tan

cos

sin 
adj
hyp
hyp
What kind of triangles do we use this for?
right triangles
What if it’s not a right triangle? GASP!!
What do we do then??
The Law of Sines:
B
sin A sin B  sin C 


a
b
c
c
A
a
b
C
Note:
 capital letters always stand
for __________!
angles
Use the Law of Sines ONLY when:
 lower-case letters always
stand for ________!
sides
 you know an angle and its opposite side
 you DON’T have a right triangle AND
The Law of Sines
• The Law of Sines says that in any triangle the
lengths of the sides are proportional to the
sines of the corresponding opposite angles.
Let’s do some problems! 
Ex. 1: Use the Law of Sines to find each missing angle or side. Round any
decimal answers to the nearest tenth.
63°
29
A
sin A sin B  sin C 


a
b
c
a
79˚
38˚
42
C
sin 63 sin C 

42
29
29 sin 63  42 sin C
29 sin 63
 sin C 
42
.6152 ...  sin C 
38.0  C
sin 63 sin 79

42
a
a sin 63  42 sin 79
42 sin 79
a
sin 63
a  46.3
Ex. 2: Use the Law of Sines to find each missing angle or side.
Round any decimal answers to the nearest tenth.
T
sin A sin B  sin C 


a
b
c
s
40°
sin 51 sin 40

4.8
r
51˚ r
89°
4.8
sin 51 sin 89

4.8
s
r sin 51  4.8 sin 40
4.8 sin 40
r
sin 51
s sin 51  4.8 sin 89
4.8 sin 89
s
sin 51
r  4.0
s  6.2
Ex. 3: Draw ΔABC and mark it with the given information. Solve the
triangle. Round any decimal answers to the nearest tenth.
B
a.
a  7, mA  37, mB  76
sin A sin B  sin C 


a
b
c
sin 37 sin 76

7
b
C  67
A
c
76˚ 7
37˚
67˚
b
sin 37 sin 67

7
c
b sin 37  7 sin 76 c sin 37  7 sin 67
7 sin 67
7 sin 76
c
b
sin 37
sin 37
b  11.3
c  10.7
C
Tracking a Satellite
• A satellite orbiting the earth passes directly
overhead at observation stations in Phoenix
and Los Angeles, 340 mi apart.
At an instant when the satellite is between
these two stations, its angle of elevation is
simultaneously observed to be 60 at Phoenix
and 75 at Los Angeles.
How far is the satellite from Los Angeles?
Solution
• We need to find the distance b in Figure 4.
Figure 4
• Since the sum of the angles in any triangle is
180, we see that C = 180 – (75 + 60) = 45
(see Figure 4) so we have
Law of Sines
Solution
cont’d
Substitute
Solve for b
• The distance of the satellite from Los Angeles
is approximately 416 mi.
• Because of prevailing winds, a tree grew so
that it was leaning 6° from the vertical. At a
point 100 feet from the tree, the angle of
elevation to the top of the tree is 22.8°. Find
the height of the tree.
The Ambiguous Case
HOW MANY TRIANGLES
Suppose we are given two angles and the
included side, then it’s clear that one and only
one triangle can be formed (see Figure 2(a)).
ASA or SAA
Figure 2(a)
• Similarly, if two sides and the included angle are known,
then a unique triangle is determined (Figure 2(c)).
SAS
Figure 2 (c)
• But if we know all three angles and no sides, we cannot
uniquely determine the triangle because many triangles
can have the same three angles.
• In general, a triangle is determined by three of its
six parts (angles and sides) as long as at least one
of these three parts is a side. So the possibilities
are:
(a) ASA or SAA
(b) SSA
(c) SAS
(d) SSS
Case 1 One side and two angles (ASA or SAA)
Case 2 Two sides and the angle opposite one of those
sides (SSA)
Case 3 Two sides and the included angle (SAS)
Case 4 Three sides (SSS)
(a) ASA or SAA
(b) SSA
(c) SAS
(d) SSS
The Ambiguous Case
• In Example 1 a unique triangle was
determined by the information given.
This is always true of Case 1 (ASA or SAA). But
in Case 2 (SSA) there may be two triangles,
one triangle, or no triangle with the given
properties.
For this reason, Case 2 is sometimes b called
the ambiguous case.
The Ambiguous Case
• To see why this is so, we show in Figure 6 the
possibilities when angle A and sides a and b are
given. In part (a) no solution is possible, since side a
is too short to complete the triangle. In part (b) the
solution is a right triangle. In part (c) two solutions
are possible, and in part (d) there is a unique triangle
with the given properties.
(a)
(b)
(c)
The ambiguous case
Figure 6
(d)
The Ambiguous Case
• We illustrate the possibilities of Case 2 in the
next examples.
SSA, the One-Solution Case
• Solve triangle ABC, where A = 45, a = 7 ,
and b = 7.
We first sketch the triangle with the
information we have (see Figure 7). Our sketch
is necessarily tentative, since we don’t yet
know the other angles. Nevertheless, we can
now see the possibilities. Since little
side a is greater than side
b, we expect only 1
Triangle
Figure 7
cont’d
• We first find B.
Law of Sines
Solve for sin B
cont’d
• Which angles B have sin B = From the preceding
section we know that there are two such angles
smaller than 180 (they are 30 and 150).
• Which of these angles is compatible with what we
know about triangle ABC? Since A = 45, we cannot
have
B = 150, because 45 + 150 > 180.
• So B = 30, and the remaining angle is
C = 180 – (30 + 45) = 105.
cont’d
• Now we can find side c.
Law of Sines
Solve for c
•
SSA, the Two-Solution Case
• Solve triangle ABC if A = 43.1, a = 186.2,
and b = 248.6.
From the given information we sketch the
triangle shown in Figure 8. Note that side a
may be drawn in two possible positions to
complete the triangle.
Figure 8
Example 4 – Solution
cont’d
• From the Law of Sines
There are two possible angles B between 0 and
180 such that sin B = 0.91225.
Using a calculator, we find that one of the angles
is sin–1(0.91225)  65.8.
The other angle is approximately 180 – 65.8 =
114.2
cont’d
• We denote these two angles by B1 and B2 so
that
B1  65.8 and B2  114.2
• Thus two triangles satisfy the given
conditions: triangle A1B1C1 and triangle
A2B2C2.
Find C
Solve triangle A1B1C1:
C1  180 – (43.1 + 65.8) = 71.1then solve
1
Example 4 – Solution
• Solve triangle A2B2C2:
Find C2
C2  180 – (43.1 + 114.2) = 22.7
• Thus
Law of Sines
cont’d
Solution
cont’d
• Triangles A1B1C1 and A2B2C2 are shown in
Figure 9.
Figure 9
SSA, the No-Solution Case
• Solve triangle ABC, where A = 42, a = 70,
and b = 122.
• Solution:
• To organize the given information, we sketch
the diagram in Figure 10 and use it to
determine how many triangles we expect.
Figure 10
Example 5 – Solution
• Since solving for h (height) we obtain 81.63
there is no possible triangle for this
information.
cont’d