y 11 3 1 3
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 451
Chapter 8: Further Applications of Trigonometry
In this chapter, we will explore additional applications of trigonometry. We will begin
with an extension of the right triangle trigonometry we explored in Chapter 5 to situations
involving non-right triangles. We will explore the polar coordinate system and
parametric equations as new ways of describing curves in the plane. In the process, we
will introduce vectors and an alternative way of writing complex numbers, two important
mathematical tools we use when analyzing and modeling the world around us.
Section 8.1 Non-right Triangles: Law of Sines and Cosines ...................................... 451
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines
Although right triangles allow us to solve many applications, it is more common to find
scenarios where the triangle we are interested in does not have a right angle.
Two radar stations located 20 miles apart
both detect a UFO located between them.
The angle of elevation measured by the
first station is 35 degrees. The angle of
15°
35°
elevation measured by the second station is
20 miles
15 degrees. What is the altitude of the
UFO?
We see that the triangle formed by the UFO and the two stations is not a right triangle.
Of course, in any triangle we could draw an altitude, a perpendicular line from one
vertex to the opposite side, forming two right triangles, but it would be nice to have
methods for working directly with non-right triangles. In this section we will expand
upon the right triangle trigonometry we learned in Chapter 5, and adapt it to non-right
triangles.
Law of Sines
Given an arbitrary non-right triangle, we can drop an altitude, which we temporarily label
h, to create two right triangles.
Using the right triangle relationships,
h
sin(α) = and
h
sin(β ) = .
a
b
b
a
h
α
β
This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2011.
This material is licensed under a Creative Commons CC-BY-SA license.
452 Chapter 8
Solving both equations for h, we get b sin(α) = h and a sin(β ) = h . Since the h is the same in both equations, we establish b sin(α) = a sin(β ) . Dividing, we conclude that
sin(α) sin(β )
=
a
b
Had we drawn the altitude to be perpendicular to side b or a, we could similarly establish
sin(α) sin(γ )
β
γ
=
and sin( ) sin( )
=
a
c
b
c
Collectively, these relationships are called the Law of Sines.
Law of Sines
Given a triangle with angles and sides opposite labeled as shown, the ratio of sine of
angle to length of the opposite side will always be equal, or, symbolically,
sin(α) sin(β ) sin(γ )
=
=
a
b
c
γ
a
b
For clarity, we call side a the corresponding side of angle α.
Similarly, we call angle α, the corresponding angle of side a.
α
β
Likewise for side b and angle β, and for side c and angle γ.
c
When we use the law of sines, we use any pair of ratios as an equation. In the most
straightforward case, we know two angles and one of the corresponding sides.
Example 1
In the triangle shown here, solve for the
unknown sides and angle.
γ
b
10
Solving for the unknown angle is relatively
easy, since the three angles must add to 180
50°
30°
degrees. From this, we can determine that
c
γ = 180° – 50° – 30° = 100°.
To find an unknown side, we need to know the corresponding angle, and we also need
another known ratio.
Since we know the angle 50° and its corresponding side, we can use this for one of the
two ratios. To look for side b, we would use its corresponding angle, 30°.
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 453
sin 50
(
)
°
sin 30
(
)
°
=
Multiply both sides by b
10
b
sin 50
(
)
°
b
= sin 30
(
)
°
Divide, or multiply by the reciprocal, to solve for b
10
10
b = sin(30 )
°
≈ 527
.
6
sin(50 )
°
Similarly, to solve for side c, we set up the equation
°
°
sin(50 ) sin 100
(
)
=
10
c 10
c = sin 100
(
)
°
≈
856
.
12
sin(50 )
°
Example 2
Find the elevation of the UFO from the beginning of the section.
To find the elevation of the UFO, we first
find the distance from one station to the
a
UFO, such as the side a in the picture,
h
then use right triangle relationships to
15°
35°
find the height of the UFO, h.
20 miles
Since the angles in the triangle add to 180 degrees, the unknown angle of the triangle
must be 180° – 15° – 35° = 130°. This angle is opposite the side of length 20, allowing
us to set up a Law of Sines relationship:
sin 130
(
)
°
sin 35
(
)
°
=
Multiply by a
20
a
sin 130
(
)
°
a
= sin 35
(
)
°
Divide, or multiply by the reciprocal, to solve for a
20
20sin(35 )
°
a =
≈
975
.
14
Simplify
sin 130
(
)
°
The distance from one station to the UFO is about 15 miles.
Now that we know a, we can use right triangle relationships to solve for h.
sin 15
( )
opposite
h
h
° =
= =
Solve for h
hypotenuse a
975
.
14
h = 975
.
14
sin 15
( )
° ≈ 876
.
3
The UFO is at an altitude of 3.876 miles.
454 Chapter 8
In addition to solving triangles in which two angles are known, the law of sines can be
used to solve for an angle when two sides and one corresponding angle are known.
Example 3
In the triangle shown here, solve for the unknown sides and
angles.
12
β
In choosing which pair of ratios from the Law of Sines to
α
use, we always want to pick a pair where we know three of
a
the four pieces of information in the equation. In this case,
85°
we know the angle 85° and its corresponding side, so we
9
will use that ratio. Since our only other known information
is the side with length 9, we will use that side and solve for its corresponding angle.
sin 85
(
)
°
sin(β )
=
Isolate the unknown
12
9
9sin 85
(
)
° = sin(β)
Use the inverse sine to find a first solution
12
Remember when we use the inverse function that there are two possible answers.
9sin( °
85 )
=
−
sin 1
β
≈
.
48
°
3438 By symmetry we find the second possible solution
12
β =
°
180 −
°
3438
.
48
=
°
6562
.
131
In this second case, if β ≈ 132°, then α would be α = 180° – 85° – 132° = –37°, which doesn’t make sense, so the only possibility for this triangle is β = 48.3438°.
With a second angle, we can now easily find the third angle, since the angles must add
to 180°, so α = 180° – 85° – 48.3438° = 46.6562°.
Now that we know α, we can proceed as in earlier examples to find the unknown side a.
sin 85
(
)
°
sin( .
46 6562 )
°
=
12
a
12sin(46 6562
.
)
°
a =
≈ 8 7603
.
sin 85
(
)
°
Notice that in the problem above, when we use Law of Sines to solve for an unknown
angle, there can be two possible solutions. This is called the ambiguous case, and can
arise when we know two sides and a non-included angle. In the ambiguous case we may
find that a particular set of given information can lead to 2, 1 or no solution at all.
However, when an accurate picture of the triangle or suitable context is available, we can
determine which angle is desired.
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 455
Try it Now
1. Given α = 80 ,° a =
a
,
120 n
d b =121, find the corresponding and missing side and
angles. If there is more than one possible solution, show both.
Example 4
Find all possible triangles if one side has length 4 opposite an angle of 50° and a second
side has length 10.
Using the given information, we can look for the angle opposite the side of length 10.
sin 50
(
)
°
sin(α)
=
4
10
10sin 50
(
)
°
sin(α) =
≈ 915
.
1
4
Since the range of the sine function is [-1, 1], it is impossible for the sine value to be
1.915. There are no triangles that can be drawn with the provided dimensions.
Example 5
Find all possible triangles if one side has length 6 opposite an angle of 50° and a second
side has length 4.
Using the given information, we can look for the angle opposite the side of length 4.
sin 50
(
)
°
sin(α)
=
6
4
4sin 50
(
)
°
sin(α) =
≈ 511
.
0
Use the inverse to find one solution
6
=
−
sin 1
α
(
)
511
.
0
≈
°
710
.
30
By symmetry there is a second possible solution
α =
°
180 −
°
710
.
30
=
°
290
.
149
If we use the angle .
30
°
710 , the third angle would be
°
180 − °
50 −
°
710
.
30
=
°
290
.
99
.
We can then use Law of Sines again to find the third side.
sin 50
(
)
°
sin 99
( 290
.
)
°
=
Solve for c
6
c
c = 7.730
If we used the angle α = 149.290°, the third angle would be 180° – 50° – 149.290° =
–19.29°, which is impossible, so the previous triangle is the only possible one.
Try it Now
2. Given α = 80 ,° a =
a
,
100 n
d b = 10 find the missing side and angles. If there is more
than one possible solution, show both.
456 Chapter 8
Law of Cosines
Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and
travels another 8 miles. How far from port is the boat?
8 mi 20Ůnfortunately, while the Law of Sines lets us address many non-right
triangle cases, it does not allow us to address triangles where the one
known angle is included between two known sides, which means it is
10 mi
not a corresponding angle for a known side. For this, we need another
tool.
Given an arbitrary non-right triangle, we can
drop an altitude, which we temporarily label
h, to create two right triangles. We will
β
c
a
divide the base b into two pieces, one of
h
which we will temporarily label x. From
α
γ
this picture, we can establish the right
x
b - x
triangle relationship
b
x
cos(α) = , or equivalently, x = c cos(α )
c
Using the Pythagorean Theorem, we can establish
( b − x)2
2
2
+ h = a and
2
2
2
x + h = c
Both of these equations can be solved for 2
h
2
2
h = a − ( b − x)2
and
2
2
2
h = c − x
Since the left side of each equation is 2
h , the right sides must be equal
2
2
2
c − x = a − ( b − x)2
Multiply out the right
2
2
2
c − x = a − ( 2
2
b − 2 bx + x )
Simplify
2
2
2
2
2
c − x = a − b + 2 bx − x
c 2 = a 2 − b 2 + bx
2
Isolate 2
a
a 2 = c 2 + b 2 − bx
2
Substitute in c cos(α) = x from above
2
2
2
a = c + b − 2 bc cos(α)
This result is called the Law of Cosines. Depending upon which side we dropped the
altitude down from, we could have established this relationship using any of the angles.
The important thing to note is that the right side of the equation involves an angle and the
sides adjacent to that angle – the left side of the equation involves the side opposite that
angle.
Section 8.1 Non-Right Triangles: Laws of Sines and Cosines 457
Law of Cosines
Given a triangle with angles and opposite sides labeled as shown,
2
2
2
a = c + b − 2 bc cos(α)
γ
a
2
2
2
b = a + c − 2 ac cos(β )
b
2
2
2
c = a + b − 2 ab cos(γ )
α
β
c
Notice that if one of the angles of the triangle is 90 degrees, cos(90°) = 0, so the formula
2
2
2
c = a + b − 2 ab co 90
s(
)
°
Simplifies to
2
2
2
c = a + b
You should recognize this as the Pythagorean Theorem. Indeed, the Law of Cosines is
sometimes called the Generalized Pythagorean Theorem, since it extends the
Pythagorean Theorem to non-right triangles.
Example 6
Returning to our question from earlier, suppose a boat leaves port,
travels 10 miles, turns 20 degrees, and travels another 8 miles. How far
from port is the boat?
8 mi
20°
The boat turned 20 degrees, so the obtuse angle of the non-right triangle
shown in the picture is the supplemental angle, 180° - 20° = 160°.
10 mi
With this, we can utilize the Law of Cosines to find the missing side of
the obtuse triangle – the distance from the boat to port.
2 x