§1.5
Notice that reflection around the y-axis is equivalent to reflection around the x-axis ( θ →
− θ) followed by a rotation of 180◦ (− θ → − θ + 180◦ = 180◦ − θ), as in Figure 1.5.7.
y
(− x, y)
( x, y)
180◦ − θ
r
r
θ
x
− θ
r
( x, − y)
Figure 1.5.7
Reflection of θ around the y-axis = 180◦ − θ
It may seem that these geometrical operations and formulas are not necessary for evalu-
ating the trigonometric functions, since we could just use a calculator. However, there are
two reasons for why they are useful. First, the formulas work for any angles, so they are
often used to prove general formulas in mathematics and other fields, as we will see later
in the text. Second, they can help in determining which angles have a given trigonometric
function value.
Example 1.27
Find all angles 0◦ ≤ θ < 360◦ such that sin θ = −0.682.
✄
Solution: Using the sin−1 button on a calculator with −0.682 as the input, we get θ = −43◦, which
✂
✁
is not between 0◦ and 360◦.7 Since θ = −43◦ is in QIV, its reflection 180◦ − θ around the y-axis will be in QIII and have the same sine value. But 180◦ − θ = 180◦ − (−43◦) = 223◦ (see Figure 1.5.8). Also, we
know that −43◦ and −43◦ +360◦ = 317◦ have the same trigonometric function values. So since angles
in QI and QII have positive sine values, we see that the only angles between 0◦ and 360◦ with a sine
of −0.682 are θ = 223◦ and 317◦ .
y
x
180◦ − θ = 223◦
θ = −43◦
r
r
(− x, y)
( x, y)
Figure 1.5.8
Reflection around the y-axis: −43◦ and 223◦
✄
7In Chapter 5 we will discuss why the sin−1 button returns that value.
✂
✁
Rotations and Reflections of Angles • Section 1.5
37
Exercises
1. Let θ = 32◦. Find the angle between 0◦ and 360◦ which is the
(a) reflection of θ around the x-axis
(b) reflection of θ around the y-axis
(c) reflection of θ around the origin
2. Repeat Exercise 1 with θ = 248◦.
3. Repeat Exercise 1 with θ = −248◦.
4. We proved formulas (1.4)-(1.6) for any angle θ in QI. Mimic that proof to show that the formulas hold for θ in QII.
5. Verify formulas (1.4)-(1.6) for θ on the coordinate axes, i.e. for θ = 0◦, 90◦, 180◦, 270◦.
6. In Example 1.26 we used the formulas involving θ + 90◦ to prove that the slopes of perpendicu-
lar lines are negative reciprocals. Show that this result can also be proved using the formulas
involving θ − 90◦. ( Hint: Only the last paragraph in that example needs to be modified. )
For Exercises 7 - 14, find all angles 0◦ ≤ θ < 360◦ which satisfy the given equation: 7. sin θ = 0.4226
8. sin θ = 0.1909
9. cos θ = 0.4226
10. sin θ = 0
11. tan θ = 0.7813
12. sin θ = −0.6294
13. cos θ = −0.9816
14. tan θ = −9.514
y
15. In our proof of the Pythagorean Theorem in Section 1.2, we
claimed that in a right triangle △ ABC it was possible to draw
B (0, b)
a line segment CD from the right angle vertex C to a point D
D
on the hypotenuse AB such that CD ⊥ AB. Use the picture on
the right to prove that claim. ( Hint: Notice how △ ABC is placed
on the x y-coordinate plane. What is the slope of the hypotenuse?
x
C (0, 0)
A ( a, 0)
What would be the slope of a line perpendicular to it? ) Also, find
the ( x, y) coordinates of the point D in terms of a and b.
16. It can be proved without using trigonometric functions that the slopes of perpendicular lines
are negative reciprocals. Let y = m 1 x + b 1 and y = m 2 x + b 2 be perpendicular lines (with nonzero slopes), as in the picture below. Use the picture to show that m 2 = − 1 .
m 1
( Hint: Think of similar triangles and the definition of slope. )
y
y = m 2 x + b 2
y = m 1 x + b 1
( x 2, y 2)
( x 3, y 2)
( x 2, y 1)
( x 3, y 1)
( x 1, y 1)
x
17. Prove formulas (1.19)-(1.21) by using formulas (1.10)−(1.12) and (1.13)−(1.15).
2 General Triangles
In Section 1.3 we saw how to solve a right triangle: given two sides, or one side and one
acute angle, we could find the remaining sides and angles. In each case we were actually
given three pieces of information, since we already knew one angle was 90◦.
For a general triangle, which may or may not have a right angle, we will again need three
pieces of information. The four cases are:
Case 1:
One side and two angles
Case 2:
Two sides and one opposite angle
Case 3:
Two sides and the angle between them
Case 4:
Three sides
Note that if we were given all three angles we could not determine the sides uniquely; by
similarity an infinite number of triangles have the same angles.
In this chapter we will learn how to solve a general triangle in all four of the above cases.
Though the methods described will work for right triangles, they are mostly used to solve
oblique triangles, that is, triangles which do not have a right angle. There are two types
of oblique triangles: an acute triangle has all acute angles, and an obtuse triangle has
one obtuse angle.
As we will see, Cases 1 and 2 can be solved using the law of sines, Case 3 can be solved
using either the law of cosines or the law of tangents, and Case 4 can be solved using the law
of cosines.
2.1 The Law of Sines
Theorem 2.1. Law of Sines: If a triangle has sides of lengths a, b, and c opposite the
angles A, B, and C, respectively, then
a
b
c
=
=
.
(2.1)
sin A
sin B
sin C
Note that by taking reciprocals, equation (2.1) can be written as
sin A
sin B
sin C
=
=
,
(2.2)
a
b
c
and it can also be written as a collection of three equations:
a
sin A
a
sin A
b
sin B
=
,
=
,
=
(2.3)
b
sin B
c
sin C
c
sin C
38
The Law of Sines • Section 2.1
39
Another way of stating the Law of Sines is: The sides of a triangle are proportional to the
sines of their opposite angles.
To prove the Law of Sines, let △ ABC be an oblique triangle. Then △ ABC can be acute, as
in Figure 2.1.1(a), or it can be obtuse, as in Figure 2.1.1(b). In each case, draw the altitude 1
from the vertex at C to the side AB. In Figure 2.1.1(a) the altitude lies inside the triangle,
while in Figure 2.1.1(b) the altitude lies outside the triangle.
C
C
b
a
b
h
h
a
180◦ − B
c
A
B
c
A
B
(a) Acute triangle
(b) Obtuse triangle
Figure 2.1.1
Proof of the Law of Sines for an oblique triangle △ ABC
Let h be the height of the altitude. For each triangle in Figure 2.1.1, we see that
h = sin A
(2.4)
b
and
h = sin B
(2.5)
a
(in Figure 2.1.1(b), ha = sin (180◦ − B) = sin B by formula (1.19) in Section 1.5). Thus, solving for h in equation (2.5) and substituting that into equation (2.4) gives
a sin B = sin A ,
(2.6)
b
and so putting a and A on the left side and b and B on the right side, we get
a
b
=
.
(2.7)
sin A
sin B
By a similar argument, drawing the altitude from A to BC gives
b
c
=
,
(2.8)
sin B
sin C
so putting the last two equations together proves the theorem.
QED
Note that we did not prove the Law of Sines for right triangles, since it turns out (see
Exercise 12) to be trivially true for that case.
1Recall from geometry that an altitude of a triangle is a perpendicular line segment from any vertex to the line
containing the side opposite the vertex.
40
Chapter 2 • General Triangles
§2.1
Example 2.1
Case 1: One side and two angles.
C = 75◦
Solve the triangle △ ABC given a = 10, A = 41◦, and C = 75◦.
b
a = 10
Solution: We can find the third angle by subtracting the other two angles
from 180◦, then use the law of sines to find the two unknown sides. In this
example we need to find B, b, and c. First, we see that
c
A = 41◦
B
B = 180◦ − A − C = 180◦ − 41◦ − 75◦
⇒
B = 64◦ .
So by the Law of Sines we have
b
a
a sin B
10 sin 64◦
=
⇒
b =
=
⇒
b = 13.7 , and
sin B
sin A
sin A
sin 41◦
c
a
a sin C
10 sin 75◦
=
⇒
c =
=
⇒
c = 14.7 .
sin C
sin A
sin A
sin 41◦
Example 2.2
Case 2: Two sides and one opposite angle.
C
Solve the triangle △ ABC given a = 18, A = 25◦, and b = 30.
b = 30
a = 18
Solution: In this example we know the side a and its opposite angle A,
and we know the side b. We can use the Law of Sines to find the other
c
A = 25◦
B
opposite angle B, then find the third angle C by subtracting A and B from
180◦, then use the law of sines to find the third side c. By the Law of Sines, we have
sin B
sin A
b sin A
30 sin 25◦
=
⇒
sin B =
=
⇒
sin B = 0.7044 .
b
a
a
18
✄
Using the sin−1 button on a calculator gives B = 44.8◦. However, recall from Section 1.5 that
✂
✁
sin (180◦ − B) = sin B. So there is a second possible solution for B, namely 180◦ − 44.8◦ = 135.2◦.
Thus, we have to solve twice for C and c : once for B = 44.8◦ and once for B = 135.2◦:
B = 44.8◦
B = 135.2◦
C = 180◦ − A − B = 180◦ − 25◦ − 44.8◦ = 110.2◦
C = 180◦ − A − B = 180◦ − 25◦ − 135.2◦ = 19.8◦
c
a
a sin C
18 sin 110.2◦
c
a
a sin C
18 sin 19.8◦
=
⇒ c =
=
=
⇒ c =
=
sin C
sin A
sin A
sin 25◦
sin C
sin A
sin A
sin 25◦
⇒ c = 40
⇒ c = 14.4
Hence, B = 44.8◦, C = 110.2◦, c = 40 and B = 135.2◦, C = 19.8◦, c = 14.4 are the two possible sets of solutions. This means that there are two possible triangles, as shown in Figure 2.1.2.
C = 110.2◦
C = 19.8◦
b = 30
a = 18
b = 30
a = 18
A = 25◦
B = 44.8◦
A = 25◦
B = 135.2◦
c = 40
c = 14.4
(a) B = 44.8◦
(b) B = 135.2◦
Figure 2.1.2
Two possible solutions
The Law of Sines • Section 2.1
41
In Example 2.2 we saw what is known as the ambiguous case. That is, there may be more
than one solution. It is also possible for there to be exactly one solution or no solution at all.
Example 2.3
Case 2: Two sides and one opposite angle.
Solve the triangle △ ABC given a = 5, A = 30◦, and b = 12.
Solution: By the Law of Sines, we have
sin B
sin A
b sin A
12 sin 30◦
=
⇒
sin B =
=
⇒
sin B = 1.2 ,
b
a
a
5
which is impossible since |sin B| ≤ 1 for any angle B. Thus, there is no solution .
There is a way to determine how many solutions a triangle has in Case 2. For a triangle
△ ABC, suppose that we know the sides a and b and the angle A. Draw the angle A and
the side b, and imagine that the side a is attached at the vertex at C so that it can “swing”
freely, as indicated by the dashed arc in Figure 2.1.3 below.
C
C
C
a
b
h
b
b
a
h a
h
a
A
B
c
A
B
A
B
B
(a) a < h: No solution
(b) a = h: One solution
(c) h < a < b: Two solutions
C
b
a
b
c
A
B
(d) a ≥ b: One solution