Download Math 3329-Uniform Geometries — Lecture 07 C B A a a b b c D

Math 3329-Uniform Geometries — Lecture 07
A
b
C
c
b0
D
a0
B
a
Figure 1. If we slide vertex C to the left, we can make the angle C acute. Note that CB = a
1. Law of Sines
Take a look back at Figure 1. Note that
b0
.
b
Using the right triangle on the other side of the figure, we see that
b0
(2)
sin(B) = .
c
Solving equations (1) and (2) for b0 , we get
(1)
sin(C) =
(3)
b sin(C) = c sin(B).
We could also write this as
b
c
=
.
sin(B)
sin(C)
(4)
A
b0
D
c
b
a0
a
C
B
Figure 2. If we slide vertex C to the right, the angle C is now obtuse.
A
c
b
C
a
B
Figure 3. The Pythagorean theorem states that if C = 90◦ , then c2 = a2 + b2 .
We get the same relationships for the triangles in Figure 2 and Figure 3. For the right triangle in Figure 3,
we get
(5)
sin(C) = 1,
1
2
since sin(90◦) = 1, and
b
sin(B) = .
c
(6)
Equation (6) becomes
(7)
c=
b
,
sin(B)
and since sin(C) = 1, we can write this as
c
c
b
=
=
.
1
sin(C)
sin(B)
(8)
We can take any triangle and use any of the three sides as the base, and the triangle will look something
a
like Figure 3, Figure 2, or Figure 1, so there must not be anything special about B and C, and sin(A)
must
be equal to the things in equation (4) as well. This relationship is known as the Law of Sines.
Law of Sines. For any triangle with sides a, b, and c with angles A, B, and C opposite each side,
a
b
c
(9)
=
=
.
sin(A)
sin(B)
sin(C)
1.1. Quiz.
–1– Find equations similar to (1), (2) and (4) but for Figure 2. You may need the trig identities
sin(180◦ + θ) = − sin(θ), and sin(θ) = − sin(−θ).
C0
b0
C
b
A
c0
c
a
B
A0
a0
B0
Figure 4. These are similar triangles.
2. Similarity
We will want to say that two objects that have the same shape, but not necessarily the same size, are
similar. For two triangles, this will mean that corresponding angles are congruent (have the same measure).
In Figure 4, for example, A = A0 , B = B 0 , and C = C 0 (using these letters as the measures of the angles),
so these triangles are similar. In this case the sides of triangle 4A0 B 0 C 0 are twice as long as the sides of
4ABC. In general, for two similar triangles, there is no special relationship between the lengths of a and
a0 . One can be a little bigger than the other, or a lot bigger. There is a useful relationship, if we consider
all parts of the triangles together, however. The Law of Sines gives us the following, for example.
a
sin A
sin A0
a0
=
=
= 0.
0
b
sin B
sin B
b
This says that the ratio between sides a and b is the same as the ratio between the sides a0 and b0 . In other
words, if a is k times longer than b, then a0 will also be k times longer than b0 . The same holds for any pair
of sides, so
(10)
(11)
a
a0 a
a0
b
b0
= 0,
= 0 , and
= 0.
b
b
c
c
c
c
3
We can also flip these equations around to get
b
a
c
b
c
a
= 0 , 0 = 0 , and 0 = 0 ,
(12)
a0
b
a
c
b
c
or more compactly
a
b
c
= 0 = 0.
(13)
a0
b
c
3. Congruence
If two similar triangles are the same size, we’ll say that they are congruent. Congruent triangles are the
same size and shape. This means that all corresponding parts are the same, corresponding sides have the
same length, corresponding angles have the same measure, and the areas are the same, as we’ve said before.
Suppose that we have a triangle with angles A, B, and C and opposite sides a, b, and c. We may try to
build another triangle with angles A0 , B 0 , and C 0 and opposite sides a0 , b0 , and c0 so that A = A0 , B = B 0 ,
and C = C 0 . The length of the side a0 , for example, could be anything for all we know. Given a value for a0 ,
however, forces a value for b0
(14)
b0 =
a0 b
.
a
Similarly,
a0 c
,
a
and so fixing the size of one side of the second triangle forces unique values for the other two sides. This is
a property of similar triangles.
(15)
c0 =
Basic Principle 1. (AAA) If we have a pair of triangles with corresponding angles congruent, then the
triangles must be similar.
If all three angles match the three angles of another triangle, then they are similar. Certainly, this is not
enough to guarantee congruence between the two triangles. Note however, if two pairs of corresponding
angles have the same measure (i.e., are congruent), then the third pair must also, since the angle sum of a
triangle is always 180◦.
If we have two pairs of congruent angles, this guarantees similarity. If in addition, one pair of corresponding
sides have the same length (e.g., a = a0 ), by equation (13), all pairs of sides must have the same length. We
have the following theorem.
Theorem 1. (AAS, ASA, or AAS) If we have two triangles, and two pairs of corresponding angles are
congruent, and one pair of corresponding sides are congruent, then the triangles are congruent.
We have already mentioned the SAS criterion for the congruence of triangles several times. This states that
if two sides and the included angle in one triangle are congruent to two sides and the included angle of
another triangle, then the two triangles must be congruent. It is called SAS, because the angle must be
between the two sides, side-angle-side (e.g. a, C, and b). If these three pieces of information are sufficient to
guarantee congruence, they must also determine all the measurements of the triangle. We can use the Law
of Cosines to do this. For example, given a, C, and b, we can find c using
p
(16)
c = a2 + b2 − 2ab cos(C).
We can then find A, for example, using the equation
(17)
a2 = b2 + c2 − 2bc cos(A).
In fact, once you have all three sides of a triangle, you can use the Law of Cosines to find unique values for
any of the angles. As long as the angles are unique, having three sides congruent is sufficient to guarantee
that the triangles are congruent.
4
Theorem 2. (SAS and SSS) If two sides and the included angle (SAS) match on two triangles, then they
are congruent. If all three sides match (SSS), then the two triangles are congruent.
OK. We have SSS, SAS, AAS, ASA, and SAA criteria for congruence. AAA only guarantees similarity. The
two remaining possibilities, are ASS and SSA. These guarantee neither congruence nor similarity.
A
b
C
c
b0
C0
B
Figure 5. Triangles 4ABC and 4ABC 0 , satisfy SSA, but are not congruent.
3.1. Quiz.
–1– Sketch a graph of the cosine function in degrees over −180 ≤ x ≤ 360. It should
pass through
√
3
(−180, −1), (−90, 0), (0, 1), (90, 0) and (180, −1). Now draw a horizontal line y = 2 . You should
√
see three points of intersection. These correspond to solutions to the equation cos(x) = 23 . What
are the three values of x? (In general, for 0 < k < 1, the equation cos(x) = k will have one solution
between −90◦ and 0◦ , one solution between 0◦ and 90◦ , and one solution between 270◦ and 360◦.)
–2– For −1 < k < 0, the equation cos(x) = k will have three solutions over −180 ≤ x ≤ 360. Where will
each of these be?
–3– The angle sum of a triangle is 180◦. We can have triangles where one angle is almost 180◦, and the
other two are really small. Draw a triangle like this.
–4– Suppose θ is the measure of an angle from some triangle. Describe the range of values that θ can
take, if the triangle is not degenerate (it doesn’t have zero area, for example).
–5– For equation (17), the angle A must be between 0◦ and 180◦ . Can there be more than one solution
over this range? This will guarantee uniqueness for the SSS criterion.
–6– In Figure 5, are triangles 4ABC and 4ABC 0 congruent?
–7– Assume that b = b0 . Do 4ABC and 4ABC 0 have another pair of congruent sides? A pair of
congruent angles? What does this say about SSA?
–8– Note that we are using SSA to mean that if we look at a side of a triangle and start running around
the triangle in one direction, take the next side we see, and then the first angle after that, we
want these three things to be the same as the corresponding things of the other triangle. Which
combinations of letters (S’s and A’s) guarantee congruence?
–9– Suppose 4ABC and 4A0 B 0 C 0 are similar (with A corresponding to A0 , etc.). If a = 3, b = 2, b0 = 6,
and c0 = 12, find the lengths of the other sides.
–10– In the xy-plane, let A = (0, 0), B = (2, 0), and C = (3, 4). If A0 = (0, 0), B 0 = (x, 0), and 4ABC
and 4A0 B 0 C 0 are similar, find the coordinates of C 0 in terms of x.
3.2. Homework.
–1– Because SSS is a congruence criterion, given all three sides of a triangle should allow you to find all
three angles as well. Suppose a = 3, b = 2, and c = 4. Find A, B, and C using the Law of Cosines
and/or the Law of Sines. Round degrees to two decimal places (e.g. 35.23◦.
5
–2– It’s possible to choose numbers for a, b, and c that do not correspond to an actual triangle. Find a
triple of positive numbers that won’t work. (Hint: Imagine a triangle of locations on a map. To go
from vertex A to C, which is shorter: traveling along the side AC or taking a detour through B?)
–3– There is no SSA congruence criterion. This is equivalent to the fact that given information like the
angle A, the side b, and the side a, it is possible to find multiple solutions when trying to determine
the other side and angles. Find all solutions given A = 20◦, b = 10, and a = 2. (You should be able
to find two triples (B, C, c).)
–4– Now try A = 60◦, b = 2, and a = 10.
–5– Also try A = 110◦, b = 2, and a = 5.
–6– Find conditions on A that will guarantee unique solutions.
–7– Find conditions on b and a that will guarantee unique solutions.
4. Fixing the homework
My problem 3 isn’t what I wanted, and that may have messed up the problems that followed. Let’s work
through what I wanted you to see.
C
b
a
c
A
B
B
B
Figure 6. Creating a triangle given angle ∠A, side b, and side a.
Suppose we are given acute angle ∠A, side b, and side a, and we want to build a triangle. Looking at Figure
6, we start with angle ∠A. We can mark off a distance b to get the point C. From there we attach a side
a at C, and attach it to the other side of angle ∠A at a point B. We know that this may happen in more
than one way. This depends on the length of a, so we can figure out what happens for varying lengths of a.
Before doing this, note that the Law of Cosines tells us that the side c must satisfy the equation
a2 = b2 + c2 − 2bc cos(A).
(18)
Since a, b, and A are known, we can solve for c using the quadratic formula. Since
0 = c2 − 2bc cos(A) + (b2 − a2 ),
(19)
we know the solutions are
(20)
c=
2b cos(A) ±
p
4b2 cos2 (A) − 4(b2 − a2 )
2
OK. Let’s look at what happens when a is really small, and as it gets longer. If a is really short, it won’t
reach the other side of the angle, and we can’t make the triangle. The shortest distance from C to a possible
point B is straight down, and this would make a right triangle. This shortest “good” a is
(21)
a = b sin(A),
since B would be a right angle, and sine is opposite over hypotenuse. If we plug this value into equation
(20), the stuff under the radical would be
(22)
4b2 cos2 (A) − 4(b2 − a2 ) = 4b2 cos2 (A) − 4(b2 − b2 sin2 (A)) = 4b2 cos2 (A) − 4b2 + 4b2 sin2 (A) = 0.
What’s left over is positive (since A is acute), so we have exactly one solution
(23)
c = b cos(A),
6
which is what we should have expected. Note that if a were any shorter, then 4(b2 − a2 ) would be a little
bigger, and the stuff under the radical would be negative. This would mean that c would have to be complex,
and therefore couldn’t be a side of a normal triangle.
If a is a little bit bigger than b sin(A), then the side a could tilt back towards the vertex A (making a shorter
c < b cos(A)), or tilt away (making a longer c > b cos(A)). This would correspond to the stuff under the
radical being small but positive, and we would get two postive solutions for c.
As we make a longer and longer, we’ll eventually get to the point, where one of the solutions for c is zero or
negative. That’s what you’re looking at in the new homework.
4.1. Homework.
–1– As a starts small and gets bigger, we start out with no solutions, then one, and then two solutions.
What’s the smallest value for a that brings us back to only one solution that corresponds to an
actual triangle?
–2– Plug this into equation (20).
–3– Suppose a is larger than the number in problem 1. What happens when you plug this into equation
(20)? You’ll be saying stuff like, “this thing will be negative, so that thing will be positive, because
. . .”
References
[Bonola]
Roberto Bonola (1955). Non-Euclidean Geometry (H.S. Carslaw, Trans.). Dover Publications, New York. (Original
translation, 1912, and original work published in 1906.)
[Descartes] Rene Descartes (1954). The Geometry of Rene Descartes (D.E. Smith and M.L. Latham, Trans.). Dover Publications, New York. (Original translation, 1925, and original work published in 1637.)
[Euclid]
Euclid (1956). The Thirteen Books of Euclid’s Elements (2nd Ed., Vol. 1, T.L. Heath, Trans.). Dover Publications,
New York. (Original work published n.d.)
[Eves]
Howard Eves (1990). An Introduction to the History of Mathematics (6th Ed.). Harcourt Brace Jovanovich, Orlando, FL.
[Federico] P.J. Federico (1982). Descartes on Polyhdra: A study of the De Solidorum Elementis. Springer-Verlag, New York.
[Henderson] David W. Henderson (2001). Experiencing Geometry: In Euclidean, Spherical, and Hyperbolic Spaces 2nd Ed.
Prentice Hall, Upper Saddle River, NJ.
[Henle]
Michael Henle (2001). Modern Geometries: Non-Euclidean, Projective, and Discrete 2nd Ed. Prentice Hall, Upper
Saddle River, NJ.
[Hilbert]
David Hilbert (1971). Foundations of Geometry (2nd Ed., L. Unger, Trans.). Open Court, La Salle, IL. (10th
German edition published in 1968.)
[Hilbert2]
D. Hilbert and S. Cohn-Vossen (1956). Geometry and the Imagination (P. Nemenyi, Trans.). Chelsea, New York.
(Original work, Anschauliche Geometrie, published in 1932.)
[Motz]
Lloyd Motz and Jefferson Hane Weaver (1993). The Story of Mathematics. Avon Books, New York.
[Weeks]
Jeffrey R. Weeks (1985). The Shape of Space. Marcel Dekker, New York.