Transcript (a) (c)

a b c
2
c
a
b
2
2
This is a right triangle:
We call it a right triangle
because it contains a
right angle.
The measure of a right
o
angle is 90
90o
The little square in the
angle tells you it is a
right angle.
90o
About 2,500 years ago, a
Greek mathematician named
Pythagorus discovered a
special relationship between
the sides of right triangles.
Pythagorus realized that if
you have a right triangle,
5
3
4
and you square the lengths
of the two sides that make
up the right angle,
5
3
2
4
3
4
2
and add them together,
5
3
3 4
2
4
2
you get the same number
you would get by squaring
the other side.
5
3
3 4 5
2
4
2
2
Is that correct?
2 ?
3 4 5
2
?
2
9  16  25
It is. And it is true for any
right triangle.
6  8  10
2
2
2
10
8
36  64  100
6
The two sides which
come together in a right
angle are called
The two sides which
come together in a right
angle are called
The two sides which
come together in a right
angle are called
The lengths of the legs are
usually called a and b.
a
b
The side across from the
right angle is called the
a
b
And the length of the
hypotenuse
is usually labeled c.
a
c
b
The relationship Pythagorus
discovered is now called
The Pythagorean Theorem:
a
c
b
The Pythagorean Theorem
says, given the right triangle
with legs a and b and
hypotenuse c,
a
c
b
then a  b  c .
2
a
2
c
b
2
You
Suppose
can use
youThe
drive
Pythagorean
directly
Theorem
west for 48
to miles,
solve many kinds
of problems.
48
Then turn south and drive for
36 miles.
48
36
How far are you from where
you started?
48
36
?
Using The Pythagorean
Theorem,
2
2
48 + 36 = c
2
36
48
c
Why?
Can
you see that we have a
right triangle?
2
2
48 + 36 = c
2
36
48
c
Which sides
side isare
thethe
hypotenuse?
legs?
2
2
48 + 36 = c
2
36
48
c
Then all we need to do is
calculate:
48  36  2304 1296 
2
2
3600  c
2
2
Andsince
So,
you end
c isup
3600,
60 miles
c is 60.
from
where you started.
48
36
60
Find the length of a diagonal
of the rectangle:
15"
?
8"
Find the length of a diagonal
of the rectangle:
15"
b=8
c
?
a = 15
8"
15
225
acc 
17
8b
64
289 c
2
b=8
c
a = 15
2
2
Find the length of a diagonal
of the rectangle:
15"
17
8"
Practice using
The Pythagorean Theorem
to solve these right triangles:
c = 13
5
12
b
10
26
b = 24
a b c
2
2
2
10  b  26
2
100  b  676
2
b  676  100
2
b  576
2
2
2
10 (a)
26
(c)
b  24
12
b= 9
15
7.2 The Converse of the Pythagorean Theorem
• If c2 = a2 + b2, then ∆ABC is a right triangle.
B
c
a
C
b
A
Verify a Right Triangle
• Is ∆ABC a right triangle?
C
16
12
A
B
c  a b
2
2
20
2
20  12  16
2
2
2
400  144  256
400  400
• Yes, it is a right triangle.
Classifying Triangles
In
ABC with longest side c:
C
If c2 < a2 + b2,
a
B
If c2 = a2 + b2, then
B
If c2 > a2 + b2, then
B
A
c
ABC is right.
c
a
C
b
then
A
b
ABC is obtuse.
c
a
C
b
A
ABC is acute.
Acute Triangles
• Show that the triangle is an acute triangle.
35
5
c a b
2

35
2

2
2
 4 5
2
2
35  16  25
35  41
• Because c2 < a2 + b2, the triangle is acute.
4
Obtuse Triangles
• Show that the triangle is an obtuse triangle.
15
12
8
c a b
2
2
2
15  8  12
2
2
2
225  64  144
225  208
• Because c2 > a2 + b2, the triangle is obtuse.
Classify Triangles
• Classify the triangle as acute, right, or obtuse.
8
6
5
c a b
2
2
2
8 5 6
2
2
2
64  25  36
64  61
• Because c2 > a2 + b2, the triangle is obtuse.
Classify Triangles
• Classify the triangle with the given side lengths as acute,
right, or obtuse.
• A. 4, 6, 7
c a b
2
2
2
7 4 6
2
2
2
49  16  36
49  52
• Because c2 < a2 + b2, the triangle is acute.
Classify Triangles
• Classify the triangle with the given side lengths as acute,
right, or obtuse.
• B. 12, 35, 37
c a b
2
2
2
37  12  35
1369  144  1225
2
2
2
1369  1369
• Because c2 = a2 + b2, the triangle is right.
7.3 Similar Right Triangles
Geometry
Mr. Lopiccolo
Objectives/Assignment
• Solve problems involving similar right
triangles formed by the altitude drawn to the
hypotenuse of a right triangle.
• Use a geometric mean to solve problems such
as estimating a climbing distance.
• Assignment: pp. 453-454 (4-26) even
Proportions in right triangles
• In Lesson 6.4, you learned
that two triangles are
similar if two of their
corresponding angles are
congruent. For example P
∆PQR ~ ∆STU. Recall
that the corresponding
side lengths of similar
triangles are in
proportion.
S
U
R
T
Q
Activity: Investigating similar right
triangles. Do in pairs or threes
1.
2.
3.
4.
Cut an index card along one
of its diagonals.
On one of the right triangles,
draw an altitude from the
right angle to the hypotenuse.
Cut along the altitude to form
two right triangles.
You should now have three
right triangles. Compare the
triangles. What special
property do they share?
Explain.
Tape your group’s triangles
to a piece of paper and place
in labwork.
Theorem 7.5
• If the altitude is drawn
to the hypotenuse of a
right triangle, then the
two triangles formed
are similar to the
original triangle and to
A
each other.
C
D
∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD
B
A plan for proving thm. 7.5 is shown
below:
• Given: ∆ABC is a right triangle; altitude CD is drawn
to hypotenuse AB.
• Prove: ∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD
• Plan for proof: First prove that ∆CBD ~ ∆ABC. Each
triangle has a right triangle and each includes B. The
triangles are similar by the AA Similarity Postulate. You
can use similar reasoning to show that ∆ACD ~ ∆ABC. To
show that ∆CBD ~ ∆ACD, begin by showing that ACD
 B because they are both complementary to DCB.
Then you can use the AA Similarity Postulate. C
A
D
B
Ex. 1: Finding the Height of a Roof
• Roof Height. A roof has a
cross section that is a right
angle. The diagram shows
the approximate
dimensions of this cross
section.
• A. Identify the similar
triangles.
• B. Find the height h of the
roof.
Solution:
Y
• You may find it helpful to
sketch the three similar
3.1 m
h
triangles so that the
corresponding angles and X
W
sides have the same
orientation. Mark the
congruent angles. Notice
that some sides appear in
more than one triangle. For 5.5 m
instance XY is the
hypotenuse in ∆XYW and
the shorter leg in ∆XZY.
Y
h
∆XYW ~ ∆YZW ~ ∆XZY.
Z
6.3 m
Z
X
W
3.1 m
5.5 m
Y
Solution for b.
• Use the fact that ∆XYW ~ ∆XZY to write a
proportion.
YW
ZY
=
XY
XZ
Corresponding side lengths are in
proportion.
h
5.5
=
3.1
6.3
Substitute values.
6.3h = 5.5(3.1)
h ≈ 2.7
Cross Product property
Solve for unknown h.
The height of the roof is about 2.7 meters.
Using a geometric mean to solve
problems
• In right ∆ABC,
altitude CD is drawn
A
to the hypotenuse,
forming two smaller
right triangles that
C
are similar to ∆ABC
From Theorem 9.1,
you know that ∆CBD
~ ∆ACD ~ ∆ABC.
C
D
B
B
C
D
D
A
A
B
C
Write this down!
C
A
D
B
B
C
D
C
A
D
B
Notice that CD is the longer leg
of ∆CBD and the shorter leg
of ∆ACD. When you write a
proportion comparing the
legs lengths of ∆CBD and
∆ACD, you can see that CD
is the geometric mean of BD
and AD.
Longer leg of ∆CBD.
Shorter leg of ∆CBD.
BD
CD
A
C
Shorter leg of ∆ACD
=
CD
AD
Longer leg of ∆ACD.
Copy this down!
C
A
D
B
B
C
D
C
A
Sides CB and AC also appear in
more than one triangle. Their
side lengths are also geometric
means, as shown by the
proportions below:
D
B
AB
CB
Hypotenuse of ∆CBD
A
Shorter leg of ∆ABC.
Hypotenuse of ∆ABC.
C
=
CB
DB
Shorter leg of ∆CBD.
Copy this down!
C
A
D
B
B
C
D
C
A
Sides CB and AC also appear in
more than one triangle. Their
side lengths are also geometric
means, as shown by the
proportions below:
D
B
AB
AC
Hypotenuse of ∆ACD
A
Longer leg of ∆ABC.
Hypotenuse of ∆ABC.
C
=
AC
AD
Longer leg of ∆ACD.
Geometric Mean Theorems
• Theorem 7.6: In a right triangle, the
altitude from the right angle to the
hypotenuse divides the hypotenuse into
two segments. The length of the altitude
is the geometric mean of the lengths of
the two segments
• Theorem 7.7: In a right triangle, the
altitude from the right angle to the
hypotenuse divides the hypotenuse into
two segments. The length of each leg of
the right triangle is the geometric mean
of the lengths of the hypotenuse and the
segment of the hypotenuse that is
adjacent to the leg.
A
C
D
BD
=
CD
CD
AD
AB
CB
=
CB
DB
AB
AC
=
AC
AD
B
What does that mean?
2
x
6
6
=
x
18 = x2
√18 = x
5
y
3
x
5+2
3
y
7
y
=
y
=
2
y
2
√9 ∙ √2 = x
14 = y2
3 √2 = x
√14 = y
Ex. 3: Using Indirect Measurement.
• MONORAIL TRACK. To
estimate the height of a
monorail track, your friend
holds a cardboard square at eye
level. Your friend lines up the
top edge of the square with the
track and the bottom edge with
the ground. You measure the
distance from the ground to
your friend’s eye and the
distance from your friend to the
track.
In the diagram, XY = h – 5.75 is the difference between the track
height h and your friend’s eye level. Use Theorem 9.2 to write a
proportion involving XY. Then you can solve for h.
*You will be able to find the
lengths of sides of special
right triangles
45-45-90
And
30-60-90
45  45  90



Leg:Leg:Hypotenuse
30  60  90



1:1: 2
x: x: x 2
1: 3 : 2
x : x 3 : 2x
Short Leg:Long Leg:Hypotenuse
In a 45-45-90 triangle…
We will use a reference triangle
to set up a proportion then solve.
45-45-90 Right Triangle
45
2
1
45
1
This is our reference triangle for
the 45-45-90.
45-45-90 Right Triangle
45
x 2
x
45
x
EX: 1 Solve for x
3
Let’s set up a proportion by
using our reference triangle.
x
2
1
3
3
1

x
2
x3 2
1
EX: 2 Solve for x
5
x
2
1
5
5
1

x
2
x5 2
1
EX: 3 Solve for x
45
3
3
2

x
1
2
1
1
x
3
x
2
3
2
x

2
2
3 2
x
2
30-60-90 Right Triangle
60
2
1
30
3
This is our reference triangle for
the 30-60-90 triangle.
We will use a reference triangle
to set up a proportion then solve.
30-60-90 Right Triangle
60
x
2x
30
x 3
Ex: 1
60
8
60
x
2
1
30
y
x
1

8
2
2x  8
x4
30
y
3

8
3
2
8 3  2y
4 3y
Ex: 2
Solve for x
30
60
x
1
2
24
60
30
24
2

x
1
2x = 24
x = 12
3
Ex: 3
30
60
14
2
1
y
30
3
60
x
14
2

x
1
2x = 14
x=7
14
2

y
3
2y = 14 3
y=7 3
Ex: 4
x
5 3
60
1
60
2
30
y
30
3
5 3
3

x
1
x=5

3
5 3
y
2
y = 10
• To find trigonometric ratios using right triangles, and
• To solve problems using trigonometric ratios.
Great Chief Soh Cah Toa
A young brave, frustrated by his inability to understand the geometric
constructions of his tribe's battle dress, kicked out in anger against a
stone and crushed his big toe. Fortunately, he learned from this
experience, and began to use study and concentration to solve his
problems rather than violence. This was especially effective in his study
of math, and he went on to become the wisest man of his tribe. He
studied many aspects of trigonometry; and even today we remember
many of the functions by his name. When he became an adult, the tribal
priest gave him a name that reflected his special nature -- one that
reminded them of his great discoveries and of the event which changed
his life. Because he was troubled throughout his life by the problematic
foot, he was constantly at the edge of the river, soaking his aches in the
cooling waters. For that behavior, he was named
Chief Soh Cah Toa.
•Sine: Opposite side over hypotenuse.
Cosine: Adjacent side over hypotenuse.
•Tangent: Opposite side over adjacent.
Find the sin S, cos S, tan S, sin E, cos E and tan E.
Express each ratio as a fraction and as a decimal.
M
Sin S = ME/SE = 3/5 or 0.6
Cos S = SM/SE = 4/5 or 0.8
Tan S = ME/SM = ¾ or 0.75
Sin E = SM/SE = 4/5 or 0.8 S
Cos E = ME/SE = 3/5 or 0.6
Tan E = SM/ME = 4/3 or 1.3
4
3
E
5
Find each value using a calculator. Round to
the nearest ten thousandths.
1. Cos 41
2. Sin 78
A plane is one mile about sea level when it begins
to climb at a constant angle of 2 for the next 70
ground miles. How far above sea level is the
plane after its climb?
2
h
1 mi
Sea level
70 mi
Tan 2 = h/70
70 tan 2 = h
h = 2.44
37
Show that the sin and cos
give you the same
hypotenuse.
4
53
3
-1
SINE
Pronounced
“sign inverse”
-1
COSINE
Pronounced
“co-sign
inverse”
-1
TANGENT
Pronounced
“tan-gent
inverse”
Greek Letter q
Prounounced
“theta”
Represents an unknown angle
Opp Leg
Sin 
Hyp
Adj Leg
Cos 
Hyp
Opp Leg
Tan 
Adj Leg
hypotenuse
q
adjacent
opposite
opposite
We need a way
to remember
all of these
ratios…
Some
Old
Hippie
Came
A
Hoppin’
Through
Our
Old Hippie Apartment
SOHCAHTOA
Old Hippie
Sin
Opp
Hyp
Cos
Adj
Hyp
Tan
Opp
Adj
Finding an angle.
(Figuring out which ratio to use and getting to
use the 2nd button and one of the trig buttons.)
Ex. 1: Find q. Round to four decimal places.
nd
2
17.2
q
9
17.2
tan q 
9
tan 17.2 
9
)
q  62.3789
Shrink yourself down and stand where
the angle is.
Now, figure out which trig ratio you have
and set up the problem.
Make sure you are in degree mode (not radians).
Ex. 2: Find q. Round to three decimal places.
7
q
23
nd
2
7
cos q 
23

cos 7
23
q  72.281
Make sure you are in degree mode (not radians).
)
Ex. 3: Find q. Round to three decimal places.
q
200
sin q 
400
200
nd
2
sin
200

400 )
q  30
Make sure you are in degree mode (not radians).
When we are trying to find a side
we use sin, cos, or tan.
When we are trying to find an angle
we use sin-1, cos-1, or tan-1.