The

a

b

c

222 cba

This is a right triangle:

We call it a right triangle because it contains a right angle.

The measure of a right angle is 90o

90o

The little square

90o

in theangle tells you it is aright angle.

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,

3

4

5

and you square the lengths of the two sides that make up the right angle,

24233

4

5

and add them together,

3

4

5

2423 22 43

22 43

you get the same number you would get by squaring the other side.

222 543 3

4

5

Is that correct?

222 543 ?

25169 ?

It is. And it is true for any right triangle.

8

6

10222 1086

1006436

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

a

b

is called the

And the length of the hypotenuse

is usually labeled c.

a

b

c

The relationship Pythagorus discovered is now called The Pythagorean Theorem:

a

b

c

The Pythagorean Theorem says, given the right triangle with legs a and b and hypotenuse c,

a

b

c

then

a

b

c

.222 cba

You can use The Pythagorean Theorem to solve many kinds of problems.

Suppose you drive directly west for 48 miles,

48

Then turn south and drive for 36 miles.

48

36

How far are you from where you started?

48

36?

482

Using The Pythagorean Theorem,

48

36c

362+ = c2

Why? Can you see that we have a right triangle?

48

36c

482 362+ = c2

Which side is the hypotenuse? Which sides are the legs?

48

36c

482 362+ = c2

22 3648

Then all we need to do is calculate:

12962304

3600 2c

And you end up 60 miles from where you started.

48

3660

So, since c2 is 3600, c is 60.So, since c2 is 3600, c is

Find the length of a diagonal of the rectangle:

15"

8"?


15"

8"?

b = 8

a = 15

c

222 cba 222 815 c 264225 c 2892 c 17c

b = 8

a = 15

c


15"

8"17

Practice using The Pythagorean Theorem to solve these right triangles:

5

12

c = 13

10

b

26

10

b

26

= 24

(a)

(c)

222 cba 222 2610 b

676100 2 b1006762 b

5762 b24b

12

b

15

= 9

7.2 The Converse of the Pythagorean Theorem

• If c2 = a2 + b2, then ∆ABC is a right triangle.

C A

B

b

ac

Verify a Right Triangle• Is ∆ABC a right triangle?

• Yes, it is a right triangle.

C

AB

12 16

20

2 2 2c a b 2 2 220 12 16

400 144 256 400 400

Classifying TrianglesIn ABC with longest side c:

If c2 < a2 + b2, then ABC is acute.

If c2 = a2 + b2, then ABC is right.

If c2 > a2 + b2, then ABC is obtuse.

C

B A

a b

c

C A

B

b

ac

C A

B

b

a c

Acute Triangles• Show that the triangle is an acute triangle.

• Because c2 < a2 + b2, the triangle is acute.

35

2 2 2c a b

35 16 25 35 41

22 235 4 5

5 4

Obtuse Triangles• Show that the triangle is an obtuse triangle.

• Because c2 > a2 + b2, the triangle is obtuse.

15

2 2 2c a b

225 64 144 225 208

2 2 215 8 12

12 8

Classify Triangles• Classify the triangle as acute, right, or obtuse.

• Because c2 > a2 + b2, the triangle is obtuse.

8

2 2 2c a b

64 25 36 64 61

2 2 28 5 6

6 5

Classify Triangles• Classify the triangle with the given side lengths as acute,

right, or obtuse.

• A. 4, 6, 7

• Because c2 < a2 + b2, the triangle is acute.

2 2 2c a b

49 16 36 49 52

2 2 27 4 6

Classify Triangles• Classify the triangle with the given side lengths as acute,

right, or obtuse.

• B. 12, 35, 37

• Because c2 = a2 + b2, the triangle is right.

2 2 2c a b

1369 144 1225

1369 1369

2 2 237 12 35

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 ∆PQR ~ ∆STU. Recall that the corresponding side lengths of similar triangles are in proportion.

P

R Q

S

U T

Activity: Investigating similar right triangles. Do in pairs or threes

1. Cut an index card along one of its diagonals.

2. 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.

3. You should now have three right triangles. Compare the triangles. What special property do they share? Explain.

4. 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 each other. A B

C

D

∆CBD ~ ∆ABC, ∆ACD ~ ∆ABC, ∆CBD ~ ∆ACD

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.

A B

C

D

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:• You may find it helpful to

sketch the three similar triangles so that the corresponding angles and sides have the same orientation. Mark the congruent angles. Notice that some sides appear in more than one triangle. For instance XY is the hypotenuse in ∆XYW and the shorter leg in ∆XZY.

h3.1 m

Y

X W

h

5.5 m

Z

Y W

5.5 m

3.1 m

6.3 m

Z

X Y

∆XYW ~ ∆YZW ~ ∆XZY.

Solution for b.

• Use the fact that ∆XYW ~ ∆XZY to write a proportion.

YW

ZY= XY

XZ

h

5.5= 3.1

6.3

6.3h = 5.5(3.1)

h ≈ 2.7

The height of the roof is about 2.7 meters.

Corresponding side lengths are in proportion.

Substitute values.

Cross Product property

Solve for unknown h.

Using a geometric mean to solve problems

• In right ∆ABC, altitude CD is drawn to the hypotenuse, forming two smaller right triangles that are similar to ∆ABC From Theorem 9.1, you know that ∆CBD ~ ∆ACD ~ ∆ABC.

A

C

BD

C D

B

A D

C

A C

B

Write this down!

A

C

BD

C D

B

A D

C

A C

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.

BD

CD= CD

AD

Shorter leg of ∆CBD.

Shorter leg of ∆ACD

Longer leg of ∆CBD.

Longer leg of ∆ACD.

Copy this down!

A

C

BD

C D

B

A D

C

A C

B

Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below:

AB

CB= CB

DB

Hypotenuse of ∆ABC.

Hypotenuse of ∆CBD

Shorter leg of ∆ABC.

Shorter leg of ∆CBD.

Copy this down!

A

C

BD

C D

B

A D

C

A C

B

Sides CB and AC also appear in more than one triangle. Their side lengths are also geometric means, as shown by the proportions below:

AB

AC= AC

AD

Hypotenuse of ∆ABC.

Hypotenuse of ∆ACD

Longer leg of ∆ABC.

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

BD

BD

CD= CD

AD

AB

CB= CB

DB

AB

AC= AC

AD

What does that mean?

6

x= x

3

5 + 2

y= y

2

36

x y 5

2

18 = x2

√18 = x

√9 ∙ √2 = x

3 √2 = x

14 = y2

7

y= y

2

√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

904545

906030

2:1:1

2:3:1

Leg:Leg:Hypotenuse

Short Leg:Long Leg:Hypotenuse

2:: xxx

xxx 2:3:

In a 45-45-90 triangle…

We will use a reference triangle to set up a proportion then solve.

45-45-90 Right Triangle

1

1This is our reference triangle for the 45-45-90.

2

45

45


x

x

2x

45

45

3

3

x

1

1

3

1

x

EX: 1 Solve for x Let’s set up a proportion by using our reference triangle.

2

2

3 2x

5

5

x

1

1

5

1

x

2

2

5 2x

EX: 2 Solve for x

x

3

1

1 x

1

3 22

45

EX: 3 Solve for x

x 3

2x

3

2

2

2

x 3 2

2


12

30

60

This is our reference triangle for the 30-60-90 triangle.

3

We will use a reference triangle to set up a proportion then solve.


x2x

30

60

3x

60

30

x

y

8

x

1

8

2y 8

23

2 8x 4x

8 3 2y4 3 y

12

330

60

Ex: 1

60

30

x

24

Solve for x

24

2

x

1

12

330

60

2x = 24

x = 12

Ex: 2

60

30

x

14

14

2

x

1

12

330

60

2x = 14

x = 7

Ex: 3

y

14

2

y

3

2y = 14 3

y = 7 3

60

30

x

x

1

12

330

60

x = 5

Ex: 4

y

y

5 3

5 3

3 2

5 3

3

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.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

E S

4 3

5

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

Cos E = ME/SE = 3/5 or 0.6

Tan E = SM/ME = 4/3 or 1.3

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

1 mi

Sea level

h

70 mi

Tan 2 = h/7070 tan 2 = hh = 2.44

Show that the sin and cos give you the same hypotenuse.

37

53

4

3

SINE-1

Pronounced “sign

inverse”

Pronounced “co-sign inverse”

COSINE-1

Pronounced “tan-gent inverse”

TANGENT-1

Prounounced “theta”

Greek Letter

Represents an unknown angle

oppositehypotenuse

SinOpp

Hyp

Leg

adjacent

CosAdj

Hyp

Leg

TanOpp

Adj

Leg

Leg

hypotenuseopposite

adjacent

We need a way to remember all of these ratios…

Old Hippie

Old Hippie

SomeOldHippieCameAHoppin’ThroughOurApartment

SOHCAHTOA

Old Hippie

Old Hippie

SinOppHypCosAdjHypTanOppAdj

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 . Round to four decimal places.

9

17.2

Make sure you are in degree mode (not radians).

9

2.17tan

2nd tan 17.2 9

3789.62

)

Shrink yourself down and stand where the angle is.Now, figure out which trig ratio you have and set up the problem.

Ex. 2: Find . Round to three decimal places.

23

7


23

7cos

2nd cos 7 23

281.72

)

Ex. 3: Find . Round to three decimal places.

400

200


400

200sin

2nd sin 200 400

30)

When we are trying to find a sidewe use sin, cos, or tan.

When we are trying to find an angle we use sin-1, cos-1, or tan-1.

The

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