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Lesson 9.1 Using Similar Right Triangles. Students need scissors, rulers, and note cards. Today, we are going to… …use geometric mean to solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle. - PowerPoint PPT Presentation
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Lesson 9.1Using Similar Right Triangles
Students need scissors, rulers, and note cards.
Today, we are going to……use geometric mean to solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle
With a straight edge, draw one diagonal of the note card.
Draw an altitude from one vertex of the note card to the diagonal.
Cut the note card into three triangles by cutting along the segments.
C
BAA D
B
DB
Chy
pote
nuse
hyp
hyp
short leg
short legshortlo
ng
leg
lon
g le
g
lon
g
Color code all 3 sides of all 3 triangles on the front and back.
Arrange the small and
medium triangles on top of the
large triangle like this.?
If the altitude is drawn to the hypotenuse of a right
triangle, then…
Theorems 9.1 – 9.3
Theorem 9.1
…the three triangles formed are similar to each other.
BDBDCD
AD=
BD is a side of the medium and a side of the small
A
B
C D
D AC
B
x
m n
xx=mn
BDBD=
CDAD
Theorem 9.2
…the altitude is the geometric mean of the two
segments of the hypotenuse.
D AC
B
x
m n
xx=mn
____ is the geometric mean of ____ and ____
When you do these problems, always tell yourself…
D AC
B
x
8 3
xx=
83
1. Find x.
x ≈ 4.9
D AC
B
4
8 x
44=
8x
2. Find x.
x = 2
CBCD CB
CA=
A
B
C D
CB is a side of the large and
a side of the medium
D AC
B
x
m
h
xx=mh
CBCB=
CDCA
ABAD AB
AC=
A
B
C D
AB is a side of the large and
a side of the small
D AC
B
x
nh
xx=nh
ABAB=
ADAC
Theorem 9.3
…the leg of the large triangle is the geometric mean of the
“adjacent leg” and the hypotenuse.
D AC
B
x
m n
xx=mh
y
h
yy=nh
____ is the geometric mean of ____ and ____
When you do these problems, always tell yourself…
D AC
B
x
9
14
xx=9
143. Find x.
x ≈ 11.2
D AC
B
x
410
xx=4
104. Find x.
x ≈ 6.3
5. Find x, y, zx y z
9 4
xx=9
13
x ≈ 10.8
yy=94
y = 6
zz=4
13
z ≈ 7.2
33=x56. Find h.
x = 1.8
h4
3
5
x1.8
3.2h
h=
3.2
1.8
h = 2.4
Lesson 9.2 & 9.3 The Pythagorean Theorem
& Converse
Today, we are going to……prove the Pythagorean Theorem…use the Pythagorean Theorem and
its Converse to solve problems
Theorem 9.4Pythagorean Theorem
hyp2 = leg2 + leg2 (c2 = a2 + b2 )
2. Find x.
8
20
10 x
y
102 = y2 + 82
100 = y2 + 6436 = y2
14
x2 = 142 + 82
x2 = 196 + 64
x2 = 260
x ≈ 16.1
y = 6
Why can’t we use a geo mean proportion?
A Pythagorean Triple is a set of three positive
integers that satisfy the equation c 2 = a 2 + b 2.
The integers 3, 4, and 5 form a Pythagorean Triple because
52 = 32 + 42.
Theorem 9.5Converse of the
Pythagorean Theorem
If c2 = a2 + b2, then the triangle is a right triangle.
The “hypotenuse” is too short for the opposite angle to be 90˚ c2 < a2 + b2
The “hypotenuse” is too long for the opposite angle to be 90˚
c2 > a2 + b2
The hypotenuse is the perfect length for the opposite angle to be 90˚
c2 = a2 + b2
Theorem 9.6
If c2 < a2 + b2, then the triangle is an acute
triangle.
Theorem 9.7
If c2 > a2 + b2, then the triangle is an obtuse
triangle.
How do we know if 3 lengths can represent the side lengths
of a triangle?
L < M + S
What kind of triangle?
L2 = M2 + S2
L2 < M2 + S2
L2 > M2 + S2
Right
Acute
Obtuse
Can a triangle be formed?
L = M + S
L < M + S
L > M + S
No can be formed
Yes, can be formed
No can be formed
Do the lengths represent the lengths of a triangle? Is it a right triangle, acute triangle, or obtuse triangle?
3. 10, 24, 26
4. 3, 5, 7
5. 5, 8, 9
262 = 102 + 242
72 > 32 + 52
92 < 52 + 82
right triangle
obtuse triangle
acute triangle
26 < 10 + 24?
7 < 5 + 3?
9 < 8 + 5?
92 = 52 + 56 2
Do the lengths represent the lengths of a triangle? Is it a right triangle, acute triangle, or obtuse triangle?
6. 5, 56 , 9
7. 23, 44, 70
8. 12, 80, 87 872 > 122 + 802
right triangle
not a triangle
obtuse triangle
70 < 23 + 44?
87 < 12 + 80?
9 < 5 + 56 ?
Find the area of the triangle.
9.
6
9
81 = x2 + 36
92 = x2 + 62
45 = x2
6.7 = xA ≈ ½ (6)(6.7)
≈ 20.12 units2
Find the area of the triangle.
10.13
20400 = x2 + 169
202 = x2 + 132
231 = x2
15.2 = xA ≈ ½ (13)(15.2)
≈ 98.8 units2
Find the area of the triangle.
11.
49 = x2 + 4
72 = x2 + 22
45 = x2
6.7 = xA ≈ ½ (4)(6.7)
≈ 13.4 units2
7
2
x
How much ribbon is needed using Method 1?
Which method requires less ribbon?
The diagram shows the ribbon for Method 2. How much ribbon is needed to wrap the box?
(3+12+3+12) + (3+6+3+6) = 48 in.
?
?
30
18
35 in
302 + 182 =
Project ideas…A 20-foot ladder leans against a wall so that the base of the ladder is 8 feet from the base of the building. How far up on the building will the ladder reach?
A 50-meter vertical tower is braced with a cable secured at the top of the tower and tied 30 meters from the base. How long is the cable?
The library is 5 miles north of the bank. Your house is 7 miles west of the bank. Find the distance from your house to the library.
Project ideas…Playing baseball, the catcher must throw a ball to 2nd base so that the 2nd base player can tag the runner out. If there are 90 feet between home plate and 1st base and between 1st and 2nd bases, how far must the catcher throw the ball?
While flying a kite, you use 100 feet of string. You are standing 60 feet from the point on the ground directly below the kite. Find the height of the kite.
Lesson 9.4Special Right Triangles
Today, we are going to……find the side lengths of special right triangles
45˚
45˚
5
xy
1. Find x. Use the Pythagorean Theorem to find y. Leave y in simplest radical form.
x = 5y 5 2
45˚
45˚
9
xy
2. Find x. Use the Pythagorean Theorem to find y. Leave y in simplest radical form.
x = 9y 9 2
Do you notice a pattern?
45˚
45˚
5
55
45˚
45˚
9
992 2
In a 45˚- 45˚- 90˚ Triangle,
hypotenuse = leg 2
Theorem 9.845˚- 45˚- 90˚
Triangle Theorem
45˚
45˚
x
xx 2
45˚
45˚
7
xy
3. Find x and y.
x = 7y = 7 2
45˚
45˚
y
x3
4. Find x and y.
2x = 3
y = 3
45˚
45˚
y
x10
5. Find x and y.
x = 5 2
x =
2
10
2
y = 5 2
6. Label the measures of all angles.Find x. Use the Pythagorean Theorem to find y in simplest radical form. 66
6
x
y
x = 3 60˚60˚
30˚30˚
y = 3 3
7. Label the measures of all angles.Find x. Use the Pythagorean Theorem to find y in simplest radical form. 88
8
x
y
x = 4 60˚60˚
30˚30˚
y = 4 3
Do you notice a pattern?
30˚
60˚
3
63 3
60˚
30˚
8
4
34
hypotenuse = 2 short leg
long leg = short leg 3
Theorem 9.930˚-60˚-90˚
Triangle Theorem
60˚
30˚
2s
s
3s
In a 30˚-60˚-90˚ Triangle,
8. Find x and y.
10
x
y
60˚
30˚x = 5
y = 5 3
9. Find x and y.
x
10
y
60˚
30˚x = 20
y = 10 3
10. Find x and y.
y
x
12
60˚
30˚
3
x = 12
y = 24
11. Find x and y.
x
y
12
60˚
30˚
y 12
3
y = 4 3
x = 8 3
12. In regular hexagon ABCDEF, find x and y.
36012
y =
y = 30
x = 60
13. Find x and y.
x = 24
y = 12 3
14. Find x and y.
x = 8
y = 8 2
Lesson 9.5 & 9.6Trigonometric
RatiosToday, we are going to……find the sine, cosine, and tangent of an acute angle…use trigonometric ratios to solve problems
Trigonometric Ratios
sinecosinetangent
A
B
C
hypotenuse
adjacent to A
opposite A
____ is the hypotenuse____ is opposite A____ is adjacent to AABBCAC
A
B
C
hypotenuse
opposite B
adjacent to B
____ is the hypotenuse____ is opposite B____ is adjacent to BABACBC
sin A =
leg opposite A
hypotenuse
cos A =
leg adjacent to A
hypotenuse
tan A =
leg opposite A
leg adjacent to A
SOHCAHTOA
Sine is Opp / Hyp
Cosine is Adj / Hyp
Tangent is Opp / Adj
1.
sin A = = 0.38465
13
12
513
A
Opp
Hyp
Adj
cos A =
tan A = = 0.4167 5 12
= 0.9231 12 13
B
2.
sin B = = 0.923112 13
12
513
A
Adj
Hyp
Opp
cos B =
tan B = = 2.400012 5
= 0.3846 5 13
B
Find the Sine, Cosine, and Tangent.
3. sin 32° =
5. tan 32° =
0.5299
0.5299
0.6249
4. cos 58°=
6. Find x and y to the nearest tenth.
OPP
ADJ
HYPsin 42˚ =
cos 42˚ = x121
12
x
y42°
x = 12 cos 42˚
x ≈ 8.9
y121
y = 12 sin 42˚ y ≈ 8.0
7. How tall is the tree?
64.34 ft
tan 65˚ = x301
x = 30 tan 65˚
65°30 ft
xOpp
Hyp
Adj
8. Find x to the nearest tenth.
42° 12
x
x = 13.3
tan 42˚ = 12 x1
12 = x tan 42˚
x =tan 42˚
12 ? 48˚
tan 48˚ = x121
x = 12 tan 48˚
x = 13.3
Opp
Hyp
AdjAdj
Hyp
Opp
Use Inverse Sine, Inverse Cosine, and Inverse Tangent.
9. sin A =1625
10. cos A =4553
11. tan A = 0.4402
m A =
m A =
m A =
40°
32°
24°
Use Inverse Sine, Inverse Cosine, and Inverse Tangent.
12. sin-1(0.7660)= A
14. tan-1(11.4300) = A
m A =
m A =
m A =
13. cos-1 513
= A
50°
67°
85°
Calculator language:
cos-1 513
= A
cos A = 513
Human language:
15. Solve the right triangle.
30
16A
C
B
mA = 28°
mB = 62°
AB = 34
tan A = 1630
(AB)2 = 302 + 162
opp
adj
16. Solve the right triangle.
10
6A
C
B
mA = 37°
mB = 53°
AC = 8
sin A = 610
102 = (AC)2 + 62
opp
hyp
angle of elevation
angle of depression
A
B
A
B
C
C
?
20°
tan 20°=x8
x = 8 tan 20°
x = 2.9 ft
sin 45°= 3026+x
sin 45° sin 45° 30 = (26+x) sin 45°
42.43 = 26+x
16.43 = x
For the most comfortable height, the handle should be 16.43 inches.
12 miles
44˚
x
sin 28° = 60 x
x 127.8 cos 22° =
300 x
x 323.56
x 16.68
tan 44° = x 12
x 11.59
12 miles
44˚
x
cos 44° = 12 x
x 116.6
tan 25° = x 250
x 31.2
tan 58° = 50 x
x 84.8
cos 32° = x 100
x 46.7
sin 40° = 30 x
Lesson 9.7 Vectors
Today, we are going to……find the magnitude and the
direction of a vector …add two vectors
The magnitude of a vector PQ is the distance from the
initial point to the terminal point and is written | PQ |.
?
In other words, it is “the length of the vector”We use absolute value
symbols because length cannot be negative.
To find the magnitude of a vector…
Step 1: Identify the vector
component form X, Y
Step 2: Find the magnitude by
simplifying (X)2 + (Y)2
The component form of AB is
__________-5, -3
For example, to make AB, we go left 5 units and down 3 units.
To find the magnitude of AB,
we simplify (-5)2 + (-3)2
and get 34 5.8
+4
+5?
A
B
| AB | ≈ 6.4
1. Find the magnitude of the vector.
4,5
(4)2 + (5)2
First, write the component form of
the vector.
Now, find the magnitude using the
formula.
-7
+5
| AB | ≈ 8.6
A
B2. Find the magnitude of the vector.
-7,5(-7)2 + (5)2
magnitude formula?component form?
The direction of a vector is determined by the angle it
makes with a horizontal line.
45˚
45˚ southeast
30˚ southwest30˚
Identify the direction of the vector.
3.
4.
50˚ northwest
50˚
+5?
5. Find the direction of the vector
51° northeast+4
tan A = 54
m A = 51˚
opposite
adjacent
hypo
tenu
se
mark the hypotenuse
mark the opposite leg mark the adjacent leg
sin, cos, or tan?
+5
6. Find the direction of the vector
36° northwest
?
-7
tan A =
m A = 36˚
5 7
oppo
site
adjacent
hypotenuse
hypotenuse?opposite leg?
adjacent leg?
sin, cos, or tan?
Use +7 because it is length
Step 3: Direction
Step 2: Magnitude
7. Find the magnitude and direction
of AB if A (1,2) and B (4, 6).We can do this without drawing the vector!= X , Yx2 – x1 , y2 – y1
Step 1: Component Form4 – 1 , 6 – 2 = 3 , 4
(X)2 + (Y)2| AB | =
| AB | = (3)2 + (4)2 25 5=tan A =
|Y| |X|
53 northeast
tan A =
4
3
LOOK at the component form!
How do we know if the vector is north or south, east or west
without sketching it?
+,+ is right and up northeast-,+ is left and up northwest
-,- is left and down southwest+,- is right and down southeast
+,+ right,up
northeast
-,+ left ,up
northwest
left ,down-,-
southwest
right,down +,-
southeast
It might help to think about a map of the US!
8. Find the magnitude and direction
of AB if A (-3,3) and B (4, -5).= X , Y
(X)2 + (Y)2
tan A =
|Y| |X|
49 southeast
x2 – x1 , y2 – y1
| AB | =
= 7 , - 84 – –3 , – 5 – 3(7)2 + (-8)2=
| AB | = 113 10.6
tan A =
87
Component Form= X , Y
Magnitude(X)2 + (Y)2
Direction
tan A =
|Y| |X|
A north/south - east/west
x2 – x1 , y2 – y1
| AB | =
Adding Vectors in Component Form
2,-3 + 5,4 =
5,42,-3
7,1
7,19.
Pick any starting pointmove right 2 and down 3from that point, go right 5 and up 4
instead of taking this route, you could take a short cut
See a pattern?
9,-2
-3,5 6,3
Pick a starting pointmove left 3 and up 5from that point, go right 9 and down 2
short cut?
See a pattern?-3,5 + 9,-2 = 6,310.
12.
11.
3, 4 , 6,5 , 2,1u v w ------------- -
u v
u w
------------- - 9 , 1
1 , - 3
12 miles
44˚
x
sin 28° = 60 x
x 127.8 cos 22° =
300 x
x 323.56
x 16.68
tan 44° = x 12
x 11.59
12 miles
44˚
x
cos 44° = 12 x
x 116.6
tan 25° = x 250
x 31.2
tan 58° = 50 x
x 84.8
cos 32° = x 100
x 46.7
sin 40° = 30 x
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