Transcript Ch 9 Right Triangles and Trig

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
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
A
short leg
B
B
short leg
B
long
long leg
long leg
C
A
D
short
Color code all 3 sides of all
3 triangles on the front and back.
D
Arrange the
small and
medium triangles
on top of the
large triangle like
this.?
Theorems 9.1 – 9.3
If the altitude is drawn to
the hypotenuse of a right
triangle, then…
Theorem 9.1
the three triangles formed
are similar to each other.
…
B
C
BD AD
=
CD BD
D
A
BD is a side of
the medium 
and a side of
the small 
BD AD
= BD
CD
B
x
n
=
m
x
C
x
m
D
n
A
Theorem 9.2
…the altitude is the
geometric mean of the two
segments of the hypotenuse.
B
x
n
=
m
x
C
x
m
D
n
A
When you do these problems,
always tell yourself…
____ is the
geometric mean
of ____ and ____
1. Find x.
x
3
= x
8
x ≈ 4.9
C
8
B
x
D
3
A
2. Find x.
4
x
= 4
8
x=2
C
8
B
4
D
x
A
B
C
CB CA
=
CD CB
D
A
CB is a side of
the large  and
a side of the
medium 
CB CA
= CB
CD
B
x
h
=
m
x
x
C
m
D
h
A
B
C
AB
AC
=
AD AB
D
A
AB is a side of
the large  and
a side of the
small 
AB AC
= AB
AD
B
x
h
=
n
x
x
C
D
h
n
A
Theorem 9.3
…the leg of the large triangle is
the geometric mean of the
“adjacent leg” and the
hypotenuse. B
x
h
=
m
x
y
h
=
C
n
y
x
y
m
n
D
h
A
When you do these problems,
always tell yourself…
____ is the
geometric mean
of ____ and ____
3. Find x.
x
14
= x
9
x
x ≈ 11.2
C
B
D
9
14
A
4. Find x.
x
10
= x
4
B
x
x ≈ 6.3
C
D
10
4
A
5. Find x, y, z
x
x
13
=
x
9
x ≈ 10.8
y
4
=
y
9
y=6
y
z
9
4
z
13
=
z
4
z ≈ 7.2
3
5
=
3
x
x = 1.8
5
6. Find h.
h
3.2
=
1.8
h
h = 2.4
3.2
4
x1.8
h
3
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 9.4
Pythagorean Theorem
2
hyp
2
leg
2
leg
=
+
2
2
2
(c = a + b )
102 = y2 + 82
2. Find x.
2 + 64
100
=
y
Why can’t we use a
geo mean proportion? 36 = y2
y=6
x
10
8
x2 = 142 + 82
x2 = 196 + 64
y
20
14
x2 = 260
x ≈ 16.1
A Pythagorean Triple is
a set of three positive
integers that satisfy the
2
2
2
equation c = a + b .
The integers 3, 4, and 5 form a
Pythagorean Triple because
52 = 32 + 42.
Theorem 9.5
Converse of the
Pythagorean Theorem
2
c
2
a
2
b,
If = +
then the
triangle is a right triangle.
The “hypotenuse”
is toois
The hypotenuse
long for
opposite
thethe
perfect
length for
anglethe
to be
90˚
opposite
angle to
be
The
c2 >90˚
a2 +“hypotenuse”
b2
2 = a
2 + bfor
2 the
isctoo
short
opposite angle to
be 90˚
c2 < a2 + b2
Theorem 9.6
If
2
c
<
2
a
2
b,
+
then the
triangle is an acute
triangle.
Theorem 9.7
If
2
c
>
2
a
2
b,
+
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
Can a triangle
be formed?
What kind of
triangle?
L=M+S
No  can be formed
L2 = M2 + S2
Right 
L<M+S
Yes,  can be formed
L2 < M2 + S2
Acute 
L>M+S
No  can be formed
L2 > M2 + S2
Obtuse 
Do the lengths represent the lengths
of a triangle? Is it a right triangle,
acute triangle, or obtuse triangle?
262 = 102 + 242
right triangle
26 < 10 + 24?
3. 10, 24, 26
4. 3, 5, 7
7 < 5 + 3?
5. 5, 8, 9
9 < 8 + 5?
72 > 32 + 52
obtuse triangle
92 < 52 + 82
acute triangle
Do the lengths represent the lengths
of a triangle? Is it a right triangle,
acute triangle, or obtuse triangle?
6. 5, 56 , 9
9 < 5 + 56 ?
7. 23, 44, 70
70 < 23 + 44?
8. 12, 80, 87
87 < 12 + 80?
92 = 52 + 56 2
right triangle
not a triangle
872 > 122 + 802
obtuse triangle
Find the area of the triangle.
9.
92 = x2 + 62
6
81 = x2 + 36
9
A ≈ ½ (6)(6.7)
≈ 20.12 units2
45 = x2
6.7 = x
Find the area of the triangle.
10.
13
202 = x2 + 132
400 = x2 + 169
20
A ≈ ½ (13)(15.2)
≈ 98.8 units2
231 = x2
15.2 = x
Find the area of the triangle.
11.
7
x
72 = x2 + 22
49 = x2 + 4
2
A ≈ ½ (4)(6.7)
≈ 13.4 units2
45 = x2
6.7 = x
Which method requires less ribbon?
How much ribbon is needed using Method 1?
(3+12+3+12) + (3+6+3+6) = 48 in.
The diagram shows the
30
?
ribbon for Method 2.
How much ribbon is
18
?
needed to wrap the
box?
302 + 182 = 35 in
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 9.4
Special Right Triangles
Today, we are going to…
…find the side lengths of special
right triangles
1. Find x. Use the Pythagorean
Theorem to find y. Leave y in
simplest radical form.
y5 2
45˚
x=5
y
x
45˚
5
2. Find x. Use the Pythagorean
Theorem to find y. Leave y in
simplest radical form.
y9 2
45˚
x=9
y
x
45˚
9
Do you notice a pattern?
45˚
45˚
5 2
5
9 2
9
45˚
5
45˚
9
Theorem 9.8
45˚- 45˚- 90˚
Triangle Theorem
45˚
x 2
x
45˚
x
In a 45˚- 45˚- 90˚ Triangle,
hypotenuse = leg 2
3. Find x and y.
y=7 2
45˚
x=7
y
x
45˚
7
4. Find x and y.
45˚
x=3
y=3
3 2
x
45˚
y
5. Find x and y.
2
10
x=
2
45˚
10
x
x =5 2
45˚
y
y =5 2
6. Label the measures of all angles.
Find x. Use the Pythagorean
Theorem to find
30˚ 30˚
y in simplest
radical form.
6
y=3 3
x=3
6
y
x 60˚
60˚
6
7. Label the measures of all angles.
Find x. Use the Pythagorean
Theorem to find
30˚ 30˚
y in simplest
radical form.
8
y=4 3
x=4
8
y
x 60˚
60˚
8
Do you notice a pattern?
6
3 3
8
4 3
60˚
60˚
3
4
Theorem 9.9
30˚-60˚-90˚
Triangle Theorem
2s
s 3
In a 30˚-60˚-90˚
Triangle,
hypotenuse = 2 short leg
long leg = short leg 3
60˚
s
8. Find x and y.
30˚
x=5
y=5 3
y
10
60˚
x
9. Find x and y.
30˚
x = 20
y = 10 3
y
x
60˚
10
10. Find x and y.
30˚
x = 12
y
3
12
y = 24
60˚
x
11. Find x and y.
30˚
12
y
3
x=8 3
x
12
y= 4 3
60˚
y
12. In regular hexagon ABCDEF,
find x and y.
360
y=
12
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 & 9.6
Trigonometric
Ratios
Today, we are going to…
…find the sine, cosine, and tangent
of an acute angle
…use trigonometric ratios to solve
problems
Trigonometric Ratios
sine
cosine
tangent
B
opposite
A
C
hypotenuse
adjacent to A
A
AC is adjacent
BC
AB
____
the hypotenuse
opposite
to
A
A
B
adjacent
to  B
C
hypotenuse
opposite
B
A
____
BC is adjacent
AC
AB
opposite
to
B
B
the hypotenuse
sin A =
leg opposite A
hypotenuse
cos A =
leg adjacent to A
hypotenuse
tan A =
leg opposite A
leg adjacent to A
SOH
CAH
TOA
Sine is Opp / Hyp
Cosine is Adj / Hyp
Tangent is Opp / Adj
B
1.
13
5
5
sin A = 13
12
cos A =
13
5
tan A =
12
12
= 0.3846
= 0.9231
= 0.4167
A
B
2.
13
5
12
sin B = 13
5
cos B =
13
12
tan B =
5
12
= 0.9231
= 0.3846
= 2.4000
A
Find the Sine, Cosine, and Tangent.
3. sin 32° = 0.5299
4. cos 58°= 0.5299
5. tan 32° = 0.6249
6. Find x and y to the nearest tenth.
cos 42˚ = x
12
1
x = 12 cos 42˚
ADJ
OPP
x ≈ 8.9
sin 42˚ = y
12
1
y = 12 sin 42˚
HYP
y ≈ 8.0
7. How tall is the tree?
tan 65˚ = x
30
1
x
x = 30 tan 65˚
64.34 ft
65°
30 ft
8. Find x to the nearest tenth.
tan 42˚ = 12
x
1
12 = x tan 42˚
12
x=
tan 42˚
x = 13.3
x
42°
12
48˚
?
tan 48˚ = x
12
1
x = 12 tan 48˚
x = 13.3
Use Inverse Sine, Inverse
Cosine, and Inverse Tangent.
16
m A = 40°
9. sin A =
25
45
m A = 32°
10. cos A =
53
11. tan A = 0.4402
m A = 24°
Use Inverse Sine, Inverse
Cosine, and Inverse Tangent.
12. sin-1(0.7660)= A
13. cos-1
5
=A
13
m A = 50°
m A = 67°
14. tan-1(11.4300) = A m A = 85°
Calculator language:
5
=A
cos-1
13
Human language:
5
cos A =
13
15. Solve the right triangle.
mA = 28°
30
adj
C
A
opp
16
16
tan A =
30
mB = 62°
AB = 34
(AB)2 = 302 + 162
B
16. Solve the right triangle.
C
6
sin A =
10
opp
A
mA = 37°
mB = 53°
AC = 8
6
10 hyp
102 = (AC)2 + 62
B
angle of elevation
A
C
B
angle of depression C
A
B
x
tan 20°=
8
x = 8 tan 20°
x = 2.9 ft
?
30
sin 45°=
26+x
30 = (26+x) sin45°
sin 45°
sin 45°
42.43 = 26+x
16.43 = x
For the most comfortable height, the
handle should be 16.43 inches.
60
sin 28° = x
x  127.8
300
cos 22° = x
x  323.56
x
x
44˚
12
cos 44° = x
12 miles
x  16.68
44˚
x
tan 44° = 12
x  11.59
12 miles
x
tan 25° = 250
50
tan 58° = x
x  116.6
x  31.2
x
cos 32° = 100
x  84.8
30
sin 40° = x
x  46.7
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 |.
WeInuse
absolute
other
words,value
it is
symbols
because
“the length
of the length
vector”
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
For example, to make AB, we go
left 5 units and down 3 units.
The component form of AB is
-5,
-3
__________
To find the magnitude of AB,
we simplify (-5)2 + (-3)2
and get
34  5.8
1. Find the magnitude of the vector.
B
First,
the
Now, write
find the
componentusing
form the
of
magnitude
the
vector.
formula.
?
A
+5
4,5
+4
2
(4)
+
2
(5)
| AB | ≈ 6.4
2. Find the magnitude of the vector.
B
component
form?
magnitude
formula?
-7,5
+5
(-7)2 + (5)2
-7
A
| AB | ≈ 8.6
The direction of a vector is
determined by the angle it
makes with a horizontal line.
45˚
45˚ southeast
Identify the direction of the vector.
3.
30˚
30˚ southwest
4.
50˚
50˚ northwest
5. Find the direction of the vector
5
4
m A = 51˚
sin, cos,
tan
A
=
or tan?
+5
?
markthe
the
mark
hypotenuse
adjacent
leg
opposite
+4
51° northeast
opposite
6. Find the direction of the vector
sin, cos,
ortan
tan?A =
5
7
m A = 36˚
+5
?
-7
opposite leg?
hypotenuse?
36°
adjacent leg?
northwest
Use +7
because it
is length
7. Find the magnitude and direction
of AB if A (1,2) and B (4, 6).
Step 1: Component Form
We
doxthis, without
drawing
the
vector!
xcan
–
y
–
y

2
1
2
1 = X , Y
Step
4 –2:
1 ,Magnitude
6 – 2 = 3 , 4
| AB | =
2
(X)
+
2
(Y)
Step
3:
Direction
| AB | = (3)2 + (4)2|Y=| 25  5
4
tan A = tan
A
=
3
|X|
53 northeast
How do we know if the vector is
north or south, east or west
without sketching it?
LOOK at the component form!
+,+ is right and up
-,+ is left and up
-,- is left and down
+,- is right and down
northeast
northwest
southwest
southeast
northwest
northeast
-,+
 
right,down
left+,+
,down
right,up
left
,up
+,-
-,-
southwest
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).
x2 – x1 , y2 – y1 = X , Y
–
4 – 3 , – 5 – 3 = 7 , - 8
| AB | =
2
(X)
+
2
(Y)
= (7)2 + (-8)2
| AB | = 113  10.6
8
|Y|
tan A = 7
tan A =
|X|
49 southeast
Component Form
x2 – x1 , y2 – y1 = X , Y
Magnitude
| AB | =
2
(X)
+
2
(Y)
Direction
|Y|
tan A =
|X|
A north/south - east/west
Adding Vectors in
Component Form
9.
2,-3 + 5,4 = 7,1
See a pattern?
Pick
any
starting
point
from
movethat
right
point,
2 and
godown
right35 and up 4
7,1
2,-3
5,4
instead of taking this route,
you could take a short cut
10.
-3,5 + 9,-2 = 6,3
See a pattern?
from
Pickthat
aleft
starting
point,
point
9 and down 2
move
3 andgo
upright
5
9,-2
short cut?
-3,5
6,3
u  3, 4 , v  6,5 , w  2,1
11.
uv 
12.
uw 1,-3
9,1
60
sin 28° = x
x  127.8
300
cos 22° = x
x  323.56
x
x
44˚
12
cos 44° = x
12 miles
x  16.68
44˚
x
tan 44° = 12
x  11.59
12 miles
x
tan 25° = 250
x  116.6
x
cos 32° = 100
x  84.8
50
tan 58° = x
x  31.2
30
sin 40° = x
x  46.7