Transcript c 2 - Mona Shores Blogs

Chapter 9
Right Triangles
and
Trigonometry
Chapter 9 Objectives
• Prove Pythagorean Theorem
• Utilize Pythagorean converse
• Identify right, obtuse, and acute
triangles using Pythagorean converse
• Define properties of a 45-45-90 triangle
• Define properties of a 30-60-90 triangle
• Apply trigonometry to geometric
situations
• Utilize inverse trigonometry functions
Lesson 9.2
The Pythagorean Theorem
Theorem 9.4:
Pythagorean Theorem
• In a right triangle, the square of the
length of the hypotenuse is equal to the
sum of the squares of the lengths of the
legs.
• c2 = a2 + b2
– c is always the hypotenuse
– a and b are the legs in any order
a
c
b
How would
you find
the area of
the figure
at right?
Proof of Pythagorean Theorem
b
a
a
(Hint: 2 ways)
b
c
c
c
a
Side
length
squared
c
b
a
b
(a + b)2
(a + b)(a + b)
a2 + 2ab + b2
-2ab
Should give
same answer
=
a2 + b2 = c2
4(1/2ab) + c2
2ab + c2
-2ab
Area of
triangles
plus area
of inner
square
Example 1
Find the length of the missing side
I.
a = 3, b = 3
I.

c2 = a2 + b2
c2 = 32 + 32
c2 = 9 + 9
c2 = 18
c = 18
c = 32  4.2
a = 5, c = 13
II.
c2 = a2 + b2
132 = 52 + b2
169 = 25 + b2
b2 = 169 - 25
b2 = 144
b = 144
b = 12
c2 = a2 + b2
Pythagorean Triple
• A Pythagorean triple is a set of
three positive integers that satisfy
the Pythagorean Theorem.
– That means that a,b, and c will all be
nice whole numbers
• except 0
• 3,4,5
• 5,12,13
Example 2
Find the length of the missing side and determine
if the sides form a Pythagorean Triple
I.
a = 18, b = 24
I.
c2 = a2 + b2
c2 = 182 + 242
c2 = 324 + 576
c2 = 900
c = 900
c = 30
Yes, a Pythagorean Triple is formed!
c2 = a2 + b2
Lesson 9.3
Converse of the
Pythagorean Theorem
Theorem 5.13:
Triangle Inequality
• The sum of the lengths of any two
sides of a triangle is greater than
the length of the third side.
3
1
6
4
2
6
4
3
6
Add each combination of two sides to make sure
that they are longer than the third remaining side.
Theorem 9.5:
Converse of the
Pythagorean Theorem
• If c2 = a2 + b2 is true, then the triangle in
question is a right triangle.
– You need to verify the three sides of the
triangle given will make the Pythagorean
Theorem true when plugged in.
– Remember the largest number given is always
the hypotenuse
• Which is c in the Pythagorean Theorem
Example 3
Determine whether the following triangles
are right triangles or not.
I.
4,7,10
I.
c2 = a2 + b2
102 = 42 + 72
100 = 16 + 49
100  65
NO
II.
5,12,13
I.
c2 = a2 + b2
132 = 52 + 122
169 = 25 + 144
169 = 169
YES
Theorem 9.6:
Acute Triangles from
Pythagorean Theorem
• If c2 < a2 + b2, then the triangle is an acute
triangle.
– So when you check if it is a right triangle and the
answer for c2 is smaller than the answer for a2 +
b2, then the triangle must be acute
• It essentially means the hypotenuse shrunk a little!
• And the only way to make it shrink is to make the
right angle shrink as well!
a
c
b
Theorem 9.7:
Obtuse Triangles from
Pythagorean Theorem
• If c2 > a2 + b2, then the triangle is an obtuse
triangle.
– So when you check if it is a right triangle and the
answer for c2 is larger than the answer for a2 + b2,
then the triangle must be obtuse
• It essentially means the hypotenuse grew a little!
• And the only way to make it grow is to make the
right angle grow as well!
a
c
b
Example 4
Determine if the following sides create a right,
obtuse, acute, or no triangle at all.
A) 38, 77, 86
Triangle Y/N
B) 10.5, 36.5, 37.5
c
longest side
c2 = a2 + b2
862 = 382 + 772
7396 = 1444 + 5929
= 7373
7396 >
obtuse
Triangle Y/N
c2 = a2 + b2
37.52 = 10.52 + 36.52
1406.25 = 110.25 + 1332.25
= 1442.5
1406.25 <
acute
Lesson 9.5
Trigonometric Ratios
Trigonometric Ratios
• A trigonometric ratio is a ratio of
the lengths of any two sides in a
right triangle.
• You must know:
–
one angle in the triangle other than the right
angle
– one side (any side) of the triangle.
• These help find any other side of the
triangle.
Sine
• The sine is a ratio of
– side opposite the known angle, and…
– the hypotenuse
• Abbreviated
– sin
• This is used to find one of those sides.
– Use your known angle as a reference point
θ
a
c
b
sin θ =
side opposite θ
hypotenuse
b
= c
Cosine
• The cosine is a ratio of
– side adjacent the known angle, and…
– the hypotenuse
• Abbreviated
– cos
• This is used to find one of those sides.
– Use your known angle as a reference point
θ
a
c
b
cos θ =
side adjacent θ
hypotenuse
a
= c
Tangent
• The tangent is a ratio of
– side opposite the known angle, and…
– side adjacent the known angle
• Abbreviated
– tan
• This is used to find one of those sides.
– Use your known angle as a reference point
θ
a
c
b
tan θ =
side opposite θ
side adjacent θ
b
= a
SOHCAHTOA
S in
o pposite
h ypotenuse
C os
a djacent
h ypotenuse
T an
o pposite
a djacent
• This is a handy way of
remembering which
ratio involves which
components.
• Remember to start at
the known angle as the
reference point.
• Also, each combination
is a ratio
– So the sin is the
opposite side divided by
the hypotenuse
Example 5
If you do
not have a
calculator
with trig
buttons,
then turn to
p845 in
book for a
table of all
trig ratios
up to 90o.
• First determine which trig function you want to
use by identifying the known parts and the
variable side.
• Use that function on your calculator to find the
decimal equivalent for the angle.
• Set that number equal to the ratio of side lengths
and solve for the variable side using algebra.
7
x
4
sin 42o =
42o
x
7
7 (sin 42o) = x
7 (.6691) = x = 4.683
37o
x
cos 37o =
4
x
x (cos 37o) = 4
4
4
x=
= 5.008
=
o
cos 37
.7986
Get x out
of
denominat
or first by
multiplying
both sides
by x.
Lesson 9.4
Special Right Triangles
Theorem 9.8:
45-45-90 Triangle
• In a 45-45-90 triangle, the hypotenuse is
2 times as long as each leg.
– Remember each leg is the same length
• Theorem 4.7: Converse of the Base Angles
45o
2
1
45o
1
Example 6
• If you are needing to
find the length of the
hypotenuse, simply
multiply the length of
either leg by 2.
– Best way to do that
is simply write the
leg followed by 2.
I
• When you know the
hypotenuse, find the
number in front of
the 2 and that is the
length of your leg.
• If there is no 2, then
you need to divide by
2 on your calculator.
II
45o
4 2
4
45o
6 2
6
45o
4
45o
6
Find the length of the missing side(s)
Theorem 9.9:
30-60-90 Triangle
• In a 30-60-90 triangle
– the hypotenuse is twice the length of the
shortest leg, and…
– the longer leg is the 3 times as long as the
shortest leg.
60o
1
2
30o
3
Example 7
• Knowing the shortest
leg and trying to find
the hypotenuse,
simply multiply the
shortest leg by 2.
• If you need to find
the shortest leg, then
divide the hypotenuse
by 2.
60o
4
• Knowing the shortest
leg and trying to find
the longer leg,
multiply the shortest
leg by 3.
– Or just write 3
after the length of
the shortest leg.
8
30o
43
Find the length of the missing side(s)
Lesson 9.6
Solving Right Triangles
Inverse Trig Ratios
Inverse trig ratios are used to find the measure of the
angles of a triangle.
The catch is…you must know two side lengths.
Those sides determine which ratio to use based on the
same ratios we had from before.
Finding Side Lengths
Finding Angle Measures
sin
sin-1
cos
cos-1
tan
tan-1
SOHCAHTOA
Example 8
• You still base your ratio on what sides are you
working with compared to the angle you want to
find.
• Only now, your variable is θ.
• So once you find your ratio, you will then use the
inverse function of your ratio from your calculator
7
4
9
θ
17
θ
4
7
θ = sin-1
θ = sin-1 .5714
sin θ =
θ = 34.8o
4
7
9
17
θ = cos-1
cos θ =
SOHCAHTOA
9
17
θ = cos-1 .5294
θ = 58.0o
Solving a Triangle
• To solve a right triangle, you must
find
– all 3 sides
– all 3 angles
• or the other 2 angles besides the right
angle
• So your final answer when solving a
right triangle will have six parts to
the answer!
Lesson 9.7
Vectors
Vector Magnitude
• The magnitude of a vector, AB, is the distance
from the initial point A to the terminal point B.
– It is written as AB
• Because the magnitude of a vector is essentially
finding the distance between two points, we can
use the Distance Formula to calculate the
magnitude of the vector.
– AB = [(x2 – x1)2 + (y2 – y1)2]
• Or if you know the component form of the vector,
you can use the Pythagorean Theorem to
calculate the magnitude.
• x2 + y2 = AB2
Example 9
Calculate the magnitude of the vector AB.
I.
A(0,0) , B(4,5)
I.
AB = [(x2 – x1)2 + (y2 – y1)2]
AB = [(4 – 0)2 + (5 – 0)2]
AB = [42 + 52]
AB = [(16 + 25]
AB = 41  6.4
•
AB = <-4,-2>
II.
x2 + y2 = AB2
(-4)2 + (-2)2 =AB2
16 + 4 =AB2
AB2 = 20
AB = 20  4.5
Direction of a Vector
• The direction of a vector, , is determined by the
angle it makes with a horizontal line that intersects
the initial point A.
• Since the vector has a horizontal and vertical
component, we can fill in the triangle to use a
trigonometric ratio in calculating the direction of a
vector.
• Knowing the components of the vector leave us with
the inverse tangent to calculate the necessary angle
–
 = tan-1 y/x
• Two vectors are equal if they have the same magnitude and
direction.
–
However, they need not share the same initial or terminal
points.
• Two vectors are parallel if they have the same or opposite
direction.
–
The opposite direction would be 180o different than the original.
Adding Vectors
• The sum of any two vectors is
the individual sums of the
horizontal and vertical
components.
u = <a,b>
v = <c,d>
u+ v = <a+c,b+d>