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Trigonometry

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Title: Trigonometry


1
Chapter 4 Trigonometry
Objectives
  • Convert between degree and radian measure
  • Pythagorean identities
  • Find the trigonometric values of any angle
    measure
  • Unit circle
  • Graph trigonometric functions
  • Solve for lengths and angle measures of any
    type of triangle.
  • Verify identities
  • Application of trigonometry

2
Uses of trigonometry
Used to describe the motion of any object that
behaves in a circular, oscillating or periodic
manner.
Angle
Consists of two rays or half lines that originate
at a common point called the vertex.
These two rays have names
  • Terminal side The ray that rotates to
    generate the angle.
  • Initial side The ray that does not move.

Angles are commonly denoted using lower case
Greek letters.
alpha, beta, theta, gamma (respectively)
3
To better describe the formation of angles we
superimpose an xy plane on the angle with the
vertex at the origin.
y
Angles are generated by the terminal side
rotating counterclockwise or clockwise.
B
If angles are generated by the terminal side
moving counterclockwise the angle is positive.
A
x
0
If angles are generated by the terminal side
moving clockwise the angle is negative.
The direction of the arrow inside the angle will
tell you if the terminal side is generating a
positive or negative angle.
4
y
B
If angles are generated by the terminal side
moving clockwise the angle is negative.
A
x
0
5
If we have an angle of 390, this is one
revolution of the unit circle plus 30.
So, we say that 390 is coterminal with 30
Coterminal angles differ by multiples of 360
Ex 1 Determine if the following angles are
coterminal. a.) 90 and 1170 b.) 123 and
844
YES
NO
Solution
If you get a whole number they are coterminal. If
the result is not a whole number then they are
not coterminal.
6
To define the measure of an angle, we first add
the unit circle centered at the origin to the
coordinate system.
y
This is called a unit circle because the length
of the radius is 1.
1
-1
x
1
The equation of the unit circle is
-1
The domain is -1, 1
To generate angles we must consider the terminal
and initial sides.
The initial side is aligned with the x-axis and
the terminal side starts at the x-axis and
rotates to generate the angle.
7
y
This point is P(t). P is the function and t is
the input value.
x
t
For every P(t) on the unit circle we can define
its measure by using degrees or radians.
Radian Measure
For any real number t, the angle generated by
rotating counterclockwise from the positive
x-axis to the point P(t) on the unit circle is
said to have radian measure t.
8
t
3
2
y
(0,1)
1
(-1,0)
(1,0)
x
and so on
-1
(0,-1)
-2
-3
9
We can see that an angle that measures 90 is the
same measure as
We can also see that 180 is
We will use this fact to convert between degrees
and radian measure.
To convert degree measure to radian measure you
multiply the degree measure by
Ex Convert the following degree measure to
radian measure.
a.) 150 b.) 225 c.) -72
10
To convert radian measure to degree measure you
multiply the radian measure by
Ex 2 Convert the following radian measures to
degree measure.
11
3.2 The Sine and Cosine Functions
Recall The terminal side rotates to generate
the angles.
There are infinitely many points on the unit
circle that the terminal side could generate.
We will only memorize a few of them.
We will memorize all angles on the unit circle
that are in increments of 30 and 45.
12
y
Unit Circle
90
60
120
45
135
150
30
180
0
x
330
210
315
225
240
300
270
13
An ordered pair has the form (x, y)
The Sine and Cosine Functions
Suppose that the coordinates of the point P(t) on
the unit circle are (x(t), y(t)). Then the sine
of t, written sin t, and the cosine of t, written
cos t, are defined by sint t y(t) and cos t
x(t)
Our new ordered pairs are of the form P(t)
(cos t, sin t)
14
Finding the cosine and sine values of the common
angles on the unit circle.
y
90
120
60
45
135
30
150
180
0
x
330
210
315
225
We create right triangles by drawing lines
perpendicular to the x or y axis. It does not
matter which axis.
240
300
270
15
To find the cosine and sine of 30 we must use
the triangle we created.
1
60
30
1st We know the third angle is 60 by the
triangle sum theorem.
2nd We know the length of the hypotenuse since
this is a unit circle r 1
3rd Using the properties of a 30-60-90 right
triangle we can find the other two sides.
Since the base of this triangle is on the x-axis
this side would represent the cos t. The height
would represent the sin t.
Recall The side opposite the 30 angle is
half of the hypotenuse. The side
opposite the 60 angle is the product of the
short leg and the square root
of 3.
16
If the hypotenuse has length 1, then the side
opposite 30 is ½ .
The side opposite the 60 angle is
This is how you will solve for all sine and
cosine on the unit circle dealing with 30 60
90 right triangles.
So, P(t)
Finding the coordinates of a point using a 45
45 90 right triangle.
We know
  • The third angle is 45

45
  • The hypotenuse has length 1.

1
  • In a 45 45 90 right triangle the legs are
  • the same length. We can call both legs
  • the same variable since they are equal.

45
Now solve for a.
17
We rationalized the denominator in this
step because we do not leave radicals in
the denominator.
Since the legs have the same length, the cosine
and sine values are the same.
18
Pythagorean Identity
For all real numbers t, (sin t)2 (cos t)2 1
Because of the Pythagorean Identity, sine and
cosine have bounds
For all real numbers t,
The cosine function is even
The sine function is odd
For all real numbers t, cos(-t) cos t
For all real numbers t, sin(-t) - sin t
19
A Periodic Function is a function that repeats
the same thing over and over again.
Trigonometric Functions are periodic because they
repeat.
Sine and Cosine functions have a period of
From your unit circle you can see that the
ordered pairs on your unit circle do not begin to
repeat until after one complete revolution of the
unit circle 360
Reference Number
For any real number t, the reference number r
associated with t is the shortest distance along
the unit circle from t to the x-axis. The
reference number r is always in the interval
Since the answer is always in the above interval,
it is
Ex 2 What is the reference number of
20
Ex 3 Determine the values of t in
that satisfy
This is simply a quadratic function. Use what
you know.
You can view this quadratic as
We cannot work with it this way we can only have
one unknown. We need to turn both variables into
y (sine) or x (cosine).
(sin t)2 (cos t)2 1
Pythagorean Identity
Manipulate this identity so that it is something
that you can use.
We can replace (sin t)2 with (1 (cos t)2)
Now distribute
Now factor and set each factor equal to zero.
21
To make things easy, and we dont like our
leading coefficient being negative, multiply both
sides by -1.
Now factor
It may help you to view it like this to factor.
Set each factor equal to zero and solve.
(2x 1)(x 2) 0
2x 1 0 and x 2 0
Replace x with cos t since x represents cos t.
Where on the unit circle is this true?
cos t 2 will never happen it is outside the
bounds of cosine.
22
Lesson 3.3 Graphs of the Sine and Cosine
Functions
Sine Curve
We will use the values that we memorized from the
unit circle to graph one period of the sine
function.
Sin t
1
t
-1
The more points you plot the more precise your
graph will be. I will plot only the 45
increments of the angles.
23
Cosine Curve
cos t
1
t
-1
24
Ex 1 Use the graphs of y sin x and y cos
x to sketch the graphs of
Trigonometric functions are no different than any
other function when shifts are involved
minus to the right plus to the left.
Sin t
1
graph of y sin x
t
-1
Again, we must be sure the leading coefficient is
1 before we try and see the shifts.
25
Note Your shift will help guide you as to what
you should count by on the x-axis.
sin x
1
x
-1
Q What is the last value going to be that we
will write on the x-axis?
A
NOW YOU TRY THE SECOND ONE!!! ?
26
sin x
1
x
-1
You try
27
We can see from the previous 4 graphs that when
we shift one graph to the right or left that we
obtain the graph of a different equation, such as
The graph of
These are all true only because we are shifting
by
When you shift by other increments
different equalities occur.
28
When shifting by increments of pi
Ex 1 Sketch the graph of
All trig functions are of the form
y A sin(Bx C)
The name of the trig function changes only.
A is the amplitude
Dont forget!!! The leading coefficient must be
1 before you can see the shift!!!
Period is
Shift is
29
Amplitude is 2 so the height will go to 2 and
-2 on y-axis.
B 1, so the period has not changed.
Shift right units
First, I will graph y 2cosx, then shift this
graph to the right units.
Since the period did not change and the shift is
units, we will count by on the
x-axis.
cos x
y 2cos x
Notice that the zeros of this function do not
change. Amplitude is just a vertical elongation
or compression of the graph.
2
x
Now, we shift this graph!
-2
30
cos x
2
x
-2
Ex 2 Sketch the graph of
B 3, not 1 so we must find the length of the
new period.
Recall period is found by
Always reduce if necessary.
Instead of this graph being graphed from 0 to 2p,
the entire cosine curve will be graphed between 0
and
31
Recall To see a horizontal shift the leading
coefficient must be 1 and it is not.
We must factor out the 3 from the quantity.
We will shift this graph to the right
units.
The main points of the cosine curve (y cos x)
are
cos x
2
Divide all of these orginal x-coordinates by 3
and multiply the y-coordinates by 2, these will
be the locations of your new points.
x
-2
with shift
New points
32
Ex 3 Sketch the graph of
cos x
2
x
-2
33
3.4 Other Trigonometric Functions
The Tangent, Cotangent, Secant, and Cosecant
Functions
The tangent, cotangent, secant and cosecant
functions, written respectively as tan x, cot x,
sec x, and csc x are defined by the quotients
Note Tangent and secant are only defined when
cos x ? 0 cotangent is cosecant are
only defined when sin x ? 0
34
Trig. function Sign in quadrant
Trig. function I II III IV
Sin x
Cos x
Tan x
Csc x
Sec x
Cot x
- -
- -
- -
- -
- -
- -
35
Ex
Determine the values of the other trigonometric
functions.
Solution
Since we know that csc x is the reciprocal of
sin x, write the reciprocal of sin x.
Next we must find cos x because the remaining
trig. functions contain it.
What do you know that involves both sine and
cosine that will help you find cos x?
Pythagorean Identity.
36
The interval for sin x is given.
Quadrant II
Cosine is negative in QII, so
cot x
or you could have started at the beginning to
find this solution.
sec x is the reciprocal of cos x. Write it and
simplify
Note You are now ready to write your answers.
Make sure they have the correct sign
for quadrant II.
37
Ex 2
Find the values of the other trigonometric
functions.
Solution
We can see that we are in QIII. The given is the
cosine value for
38
The graph of the tangent function.
The tangent function is zero when the sine
function is zero because sine is in the numerator
of the tangent function.
The tangent function is undefined when the cosine
function is zero because the cosine function is
in the denominator of the tangent function.
The tangent function will be zero at
The tangent function is undefined at
tan t
1
t
y tan x
-1
39
Sine and cosine have periods of ,
therefore, tangent will also repeat itself on
that same period.
Ex 3 Sketch the following graphs
Solution
a.) B 2 set up your inequality
The length of your new period is .
Since we divided the period by 2 we will also
have to divide the restrictions by 2
Original restrictions
New Restrictions
40
a.) y tan 2x
41
b.)
42
This graph is the graph from part b reflected
over the x-axis.
c.)
43
c. contd.)
Now we shift it up one unit.
44
Note The zeros from the graph are not obvious.
To find them we would set the
function equal to zero.
Which implies
If we let x
We have tan x 1 where does this occur, when
x ?
We will find this later this chapter. ?
45
Now that we have the graphs of sine, cosine and
tangent we can graph the remaining trig.
Functions using the reciprocal technique.
Do you remember those properties of graphing
reciprocals?
as f(x) increases, its reciprocal
decreases!
The Cosecant function
Graph y sin x.
There, we have vertical asymptotes.
y csc x is undefined at
Where sin x 0.
Now, use the fact that as y sin x increases, y
csc x decreases and vice versa.
46
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47
Ex Graph y sec x
48
Ex Graph y cot x
Be careful with y cot x.
The restrictions for y tan x are different from
y cot x. They have different vertical
asymptotes.
49
Ex Sketch the graph of
Note There are several strategies for graphing
  • You could start with the parent function and
    build
  • on it one change at a time.
  • You can find the most important pieces of
    information
  • of this new graph and make the changes to
    the pieces,
  • and then plot your new points and new
    vertical
  • asymptotes.

This time, lets make the changes to the parent
function y csc x
50
y csc x has vertical asymptotes at
y csc 3x has vertical asymptotes at
51
Now, the amplitude changes to ½
52
Now, we shift the graph to the left units.
The vertical asymptotes go with it!!! They get
shifted too!
53
3.5 Trigonometric Identities
Using the first Pythagorean Identity we can
manipulate it and obtain two more.
Pythagorean Identities
Sum and Difference Formulas for Sine and Cosine
For every pair of real numbers and ,
we have
and
54
Ex 1 Determine the following
Using the values from the unit circle that we
have committed to memory we will find a
combination of two that equal the function we are
trying to evaluate.
What equals ?
We now label these two
Using the difference formula (we subtracted above
and for sine you do the operation of the
combination that you used to equal your
function), we will evaluate
55
Both terms have a common factor
Factor it out!
You try part b. Hint For cosine you use the
opposite formula compared to the sign you used to
obtain your function.
If you subtracted to get then
you will use the sum formula, and visa versa.
56
b.)
We can also find the tangent of functions on the
unit circle
We can find tangent of t by using this formula or
we can continue to manipulate the formula to
obtain a smaller one.
57
We maintain equality as long as we do the same
thing to both the numerator and the denominator.
Lets divide the numerator and denominator by
Now simplify
58
Ex 2 Determine
To use this formula it is a must that you can
easily find the tangent values of the measures
you have committed to memory.
YOU pick which is easier for you!!! ?
59
Double angle formulas for sine and cosine
sin 2x sin(x x) sin x cos x sin x cos x
2sin x cos x
Ex 3 Determine all the values of x in 0, 2p
that satisfy each equation.
a.) cos x sin 2x b.) sin x cos 2x c.) 1
sin x cos x
Solution
a.) We need to get everything into sin x and cos
x because that is what we know on the unit
circle.
cos x sin 2x 2sin x cos x
cos x 2sin x cos x
0 2sin x cos x cos x
cos x(2sin x 1)
60
0 cos x(2sin x 1)
cos x 0 and 2sin x 1 0
sin x ½
b.) sin x cos 2x
61
c.) 1 sin x cos x
Square both sides.
Re-order expression
In order for sine to ever be equal to 0, 2x must
be a multiple of pi.
x could be any of the following to make this
statement true
CHECK!!!
62
NOTE We squared the equation earlier, so this
could introduce extraneous
solutions. This must have happened because
the original expression, 1 sin x cos
x is only satisfied if the
following is true
When x the statement does not equal 1.
63
Half Angle Formulas
Half angle formulas come from double angle
formulas involving 2x.
Ex Double Angle formula half angle formula
Instead of doubling the angle it is cut in half.
Half Angle Formula For any real number x we have
If we replace x with 2x we have
64
Ex 4 Determine
Solution
Substitution for x
Reduce fraction
Now take cosine
Now use algebra and simplify expression
65
Now we must determine the sign.
This function is in quadrant II- sine is positive
here.
You try the next one!!! ?
66
Since is in the second quadrant,
cosine is negative.
67
Verifying Identities
Using all of the trigonometry weve learned so
far we can use this information to prove
identities. (make one side of the equation look
like the other)
Ex 5 verify the identity
It does not matter which side you start with to
make it look like the other side.
Many times we only have to manipulate one side to
look like the other.
Sometimes we work both sides together and as we
go along they will be the same.
Solution
I will start with the left side. I notice that I
can use the Pythagorean Theorem.
Remember, your goal is to turn it into tangent.
?
68
Now you try one!!!
Ex 6 Verify cot x tan x sec x csc x
69
3.6 Right Triangle Trigonometry
Trigonometric Functions of an Angle in a Right
Triangle
For the angle ? in the right triangle shown, we
have
Recall SohCahToa
c b
?
a
70
Ex 1 Suppose that an acute angle ? is known to
satisfy . Determine the
other trigonometric functions of this angle.
Lets use a right triangle and label what we know.
Solution
We must find a before we can find the others,
except for csc ?
3
5
?
a
4
The Pythagorean Theorem will help us find the
value of a.
Now that we know all of the values for a, b,
and c, we can write all of the solutions
71
Note The last example said that the angle was
acute which means it lies in the
first quadrant. If the angle was in the second,
third, or fourth, we would have
solved the problem like we just did, and then
change the signs as necessary.
Ex 2 Find the value of the six trigonometric
functions if the
Solution
Using the Pythagorean theorem we find that the
opposite side equals 4.
72
Ex 3 A climber who wants to measure the height
of a cliff is standing 35 feet from
the base of the cliff. An angle of approximately
60 is formed by the lines joining the
climbers feet with the top and bottom of the
cliff, as shown. Use this information
of approximate the height of the cliff.
Solve for x.
cliff
x
60
x 60.6 ft.
35 ft
Solution
When you have a right triangle situation and at
least one angle and one side is known, follow
these steps to find the missing piece.
1. Ask yourself What side do I have and what
side do I want?
We have the adjacent side and we want the
opposite side.
2. Which trig function involves the two answers
from question 1?
The tangent function will be used.
73
Ex 4 Two balls are against the rail at opposite
ends of a 10 foot billiard table. The
player must hit the ball on the left with the cue
ball on the right without touching
any of the other balls on the table. This is
done by banking the cue ball off the
bottom cushion, as shown. Where should the cue
ball hit the bottom cushion, and what
is the angle that its path makes with the bottom
cushion.
3
2
?
?
x 10 - x
2(10 x) 3x
Since
x 4
20 2x 3x
74
Unfortunately, we do not know a value on our unit
circle where this is true, but this is the best
we can do for now, even though we need the angle
measure, not the tangent of that angle.
We can answer where the cue ball should hit the
bottom cushion The cue ball should hit the
bottom cushion 2 feet from the bottom left or 8
feet from the bottom right.
75
Ex 5 An engineer is designing a drainage canal
that has a trapezoidal cross section,
as shown. The bottom and sides of the canal are
each L feet long, and the side makes
an angle ? with the horizontal. a.) Find an
expression for the cross-sectional area of the
canal in terms of the angle
? with the horizontal. b.) If the canal is S
feet long, approximate the angle ? that will
maximize the capacity of the
canal.
L
L
Lsin ?
L
Lcos ?
Lcos ?
a.) Area of a trapezoid
We know the length of one base (L) and can find
the height by making a right triangle.
We have angle ? and the hypotenuse we want
the opposite side
76
b.)
Capacity means volume, so a canal that is S feet
long has the capacity
To find the exact value of ? will have to wait
until the next section! ?
77
3.7 Inverse Trigonometric Functions
Recall properties of inverse functions
  • Properties of Inverse Functions
  • Suppose that f is a one-to-one function. Then
    the inverse f-1 is
  • unique, and
  • 1. The domain of f-1 is the range of f.
  • The range of f-1 is the domain of f.
  • If x is in the domain of f-1 and y is in the
    domain of f, then
  • f-1(x) y if and only if f(y) x.
  • f(f-1(x)) x when x is in the domain of f-1.
  • f-1(f(x)) x when x is in the domain of f.
  • The graph of y f-1(x) is the graph of y f(x)
    about the line
  • y x.

78
Sine
The sine function is not one to one because it is
periodic. If fails the horizontal line test for
every horizontal line between -1 and 1 on the
y-axis.
However, we can restrict the domain of the sine
function so that it is one-to-one.
The sine function is one to one on the interval
On this interval, the sine function does have an
inverse and is denoted Arcsine function or
simply y arcsin x
Note csc x (sin x)-1
The Arcsine Function The arcsine function,
denoted arcsin, has domain (-1,1) and range and
is defined by arcsin x y if and only if
sin y x
79
Arcsine Properties sin(arcsin x) x when x is
in -1, 1 and arcsin(sin x) x when x is
in
The graph of y arcsin x is the reflection of
the restricted function of y sin x reflected
about the line y x.
Notice how steep the ends of the curve are on the
inverse. This corresponds to the flatness on the
sine curve.
80
Ex 1 Find
Solution
Remember that we have a restricted interval now
for sine. So we ask ourselves Where in the
restricted interval does sin ½ ?
Recall the arcsine properties arcsin(sin x) x.
Here we have to be careful our answer must lie
in the restricted interval.
is not in the restricted interval.
Where in the restricted interval is the
the same value?
81
All of the other trigonometric functions are
defined by making domain restrictions, too.
Cosine
The Arccosine Function The arccosine function,
denoted arccos, has domain -1, 1 and range 0,
p. and is defined by arccos x y if and
only if cos y x
Arccosine Properties cos(arccos x) x when x
is in -1, 1 and arccos(cos x) x when x
is in 0, p.
82
Ex 2 Find
Solution
Where in the restricted domain does
?
At
83
Tangent
The Arctangent Function The arctangent function,
denoted arctan, has domain (-8, 8) and range
, and is defined by
arctan x y if and only
if tan y x
Arctangent Properties tan(arctan x) x for x
in (-8, 8) and arctan(tan x) x for x in
.
The graph of y arctan x is the graph of y tan
x reflected over the line y x.
84
Graph together with students on board
85
Ex 3 Find
Solution
We can see that this is of the form
We will use the difference formula for cosine.
We know the value of the very last identity.
The remaining identities must be found using a
right triangle
But this is not one of our identities
To get ? by itself we take the inverse of both
sides.
12
?
5
86
Now, using this triangle find values of
identities involving
These values are
All we have to do now is draw another triangle to
help us find
5
3
Now, find the value of your identity
?
4
87
Now evaluate
Arcsecant
The arcsecant function has the same restricted
interval as the cosine function.
Recall that y sec x has a vertical asymptote at
The Arcsecant Function The arcsecant function,
denoted arcsec, has domain (-8, -1 U 1, 8) and
range and is
defined by arcsec x y if and only if
sec y x
88
Arcsecant Properties Sec(arcsec x) x when x is
in (-8, -1 U 1, 8) and arcsec(sec x) x when
x is in
Ex Graph y arcsec x
Graph with students on board.
89
3.8 Applications of Trigonometric Functions
A Cessna Citation III business jet flying at 520
miles per hour is directly over Logan, Utah, and
heading due south toward Phoenix. Fifteen
minutes later an F-15 Fighting Eagle passes
over Logan traveling westward at 1535 miles per
hour. We would like to Determine a function that
describes the distance between the planes in
terms of the time after the F-15 passes over
Logan until it reaches the California border 20
minutes later.
Distance rate x time
d 1535t
d 520(t 0.25) 520t 130
d(t)
Recall the distance formula here involves time
in hours 20 minutes is 1/3 of an hour
t 1/3
90
It is more likely that the planes will not be
traveling in paths that are perpendicular. Lets
change the problem a little.
91
A Cessna Citation III business jet flying at 520
miles per hour is directly over Logan, Utah, and
heading due south toward Phoenix. Fifteen
minutes later an F-15 Fighting Eagle passes
over Logan traveling 24 west of south at 1535
miles per hour toward Nellis Airforce base 395
miles away.
Since this situation does not include a right
angle, we can not use the Pythagorean
Theoremdirectly.
24
1535t
520t 130
d(t)
92
Law of Cosines Suppose that a triangle has sides
of length a, b, and c and corresponding opposite
angles a, ß, and ? as shown. Then
B
ß
c a
a ?
A
C
b
We also get
93
Ex 1 A triangle has sides of length 6 and 8,
and the angle between these sides is
60. What is the length of the third side?
6 x
60
8
Solution
Using the Law of Cosines
94
Ex 2 If two sides of a non right triangle are
of lengths 15 and 25 and the included
angle measures 35, find the missing side
and one of the other angles.
15 x
35
?
25
--------------------------------------------------
----------------
95
We can now go back and solve the aircraft problem
Lets let t 0.333
96
Ex 3 A picture in an art museum is 5 feet high
and hung so that its base is 8 feet
above the ground. Find the viewing angle ?(x)
of a 6-foot tall viewer who is
standing x feet from the wall.
APB and APC are both right triangles. We can
Use the Pythagorean Theorem.
Using the Law of Cosines
97
Using a graphing calculator we can see that the
best viewing of the painting is where the maximum
occurs which is when x ? 3.7 feet. This is the
distance the viewer should stand from the wall.
When the angles and one side of a triangle are
known we can use the Law of Sines to find the
other missing parts.
The Law of Sines Suppose that a triangle has
sides of length a, b and c with corresponding
opposite angles a, ß and ? as shown. Then
B
ß
c a
a ?
A C
b
98
Ex 4 Find the missing angles.
a
If we knew a we could find ß.
ß 180 60 a
60 ß
8
Solution
We could use the Law of Cosines to find a but
it is easier to use the Law of Sines.
99
Ex 5 The aircraft carrier Carl Vinson leaves
the Pearl Harbor naval shipyard and
heads due west at 28 knots. A helicopter is
175 nautical miles from the carrier 35
south of west. a.) On what course should the
helicopter travel at its cruising
speed of 130 knots to intercept the aircraft
carrier? b.) How long will it take.
Solution
Draw a picture of the given information.
First, find ? which will give the course the
helicopter should fly.
The helicopter should fly 42 north of east.
100
Since we know ? 7 we also know the third angle
is 138.
It will take the helicopter about 1 hour and 9
minutes to intercept the aircraft carrier.
101
Ex 6 Think Pair Share A
campground lies at the west end of an east-west
road in a relatively flat, but dense,
forest. The starting point for a hike
lies 30 kilometers to the northeast of the
campground. A hiker begins at the
starting point and travels in the general
direction of the campground, reaching
the road after 25 kilometers.
Approximately how far is the campground from the
road?
B
Lets assume triangle ABC gives the correct
solution. If the hiker traveled along the line
BC' would mean the hiker was way off course BUT
IT COULD HAPPEN! This gives us an isosceles
triangle.
30
25
45 ? ?'
camp
A
C C'
This could be the value of ? or ?'
First, we need to find ?
102
Since ? is an obtuse angle it cannot be 58, this
must be the value of ?'.
? is part of a linear pair and the other angle is
the same as ?' since they are the base angles of
an isosceles triangle.
? 122
The angle at B is 180 45 122 13
AC distance of hiker from camp ? 7.95 km.
103
If the hiker was lost, but measured the distance
from the road correctly, ?B is 77, therefore
the distance would be
If the hiker was badly off course he is about
34.5 km from camp.
104
Herons Formula
Herons formula is used to find the area of a
triangle when only the lengths of the sides of
the triangle are known.
Herons Formula A triangle with sides of length
a, b and c has area given by
where P is the perimeter of the triangle, P a
b c
105
Ex 6 Find the area of the triangle.
7 9
12
P 7 9 12 28
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