Loading...

PPT – Trigonometry PowerPoint presentation | free to download - id: 5c4496-ZjgwY

The Adobe Flash plugin is needed to view this content

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

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)

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.

y

B

If angles are generated by the terminal side

moving clockwise the angle is negative.

A

x

0

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.

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.

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.

t

3

2

y

(0,1)

1

(-1,0)

(1,0)

x

and so on

-1

(0,-1)

-2

-3

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

To convert radian measure to degree measure you

multiply the radian measure by

Ex 2 Convert the following radian measures to

degree measure.

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.

y

Unit Circle

90

60

120

45

135

150

30

180

0

x

330

210

315

225

240

300

270

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)

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

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.

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.

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.

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

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

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.

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.

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.

Cosine Curve

cos t

1

t

-1

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.

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!!! ?

sin x

1

x

-1

You try

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.

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

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

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

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

Ex 3 Sketch the graph of

cos x

2

x

-2

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

Trig. function Sign in quadrant

Trig. function I II III IV

Sin x

Cos x

Tan x

Csc x

Sec x

Cot x

- -

- -

- -

- -

- -

- -

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.

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.

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

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

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

a.) y tan 2x

b.)

This graph is the graph from part b reflected

over the x-axis.

c.)

c. contd.)

Now we shift it up one unit.

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. ?

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.

(No Transcript)

Ex Graph y sec x

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.

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

y csc x has vertical asymptotes at

y csc 3x has vertical asymptotes at

Now, the amplitude changes to ½

Now, we shift the graph to the left units.

The vertical asymptotes go with it!!! They get

shifted too!

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

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

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.

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.

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

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!!! ?

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)

0 cos x(2sin x 1)

cos x 0 and 2sin x 1 0

sin x ½

b.) sin x cos 2x

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!!!

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.

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

Ex 4 Determine

Solution

Substitution for x

Reduce fraction

Now take cosine

Now use algebra and simplify expression

Now we must determine the sign.

This function is in quadrant II- sine is positive

here.

You try the next one!!! ?

Since is in the second quadrant,

cosine is negative.

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.

?

Now you try one!!!

Ex 6 Verify cot x tan x sec x csc x

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

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

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.

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.

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

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.

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

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! ?

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.

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

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.

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?

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.

Ex 2 Find

Solution

Where in the restricted domain does

?

At

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.

Graph together with students on board

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

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

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

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.

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

It is more likely that the planes will not be

traveling in paths that are perpendicular. Lets

change the problem a little.

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)

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

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

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

--------------------------------------------------

----------------

We can now go back and solve the aircraft problem

Lets let t 0.333

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

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

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.

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.

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.

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 ?

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.

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.

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

Ex 6 Find the area of the triangle.

7 9

12

P 7 9 12 28