Transcript Right Triangle Trigonometry

Pre-Calculus
Monday, April 20
Right Triangle
Trigonometry
1
Today’s Objective
Review right triangle trigonometry from
Geometry and expand it to all the
trigonometric functions
Begin learning some of the Trigonometric
identities
2
What You Should Learn
•
Evaluate trigonometric functions of acute angles.
•
Use fundamental trigonometric identities.
•
Use a calculator to evaluate trigonometric
functions.
•
Use trigonometric functions to model and solve
real-life problems.
Plan
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Questions from last week?
Notes!
Guided Practice
Homework
4
Right Triangle Trigonometry
Trigonometry is based upon ratios of the sides
of right triangles.
The ratio of sides in triangles with the same
angles is consistent. The size of the triangle
does not matter because the triangles are
similar (same shape different size).
5
The six trigonometric functions of a right triangle,
with an acute angle θ, are defined by ratios of two sides
of the triangle.
hyp
opp
θ
The sides of the right triangle are:
 the side opposite the acute angle
 the side adjacent to the acute angle
adj
θ
θ,
 and the hypotenuse of the right triangle.
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hyp
The trigonometric functions are
opp
θ
adj
sine, cosine, tangent, cotangent, secant, and cosecant.
opp
Sinθ =
cos θ= adj
tan θ= opp
hyp
hyp
adj
Csc θ=
hyp
opp
sec θ= hyp
adj
cot θ= adj
opp
Note: sine and cosecant are reciprocals, cosine and secant are reciprocals,
and tangent and cotangent are reciprocals.
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Reciprocal Functions
Another way to look at it…
sin  = 1/csc 
cos  = 1/sec 
tan  = 1/cot 
csc  = 1/sin 
sec  = 1/cos 
cot  = 1/tan 
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Given 2 sides of a right triangle you should be
able to find the value of all 6 trigonometric
functions.
Example:
5

12
9
Calculate the trigonometric functions for  .
Calculate the trigonometric functions for .
5
The six trig ratios are
Sin θ =
Cos θ =
Tanθ =
Cot θ =
Sec θ =
Cscθ =
4
5
3
5
4
3
3
4
5
3
5
4
3
sin α =
5
4
cos α =
5
3
tan α =
4
4
cot α =
3
5
sec α =
4
5
csc α =
3

4

3
What is the
relationship of
α and θ?
They are
complementary
(α = 90 – θ)
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Note sin  = cos(90   ), for 0 <  < 90
Note that  and 90   are complementary
angles.
Side a is opposite θ and also
adjacent to 90○– θ .
sin  = a and cos (90   ) = a .
b
b
So, sin  = cos (90   ).
hyp
90○– θ a
θ
b
Note : These functions of the complements are called cofunctions.
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Cofunctions
sin  = cos (90   )
sin  = cos (π/2  )
cos  = sin (90   )
cos  = sin (π/2  )
tan  = cot (90   )
tan  = cot (π/2  )
cot  = tan (90   )
cot  = tan (π/2  )
sec  = csc (90   ) csc  = sec (90   )
sec  = csc (π/2  ) csc  = sec (π/2  )
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Trigonometric Identities are trigonometric
equations that hold for all values of the variables.
We will learn many Trigonometric Identities and use
them to simplify and solve problems.
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Quotient Identities
hyp
opp
θ
adj
Sin θ = opp
hyp
cos θ = adj
hyp
tan θ =
opp
adj
opp
sin  hyp opp hyp opp




 tan 
cos  adj hyp adj adj
hyp
The same argument can be made for cot… since it is the
reciprocal function of tan.
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Quotient Identities
sin 
tan  
cos 
cos 
cot  
sin 
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Pythagorean Identities
Three additional identities that we will use are
those related to the Pythagorean Theorem:
Pythagorean Identities
sin2  + cos2  = 1
tan2  + 1 = sec2 
cot2  + 1 = csc2 
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Some old geometry favorites…
Let’s look at the trigonometric functions of a
few familiar triangles…
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Geometry of the 45-45-90 triangle
Consider an isosceles right triangle with two sides of
length 1.
45
2
1
12  12  2
45
1
The Pythagorean Theorem implies that the hypotenuse
is of length 2 .
Remember a2 + b2 = c2
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Calculate the trigonometric functions for a 45° angle.
2
1
45
1
opp
sin 45° =
hyp
1
=
2
opp
tan 45° =
adj
1
=
1
2
=
2
= 1
2
hyp
sec 45° =
=
=2
1
adj
1
2
adj
cos 45° =
=
=
2
hyp
2
adj 1
cot 45° =
= = 1
opp 1
2
hyp
csc 45° =
=
opp
1
=2
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Geometry of the 30-60-90 triangle
Consider an equilateral triangle with
each side of length 2.
30○30○
The three sides are equal, so the
angles are equal; each is 60°.
2
The perpendicular bisector
of the base bisects the
opposite angle.
60○
2
3
1
60○
2
1
Use the Pythagorean Theorem to
find the length of the altitude, 3 .
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Calculate the trigonometric functions for a 30 angle.
2
1
30
3
opp
sin 30° =
hyp
1
=
2
3
adj
cos 30° =
=
2
hyp
1
opp
tan 30° =
=
adj
3
3
=
3
adj
cot 30° =
opp
3
=
1
=3
2
hyp
sec 30° =
=
3
adj
2 3
=
3
hyp
csc 30° =
opp
2
=
1
= 2
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Calculate the trigonometric functions for a 60 angle.
2
3
60○
1
sin 60 =
opp
3
=
hyp
2
cos 60 =
tan 60 =
3
opp
=
= 3
1
adj
3
1
cot 60 = adj =
=
opp
3
3
hyp 2
sec 60 =
= = 2
adj 1
1
adj
=
2
hyp
2
2 3
hyp
csc 60 =
=
=
opp
3
3
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Some basic trig values
Sine
300
/6
450
/4
600
/3
Cosine
Tangent
1
2
3
2
3
3
2
2
2
2
1
3
2
1
2
3
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IDENTITIES WE HAVE
REVIEWED SO FAR…
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Fundamental Trigonometric Identities
Reciprocal Identities
sin  = 1/csc 
cot  = 1/tan 
cos  = 1/sec 
sec  = 1/cos 
tan  = 1/cot 
csc  = 1/sin 
Co function Identities
sin  = cos(90   )
sin  = cos (π/2  )
tan  = cot(90   )
tan  = cot (π/2  )
sec  = csc(90   )
sec  = csc (π/2  )
Quotient Identities
tan  = sin  /cos 
cos  = sin(90   )
cos  = sin (π/2  )
cot  = tan(90   )
cot  = tan (π/2  )
csc  = sec(90   )
csc  = sec (π/2  )
cot  = cos  /sin 
Pythagorean Identities
sin2  + cos2  = 1
tan2  + 1 = sec2 
cot2  + 1 = csc2 25
Example: Given sec  = 4, find the values of the
other five trigonometric functions of  .
Draw a right triangle with an angle  such
4
4
hyp
that 4 = sec  =
= .
adj 1
Use the Pythagorean Theorem to solve
for the third side of the triangle.
sin  =
15
4
cos  = 1
4
tan  = 15 = 15
1
15
θ
1
1 = 4
sin 
15
1
sec  =
=4
cos 
1
cot  =
15
csc  =
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Using the calculator
Function Keys
Reciprocal Key
Inverse Keys
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Using Trigonometry to Solve a Right
Triangle
A surveyor is standing 115 feet from the base of the
Washington Monument. The surveyor measures the
angle of elevation to the top of the monument as 78.3.
How tall is the Washington Monument?
Figure 4.33
Applications Involving Right
Triangles
The angle you are given is
the angle of elevation,
which represents the
angle from the horizontal
upward to an object.
For objects that lie below
the horizontal, it is
common to use the term
angle of depression.
Solution
where x = 115 and y is the height of the
monument. So, the height of the Washington
Monument is
y = x tan 78.3
 115(4.82882)  555 feet.
Homework
4-2 Practice 1

1-17 ODD
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Tuesday, April 21, 2015
BENCHMARK TOMORROW
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Kahoot all day long!

Here’s the deal…
1.
2.
3.
4.
5.
You need a PENCIL and PAPER
You will be GRADED for participating so make
sure your Kahoot name is your real name
If you get kicked out, you must log back in
If you do not have an ipad/smart phone, you may
work in teams of TWO or do your work on a
piece of paper and turn that in
THIS IS REQUIRED.
33
Review Exponents and Log
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Converting Logs and Exponents
https://play.kahoot.it/#/k/68a4661b-b90b4987-820e-956c7e5af6bd
Log and Inverses
https://play.kahoot.it/#/k/45a02d91-aa36485f-ba52-436768d981f7
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Review Intro to Trig
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Right Triangle Trig and Angle Measures
https://play.kahoot.it/#/k/8858d7ed-b09246c6-bc5d-b16c044665c0
Radians, Degrees, Arc Length
https://play.kahoot.it/#/k/63ac7b20-59ce4fd4-9266-954797b1b39a
35
Wednesday
BENCHMARK!
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Thursday, April 23
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Do Now – How was the benchmark? What can you improve?
Benchmark Data
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Benchmark Data 1st Period
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Analysis By Question
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Analysis By Question
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Analysis By Question
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Analysis By Question
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Analysis By Question
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Benchmark Data 2nd Period
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Analysis By Question
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Analysis By Question
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Analysis By Question
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Analysis By Question
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Analysis By Question
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Special Angle Names
Angle of Elevation
From Horizontal Up
Angle of Depression
From Horizontal Down
Angle of Elevation and
Depression
Imagine you are standing here.

The angle of elevation is measured
from the horizontal up to the object.
Angle of Elevation and
Depression
The angle of depression is measured
from the horizontal down to the
object.

Constructing a right triangle,
we are able to use trig to solve
the triangle.
Guided Practice! Follow along on
your handout!
Lighthouse & Sailboat
Suppose the angle of depression from a lighthouse
to a sailboat is 5.7o. If the lighthouse is 150 ft tall,
how far away is the sailboat?
x
5.7o
150 ft.
Construct a triangle and label the known parts.
Use a variable for the unknown value.
150 ft.
Lighthouse & Sailboat
Suppose the angle of depression from a lighthouse
to a sailboat is 5.7o. If the lighthouse is 150 ft tall,
how far away is the sailboat?
x
150 ft.
Set up an equation and solve.
5.7o
Lighthouse & Sailboat
tan(5.7 o ) 
150
x
x
x tan(5.7 )  150
o
x
150
tan(5.7o )
150 ft.
Remember to use
degree mode!
x is approximately 1,503 ft.
5.7o
River Width
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A surveyor is measuring a
river’s width. He uses a tree
and a big rock that are on the
edge of the river on opposite
sides. After turning through an
angle of 90° at the big rock,
he walks 100 meters away to
his tent. He finds the angle
from his walking path to the
tree on the opposite side to be
25°. What is the width of the
river?
Draw a diagram to describe
this situation. Label the
variable(s)
River Width
We are looking at the “opposite”
and the “adjacent” from the given
angle, so we will use tangent
d
tan(25 ) 
100
Multiply by 100 on both sides
100 tan(25 )  d
d  46.63 meters
Subway

The DuPont Circle Metrorail Station
in Washington DC has an escalator
which carries passengers from the
underground tunnel to the street
above. If the angle of elevation of
the escalator is 52° and a
passenger rides the escalator for
188 ft, find the vertical distance
between the tunnel and the street.
In other words, how far below street
level is the tunnel?
Subway
We are looking at the “opposite” and
the “hypotenuse” from the given angle
so we will use sine
h
sin(52 ) 
188
Multiply by 188 on each side
188sin(52 )  h
h  148.15 feet
Building Height
A spire sits on top of the top floor of a
building. From a point 500 ft. from the base
of a building, the angle of elevation to the
top floor of the building is 35o. The angle of
elevation to the top of the spire is 38o. How
tall is the spire?
Construct the required
triangles and label.
38o 35o
500 ft.
Building Height
Write an equation and solve.
Total height (t) = building height (b) + spire height (s)
Solve for the spire height.
s
t
Total Height
t
tan(38 ) 
500
o
b
500 tan(38 )  t
o
38o 35o
500 ft.
Building Height
Write an equation and solve.
Building Height
b
o
tan(35 ) 
500
s
500 tan(35o )  b
t
b
38o 35o
500 ft.
Building Height
Write an equation and solve.
Total height (t) = building height (b) + spire height (s)
500 tan(38o )  t
500 tan(35o )  b
s
500 tan(38 )  500 tan(35 )  s
500 tan(38o )  500 tan(35o )  s
o
The height of the
spire is
approximately 41
feet.
o
t
b
38o 35o
500 ft.
Mountain Height
A hiker measures the angle of elevation to a mountain
peak in the distance at 28o. Moving 1,500 ft closer on a
level surface, the angle of elevation is measured to be
29o. How much higher is the mountain peak than the
hiker?
Construct a diagram and label.
1st measurement
28o.
2nd measurement 1,500 ft closer is 29o.
Mountain Height
Adding labels to the diagram, we need to find h.
h ft
28o
1500 ft
29o
x ft
Write an equation for each triangle. Remember, we can
only solve right triangles. The base of the triangle with
an angle of 28o is 1500 + x.
h
tan 28 
1500  x
o
h
tan 29 
x
o
Mountain Height
Now we have two equations with two variables.
Solve by substitution.
h
tan 28 
1500  x
o
h
tan 29 
x
o
Solve each equation for h.
(1500  x) tan(28o )  h
x tan(29o )  h
Substitute.
(1500  x) tan(28o )  x tan(29o )
Mountain Height
(1500  x) tan(28o )  x tan(29o )
Solve for x. Distribute.
1500 tan(28o )  x tan(28o )  x tan(29o )
Get the x’s on one side and factor out the x.
1500 tan(28o )  x tan(29o )  x tan(28o )
1500 tan(28o )  x  tan(29o )  tan(28o ) 
Divide.
1500 tan(28o )
x
o
o
 tan(29 )  tan(28 ) 
Mountain Height
However, we were to find the height of the mountain.
Use one of the equations solved for “h” to solve for the
height.
x tan(29o )  h
1500 tan  28o  tan  29o 
tan  29   tan  28
o
o

 19,562
The height of the mountain above the hiker is 19,562 ft.