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Chapter 5
Trigonometric
Functions
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All rights reserved
© 2010
2011 Pearson Education, Inc. All rights reserved
1
SECTION 5.2
Right Triangle Trigonometry
OBJECTIVES
1
2
3
4
5
Define trigonometric functions of acute
angles.
Evaluate trigonometric functions of acute
angles.
Evaluate trigonometric functions for the
special angles 30°, 45°, and 60°.
Use fundamental identities.
Use right triangle trigonometry in applications.
TRIGONOMETRIC RATIOS AND FUNCTIONS
a = length of the side opposite 
b = length of the side adjacent to 
c = length of the hypotenuse
Six ratios can be formed with the lengths of these
a b a b c
c
sides: , , , , , and .
c c b a b
a
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RIGHT TRIANGLE DEFINITIONS OF THE
TRIGONOMETRIC FUNCTIONS OF AN
ACUTE ANGLE θ
opposite
a
sin  

hypotenuse c
hypotenuse c
csc  

opposite
a
adjacent
b
cos 

hypotenuse c
hypotenuse c
sec  

adjacent
b
opposite a
tan  

adjacent b
adjacent b
cot  

opposite a
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EXAMPLE 1
Finding the Values of Trigonometric
Functions
Find the exact values for the six trigonometric
functions of the angle  in the figure.
Solution
a b  c
2
 7  c
2
97 c
2
16  c
4c
2
2
3
2
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
2
2
5
EXAMPLE 1
Finding the Values of Trigonometric
Functions
Solution continued
Now, with c = 4, a = 3, and b = 7, we have
opp 3
sin  

hyp 4
adj
7
cos  

hyp
4
hyp 4
csc  

opp 3
hyp
4
4 7
sec  


adj
7
7
opp
3
3 7
tan  


adj
7
7
adj
7
cot  

opp
3
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EXAMPLE 3
Finding the Trigonometric Function Values
for 45°.
Use the figure to find sin 45°,
cos 45°, and tan 45°.
Solution
opposite
1
2
sin 45 


hypotenuse
2
2
adjacent
1
2
cos 45 


hypotenuse
2
2
opposite 1
tan 45 
 1
adjacent 1
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TRIGONOMETRIC FUNCTION VALUES OF
SOME COMMON ANGLES
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COFUNCTION IDENTITIES
The value of any trigonometric function of an
angle  is equal to the cofunction of the
complement of . This is true whether  is
measured in degrees or in radians.
 in degrees
cos   sin 90º   
sin   cos 90º   
tan   cot 90º   
cot   tan 90º   
sec   csc 90º   
csc   sec 90º   
If  is measured in radians, replace 90º with
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
2
.
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EXAMPLE 5
Finding Trigonometric Function Values of a
Complementary Angle
a. Given that cot 68° ≈ 0.4040, find tan 22°.
b. Given that cos 72° ≈ 0.3090, find sin 18°.
Solution
a. tan 22° = tan (90° – 68°) = cot 68° ≈ 0.4040
b. sin 18° = sin (90° – 72°) = cos 72° ≈ 0.3090
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RECIPROCAL AND QUOTIENT IDENTITIES
Reciprocal Identities
1
sin θ 
csc θ
1
cos θ 
sec θ
1
tan θ 
cot θ
1
csc θ 
sin θ
1
sec θ 
cos θ
1
cot θ 
tan θ
Quotient Identities
sin 
tan θ 
cos θ
cos 
cot θ 
sin θ
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PYTHAGOREAN IDENTITIES
sin θ  cos   1
2
2
1  tan θ  sec 
2
2
1  cot θ  csc 
2
2
The cofunction, reciprocal, quotient, and
Pythagorean identities are called the
Fundamental identities.
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APPLICATIONS
Angles that are measured between a line of
sight and a horizontal line occur in many
applications and are called angles of elevation
or angles of depression.
If the line of sight is above the horizontal line,
the angle between these two lines is called the
angle of elevation.
If the line of sight is below the horizontal line,
the angle between the two lines is called the
angle of depression.
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EXAMPLE 8
Measuring the Height of Mount Kilimanjaro
A surveyor wants to measure the height of
Mount Kilimanjaro by using the known height
of a nearby mountain. The nearby location is
at an altitude of 8720 feet, the distance
between that location and Mount
Kilimanjaro’s peak is 4.9941 miles, and the
angle of elevation from the lower location is
23.75º. See the figure on the next slide. Use
this information to find the approximate height
of Mount Kilimanjaro (in feet).
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EXAMPLE 8
Measuring the Height of Mount Kilimanjaro
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EXAMPLE 8
Measuring the Height of Mount Kilimanjaro
Solution
The sum of the side length h and the location
height of 8720 feet gives the approximate
height of Mount Kilimanjaro. Let h be
measured in miles. Use the definition of sin ,
for  = 23.75º.
opposite
h
sin  

hypotenuse 4.9941
h = (4.9941) sin θ
= (4.9941) sin 23.75°
h ≈ 2.0114
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EXAMPLE 8
Measuring the Height of Mount Kilimanjaro
Solution continued
Because 1 mile = 5280 feet,
2.0114 miles = (2.0114)(5280)
≈ 10,620 feet.
Thus, the height of Mount Kilimanjaro
≈ 10,620 + 8720 = 19,340 feet.
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EXAMPLE 9
Finding the Width of a River
To find the width of a river, a surveyor sights
straight across the river from a point A on her
side to a point B on the opposite side. See the
figure on the next slide. She then walks
200 feet upstream to a point C. The angle 
that the line of sight from point C to point B
makes with the river bank is 58º. How wide is
the river?
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EXAMPLE 9
Finding the Width of a River
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EXAMPLE 9
Finding the Width of a River
Solution
The points A, B, and C are the vertices of a
right triangle with acute angle 58º. Let w be the
width of the river.
w
 tan 58
200
w  200 tan 58
w  320.07 feet
The river is about 320 feet wide at the point A.
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