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2
Acute Angles
and Right
Triangles
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
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2 Acute Angles and Right Triangles
2.1 Trigonometric Functions of Acute Angles
2.2 Trigonometric Functions of Non-Acute
Angles
2.3 Finding Trigonometric Function Values
Using a Calculator
2.4 Solving Right Triangles
2.5 Further Applications of Right Triangles
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2.1
Trigonometric Functions of
Acute Angles
Right-Triangle-Based Definitions of the Trigonometric Functions ▪
Cofunctions ▪ Trigonometric Function Values of Special Angles
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2.1
Example 1 Finding Trigonometric Function Values of An
Acute Angle (page 46)
Find the sine, cosine, and
tangent values for angles D
and E in the figure.
A
C
B
sin A
cos A
tan A
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2.1
Example 1 Finding Trigonometric Function Values of An
Acute Angle (cont.)
Find the sine, cosine, and
tangent values for angles D
and E in the figure.
A
C
B
sin B
cos B
tan B
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2.1
Example 2 Writing Functions in Terms of Cofunctions
(page 47)
Write each function in terms of its cofunction.
(a) sin 9° = cos (90° – 9°) = cos 81°
(b) cot 76° = tan (90° – 76°) = tan 14°
(c) csc 45° = sec (90° – 45°) = sec 45°
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2.1
Example 3(a) Solving Equations Using the Cofunction
Identities (page 48)
Find one solution for the equation. Assume all angles
involved are acute angles.
(a)
Since cotangent and tangent are cofunctions, the
equation is true if the sum of the angles is 90º.
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2.1
Example 3(b) Solving Equations Using the Cofunction
Identities (page 48)
Find one solution for the equation. Assume all angles
involved are acute angles.
(b)
Since secant and cosecant are cofunctions, the
equation is true if the sum of the angles is 90º.
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2.1
Example 4 Comparing Function Values of Acute Angles
(page 49)
Determine whether each statement is true or false.
(a) tan 25° < tan 23°
In the interval from 0° to 90°, as the angle increases,
the tangent of the angle increases.
tan 25° < tan 23° is false.
(b) csc 44° < csc 40°
In the interval from 0° to 90°, as the angle increases,
the sine of the angle increases, so the cosecant of
the angle decreases.
csc 44° < csc 40° is true.
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2.1
Example 5 Finding Trigonometric Values for 30° (page 50)
Find the six trigonometric function values for a 30°
angle.
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2.1
Example 5 Finding Trigonometric Values for 30° (cont.)
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2.2
Trigonometric Functions of Non-Acute
Angles
Reference Angles ▪ Special Angles as Reference Angles ▪
Finding Angle Measures with Special Angles
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2.2
Example 1(a) Finding Reference Angles (page 55)
Find the reference angle for 294°.
294 ° lies in quadrant IV.
The reference angle is
360° – 294° = 66°.
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2.2
Example 1(b) Finding Reference Angles (page 55)
Find the reference angle for 883°.
Find a coterminal angle between 0° and 360° by
dividing 883° by 360°. The quotient is about 2.5.
883° is coterminal with 163°.
The reference angle is
180° – 163° = 17°.
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2.2
Example 2 Finding Trigonometric Functions of a
Quadrant II Angle (page 56)
Find the values of the six trigonometric functions for
225°.
The reference angle for 225° is 45°.
Since 225° lies in quadrant III, the tangent and
cotangent are positive while the sine, cosine, secant
and cosecant are negative.
2
sin225   sin 45  
2
2
cos 225   cos 45  
2
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2.2
Example 2 Finding Trigonometric Functions of a
Quadrant II Angle (page 56)
2
cos 225   cos 45  
2
tan225  tan 45  1
cot 225  cot 45  1
sec 225   sec 45   2
cscc 225   csc 45   2
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2.2
Example 3(a) Finding Trigonometric Function Values
Using Reference Angles (page 57)
Find the exact value of sin(–150°).
An angle of –150° is coterminal with an angle of
–150° + 360° = 210°.
The reference angle is
210° – 180° = 30°.
Since an angle of –150° lies in
quadrant III, its sine is negative.
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2.2
Example 3(b) Finding Trigonometric Function Values
Using Reference Angles (page 57)
Find the exact value of cot(780°).
An angle of 780° is coterminal with an angle of
780° – 2 ∙ 360° = 60°.
The reference angle is 60°.
Since an angle of 780°
lies in quadrant I, its
cotangent is positive.
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2.2
Example 4 Evaluating an Expression with Function
Values of Special Angles (page 57)
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2.2
Example 5 Using Coterminal Angles to Find Function
Values (page 57)
Evaluate each function by first expressing the
function in terms of an angle between 0° and 360°.
(a)
(b)
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2.2
Example 6 Finding Angle Measures Given an Interval
and a Function Value (page 58)
Find all values of θ, if θ is in the interval [0°, 360°) and
sin θ =
Since sin θ is negative, θ must lie in quadrants III or IV.
The absolute value of sin θ is
angle is 60°.
so the reference
The angle in quadrant III is 60° + 180° = 240°.
The angle in quadrant IV is 360° – 60° = 300°.
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2.3
Finding Trigonometric Functions Using
a Calculator
Finding Function Values Using a Calculator ▪ Finding Angle
Measures Using a Calculator
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2.3
Example 1 Finding Function Values with a Calculator
(page 62)
Approximate the value of each expression.
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2.3
Example 1 Finding Function Values with a Calculator
(cont.)
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2.3
Example 2 Using Inverse Trigonometric Functions to
Find Angles (page 62)
Use a calculator to find an angle θ in the interval
[0°, 90°] that satisfies each condition.
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2.3
Example 3 Finding Grade Resistance (page 63)
The force F in pounds when an automobile travels uphill
or downhill on a highway is called grade resistance
and is modeled by the equation
, where θ is
the grade and W is the weight of the automobile.
If the automobile is moving uphill, then θ > 0°; if it is
moving downhill, then θ < 0°.
(a) Calculate F to the nearest 10 pounds for a 5500-lb
car traveling an uphill grade with θ = 3.9°.
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2.3
Example 3 Finding Grade Resistance (cont.)
(b) Calculate F to the nearest 10 pounds for a 2800-lb
car traveling a downhill grade with θ = –4.8°.
(c) A 2400-lb car traveling uphill has a grade
resistance of 288 pounds. What is the angle of the
grade?
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2.4
Solving Right Triangles
Significant Digits ▪ Solving Triangles ▪ Angles of Elevation or
Depression
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2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (page 69)
Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm.
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2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (cont.)
Three significant digits
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2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (cont.)
Three significant digits
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2.4
Example 1 Solving a Right Triangle Given an Angle and
a Side (cont.)
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2.4
Example 2 Solving a Right Triangle Given Two Sides
(page 70)
Solve right triangle ABC, if a = 44.25 cm and
b = 55.87 cm.
Use the Pythagorean theorem to find c:
Four significant digits
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2.4
Example 2 Solving a Right Triangle Given Two Sides
(cont.)
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2.4
Example 2 Solving a Right Triangle Given Two Sides
(cont.)
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2.4
Example 3 Finding a Length When the Angle of
Elevation is Known (page 71)
The angle of depression
from the top of a tree to a
point on the ground 15.5 m
from the base of the tree is
60.4°. Find the height of
the tree.
The measure of
equals the measure of the
angle of depression because the two angles are
alternate interior angles, so
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2.4
Example 3 Finding a Length When the Angle of
Elevation is Known (cont.)
Three significant digits
The tree is about 27.3 m tall.
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2.4
Example 4 Finding an Angle of Depression (page 72)
Find the angle of
depression from the top of
a 97-ft building to the base
of a building across the
street located 42 ft away.
97
tan B 
42
97
B  tan
42
1
B  67
The angle of depression is about 67°.
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2.5
Further Applications of Right Triangles
Bearing ▪ Further Applications
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2.5
Example 1 Solving Problem Involving Bearing
(First Method) (page 83)
Radar stations A and B are on an east-west line, 8.6 km
apart. Station A detects a plane at C, on a bearing of 53°.
Stations B simultaneously detects the same plane, on a
bearing of 323°. Find the distance from B to C.
A line drawn due north is
perpendicular to an eastwest line, so right angles are
formed at A and B, and
angles CAB and CBA can be
found as shown in the figure.
Angle C is a right angle because angles CAB and CBA are
complementary.
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2.5
Example 1 Solving Problem Involving Bearing
(First Method) (cont.)
Find distance a by using the cosine function for angle A.
The distance from B to C is about 5.2 km.
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2.5
Example 2 Solving Problem Involving Bearing
(Second Method) (page 78)
A ship leave port and sails on a bearing of S 34° W for 2.5
hr. It then turns and sails on a bearing of N 56° W for 3 hr.
If the ships rate of speed is 18 knots (nautical miles per
hour), find the distance that the ship is from port.
The port is at A.
Because the NS lines are
parallel, the measure of angle
ACD is 34.
The measure of angle
ACB = 34 + 56 = 90, making ABC
a right triangle.
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2.5
Example 2 Solving Problem Involving Bearing
(Second Method) (page 78)
The ship traveled at 18 knots for 2.5 hours or 18(2.5) = 45
nautical miles from A to C.
The ship traveled at 18 knots
for 3 hours or 18(3) = 54
nautical miles from C to B.
54
45
Use the Pythagorean theorem to
find the distance from B to A.
a2  b2  c 2
542  452  c 2
4941  c 2
70  c
The ship is about 70
nautical miles from port.
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2.5
Example 3(a) Using Trigonometry to Measure a Distance
(page 79)
A method that surveyors use to determine a small
distance d between two points P and Q is called the
subtense bar method. The subtense bar with length b
is centered at Q and situated perpendicular to the line
of sight between P and Q.
(a) Find d with θ = 2°41′38″ and b = 3.5000 cm.
From the figure, we have
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2.5
Example 3(a) Using Trigonometry to Measure a Distance
(cont.)
Convert θ to decimal degrees:
Five significant digits
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2.5
Example 3(b) Using Trigonometry to Measure a Distance
(page 79)
(b) Angle θ usually cannot be measured more
accurately than to the nearest 1″. How much
change would there be in the value of d if θ were
measured 1″ larger?
Since θ is 1″ larger, θ = 2°41′39″ ≈ 2.694167.
The difference is
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2.5
Example 4 Solving a Problem Involving Angles of
Elevation (page 79)
Marla needs to find the height
of a building. From a given
point on the ground, she finds
that the angle of elevation to
the top of the building is 74.2°.
She then walks back 35 feet.
From the second point, the
angle of elevation to the top of
the building is 51.8°. Find the
height of the building.
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2.5
Example 4 Solving a Problem Involving Angles of
Elevation (cont.)
There are two unknowns, the
distance from the base of the
building, x, and the height of the
building, h.
In triangle ABC
In triangle BCD
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2.5
Example 4 Solving a Problem Involving Angles of
Elevation (cont.)
Set the two expressions for h equal and solve for x.
Since h = x tan 74.2°, substitute the expression for x to
find h.
The building is about 69 feet tall.
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2.5
Example 4 Solving a Problem Involving Angles of
Elevation (cont.)
Graphing Calculator Solution
Superimpose coordinate axes
on the figure with D at the origin.
The coordinates of A are (35, 0).
The tangent of the angle between the x-axis and the graph
of a line with equation y = mx + b is the slope of the line.
For line DB, m = tan 51.8º.
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2.5
Example 4 Solving a Problem Involving Angles of
Elevation (cont.)
Since b = 0, the equation of line
DB is
The equation of line AB is
Use the coordinates of A and
the point-slope form to find the
equation of AB:
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2.5
Example 4 Solving a Problem Involving Angles of
Elevation (cont.)
Graph y1 and y2, then find the point of intersection. The
y-coordinate gives the height, h.
The building is about 69 feet tall.
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