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2
Acute Angles
and Right
Triangles
Copyright © 2013, 2009, 2005 Pearson Education, Inc.
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Acute Angles and
2
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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2.5
Further Applications of Right
Triangles
Bearing ▪ Further Applications
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Bearing
There are two methods for expressing bearing.
When a single angle is given, such as 164°, it is
understood that the bearing is measured in a
clockwise direction from due north.
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Example 1
SOLVING A PROBLEM INVOLVING
BEARING (METHOD 1)
Radar stations A and B are on an east-west line,
3.7 km apart. Station A detects a plane at C, on a bearing of
61°. Station B simultaneously detects the same plane, on a
bearing of 331°. Find the distance from A to C.
Right angles are formed at A and B,
so 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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Caution
A correctly labeled sketch is crucial
when solving bearing applications.
Some of the necessary information is
often not directly stated in the problem
and can be determined only from the
sketch.
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Bearing
The second method for expressing bearing starts
with a north-south line and uses an acute angle to
show the direction, either east or west, from this
line.
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Example 2
SOLVING A PROBLEM INVOLVING
BEARING (METHOD 2)
A ship leaves port and sails on a bearing of N 47º E for 3.5 hr.
It then turns and sails on a bearing of S 43º E for 4.0 hr. If the
ship’s rate of speed is 22 knots (nautical miles per hour), find
the distance that the ship is from port.
Draw a sketch as shown in the figure.
Choose a point C on a bearing of
N 47° E from port at point A. Then
choose a point B on a bearing of
S 43º E from point C. Because northsouth lines are parallel, angle ACD is
47º by alternate interior angles. The
measure of angle ACB is
47º + 43º = 90º, making triangle ABC
a right triangle.
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SOLVING A PROBLEM INVOLVING
BEARING (METHOD 2) (cont.)
Example 2
Next, use the formula
relating distance, rate, and
time to find the distances
from A to C and from C to B.
a  22  4.0  88 nautical miles
b  22  3.5  77 nautical miles
Now find c, the distance from port at point A to the
ship at point B. a 2  b 2  c 2
88  77  c
2
2
2
c  882  772  120 nautical mi
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Example 3
USING TRIGONOMETRY TO MEASURE
A DISTANCE
The subtense bar method is a method that
surveyors use to determine a small distance d
between two points P and Q. The subtense bar with
length b is centered at Q and situated perpendicular
to the line of sight between P and Q. Angle θ is
measured, then the distance d can be determined.
(a) Find d with θ = 1°23′12″ and b = 2.0000 cm.
From the figure, we have
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Example 3
USING TRIGONOMETRY TO MEASURE
A DISTANCE (continued)
Let b = 2. Convert θ to decimal degrees:
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Example 3
USING TRIGONOMETRY TO MEASURE
A DISTANCE (continued)
(b) How much change would there be in the value of d
if θ were measured 1″ larger?
Since θ is 1″ larger, θ = 1°23′13″ ≈ 1.386944º.
The difference is
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Example 4
SOLVING A PROBLEM INVOLVING
ANGLES OF ELEVATION
Francisco needs to know the height of a tree. From a given
point on the ground, he finds that the angle of elevation to the
top of the tree is 36.7°. He then moves back 50 ft. From the
second point, the angle of elevation to the top of the tree is
22.2°. Find the height of the tree to the nearest foot.
The figure shows two unknowns:
x, the distance from the center of
the trunk of the tree to the point
where the first observation was
made, and h, the height of the
tree.
Since nothing is given about the length of the hypotenuse, of
either triangle ABC or triangle BCD, use a ratio that does not
involve the hypotenuse—namely, the tangent.
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Example 4
SOLVING A PROBLEM INVOLVING
ANGLES OF ELEVATION (continued)
In triangle ABC:
In triangle BCD:
Each expression equals h, so the expressions must be
equal.
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Example 4
SOLVING A PROBLEM INVOLVING
ANGLES OF ELEVATION (continued)
Since h = x tan 36.7°, we can substitute.
50 tan22.2


h
tan36.7

 tan36.7  tan22.2 
The height of the tree is approximately 45 ft.
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Example 4
SOLVING A PROBLEM INVOLVING
ANGLES OF ELEVATION (continued)
Graphing Calculator Solution
Superimpose coordinate axes
on the figure with D at the origin.
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,
m. For line DB, m = tan 22.2°.
Since b = 0, the equation of line DB is y1  (tan22.2) x.
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Example 4
SOLVING A PROBLEM INVOLVING
ANGLES OF ELEVATION (continued)
The equation of line AB is
y 2  (tan36.7)x  b.
Since b ≠ 0 here, we use the
point A(50, 0) and the pointslope form to find the equation.
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Example 4
SOLVING A PROBLEM INVOLVING
ANGLES OF ELEVATION (continued)
Graph y1 and y2, then find the point of intersection. The
y-coordinate gives the length of BC, or h.
Thus, h ≈ 45.
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