Transcript 8.7 Applications of Right Triangles

Copyright © 2011 Pearson Education, Inc.
Slide 8.7-1
Chapter 8: Trigonometric Functions and
Applications
8.1 Angles, Arcs, and Their Measures
8.2 The Unit Circle and Its Functions
8.3 Graphs of the Sine and Cosine Functions
8.4 Graphs of the Other Circular Functions
8.5 Functions of Angles and Fundamental Identities
8.6 Evaluating Trigonometric Functions
8.7 Applications of Right Triangles
8.8 Harmonic Motion
Copyright © 2011 Pearson Education, Inc.
Slide 8.7-2
8.7 Applications of Right Triangles
• Significant Digits
–
–
Represents the actual measurement.
Most values of trigonometric functions and
virtually all measurements are
approximations.
Copyright © 2011 Pearson Education, Inc.
Slide 8.7-3
8.7 Solving a Right Triangle Given an
Angle and a Side
Example Solve the right triangle ABC, with A = 34º 30 and
c = 12.7 inches.
Solution
Angle B = 90º – A = 89º 60 – 34º 30
= 55º 30.
a
sin A 
c
a

sin 34 30 
 a  12.7 sin 34 30  7.19 inches
12.7

Use given information to find b.
b

cos 34 30 
 b  12.7 cos 34 30  10.5 inches
12.7

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Slide 8.7-4
8.7 Solving a Right Triangle
Given Two Sides
Example
Solve right triangle ABC if a = 29.43 centimeters
and c = 53.58 centimeters.
Solution
Draw a sketch showing the given information.
29.43
53.58
Using the inverse sine function
on a calculator, we find A  33.32º.
sin A 
B = 90º – 33.32º  56.68º
Using the Pythagorean theorem,
b c a
2
2
2
b  53.58  29.43
2
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2
2
 b  44.77 centimeter s.
Slide 8.7-5
8.7 Angles of Elevation or Depression
The angle of elevation from point X to point Y
(above X) is the acute angle formed by ray XY and
a horizontal ray with endpoint at X.
The angle of depression from point X to point Y
(below X) is the acute angle formed by ray XY and
a horizontal ray with endpoint at X.
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Slide 8.7-6
8.7 Angles of Elevation or Depression
Solving an Applied Trigonometry Problem
1. Draw a sketch, and label it with the given
information. Label the quantity to be found
with a variable.
2. Use the sketch to write an equation relating the
given quantities to the variable.
3. Solve the equation, and check that your answer
makes sense.
Copyright © 2011 Pearson Education, Inc.
Slide 8.7-7
8.7 Solving a Problem Involving
Angle of Elevation
Example The length of the shadow of a building
34.09 meters tall is 3.62 meters. Find the elevation of
the sun.
34.09
tan B 
37.62
so
34.09

1
B  tan
 42.18 .
37.62
The angle of elevation of the sun is 42.18o.
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Slide 8.7-8
8.7 Bearing
There are two methods for expressing bearing. When a
single angle is given, it is understood that the bearing
is measure in a clockwise direction from the north.
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Slide 8.7-9
8.7 Solving a Problem Involving Bearing
(First Method)
Example Radar stations A and B are on an east-west
line, with A west of B, 3.70 km apart. Station A detects
a plane at C, on a bearing of 61.0o. Station B
simultaneously detects the same plane, on a bearing of
331.0o. Find the distance from A to C.
Copyright © 2011 Pearson Education, Inc.
Slide 8.7-10
8.7 Solving a Problem Involving Bearing
(First Method)
Solution Draw a sketch. Since a line drawn due north
is perpendicular to an east-west line, right angles are
formed at A and B, so angles CAB and CBA can be
found. Angle C is a right angle. Find distance b by
using the cosine function.
b
cos29.0 
so
3.70
b  3.70cos29.0  3.24 km
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Slide 8.7-11
8.7 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.
Either N or S always come first, followed by an acute angle, and
then E or W.
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Slide 8.7-12
8.7 Solving a Problem Involving Bearing
(Second Method)
Example The bearing
from A to C is S 52o E.
The bearing from A to B is
N 84o E. The bearing from
B to C is S 38o W. A plane
flying at 250 mph takes 2.4
hours to go from A to B.
Find the distance from A to
C.
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Slide 8.7-13
8.7 Solving a Problem Involving Bearing
(Second Method)
Solution Make a sketch.
First draw the two bearings
from point A. Then choose
a point B on the bearing
N 84o E from A, and draw
the bearing to C. Point C
will be located where the
bearing lines from A and B
intersect.
The distance from A to
B is 250(2.4) = 600 miles.
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Slide 8.7-14
8.7 Solving a Problem Involving Bearing
(Second Method)
Solution (continued)
To find b, the distance from A to C, use the sine
function.
sin46 
b
c
b
600
600sin46  b so
b  430 miles
sin46 
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Slide 8.7-15
8.7 Calculating the Distance to a Star
•
•
•
In 1838, Friedrich Bessel determined the distance to a
star called 61 Cygni using a parallax method that relied
on the measurement of very small angles.
You observe parallax when you ride in a car and see a
nearby object apparently move backward with respect to
more distance objects.
As the Earth revolved around the sun, the observed
parallax of 61 Cygni is   0.0000811º.
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Slide 8.7-16
8.7 Calculating the Distance to a Star
Example One of the nearest stars is Alpha
Centauri, which has a parallax of   0.000212º.
(a) Calculate the distance to Alpha Centauri if the
Earth-Sun distance is 93,000,000 miles.
(b) A light-year is defined to be the distance that
light travels in 1 year and equals about 5.9
trillion miles. Find the distance to Alpha
Centauri in light-years.
Copyright © 2011 Pearson Education, Inc.
Slide 8.7-17
8.7 Calculating the Distance to a Star
Solution
(a) Let d be the distance between Earth and Alpha
Centauri. From the figure on slide 8-46,
93,000,000
93,000,000
sin  
or d 
d
sin 
93,000,000
d
 2.51 1013 miles.
sin .000212
2.51 1013
 4.3 light - years.
(b) This distance equals
12
5.9  10
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Slide 8.7-18
8.7 Solving a Problem Involving
Angle of Elevation
Example 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
36.7º. He then moves back 50 feet. From the second
point, the angle of elevation is 22.2º. Find the height
of the tree.
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Slide 8.7-19
8.7 Solving a Problem Involving
Angle of Elevation
Analytic Solution There are two unknowns, the distance x
and h, the height of the tree.
h


In triangle ABC, tan 36.7 
or h  x tan 36.7 .
x
h


tan
22
.
2

or
h

(
50

x
)
tan
22
.
2
.
In triangle BCD,
50 x
Each expression equals h, so the expressions must be equal.
x tan 36.7  (50  x ) tan 22.2



x tan 36.7  50 tan 22.2  x tan 22.2



x tan 36.7  x tan 22.2  50 tan 22.2
50 tan 22.2
x
tan 36.7   tan 22.2

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
Slide 8.7-20
8.7 Solving a Problem Involving
Angle of Elevation
We saw above that h = x tan 36.7º. Substituting for x,
50 tan 22.2



h
tan
36
.
7
 45 feet.



 tan 36.7  tan 22.2 
Graphing Calculator Solution Superimpose the figure on
the coordinate axes with D at the origin.
Line DB has m = tan 22.2º with yintercept 0. So the equation of line DB
is y = tan 22.2º x.
Similarly for line AB, using the pointslope form of a line, we get the
equation y = [tan 36.7º](x – 50).
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Slide 8.7-21
8.7 Solving a Problem Involving
Angle of Elevation
Plot the lines DB and AB on the graphing calculator and find
the point of intersection.
Line DB: y = tan 22.2º x
Line AB: y = [tan 36.7º](x – 50).
Rounding the information at the bottom of the screen, we see
that h  45 feet.
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Slide 8.7-22