Transcript Right Triangle Trigonometry

Warm-up 10/29
Evaluate the following. Give exact values when possible:
 1
arcsin   
 2
 1
arccos   
 2
 1
arctan   
 2
4.8 Trigonometric Applications
and Models 2014
Objectives:
•Use right triangles to solve real-life problems.
•Use directional bearings to solve real-life problems.
•Use harmonic motion to solve real-life problems.
Terminology
• Angle of elevation –
angle from the
horizontal upward to
an object.
• Angle of depression –
angle from the
horizontal downward
to an object.
Object
Horizontal
Angle of
elevation
Observer
Angle of
depression
Observer
Horizontal
Object
Precalculus
4.8 Applications and Models
3
Example
• Solve the right triangle for
all missing sides and angles.
You Try
• Solve the right triangle for
all missing sides and angles.
Example – Solving Rt. Triangles
At a point 200 feet from the base of a building,
the angle of elevation to the bottom of a smokestack
is 35°, and the angle of elevation to the top of the
smokestack is 53°. Find the height of the smokestack.
Precalculus
4.8 Applications and Models
6
You try:
• A swimming pool is 20 meters long and 12 meters
wide. The bottom of the pool is slanted so that the
water depth is 1.3 meters at the shallow end and 4
meters at the deep end, as shown. Find the angle of
depression of the bottom of the pool.
7.69°
Trigonometry and Bearings
• In surveying and navigation, directions are
generally given in terms of bearings. A bearing
measures the acute angle that a path or line of
sight makes with a fixed north-south line.
N 35E
N
N
S 70W
35°
W
E
W
E
70°
S
S
Trig and Bearings
• You try. Draw a bearing of:
N800W
S300E
N
N
W
E
S
W
E
S
Trig and Bearings
• You try. Draw a bearing of:
N800W
S300E
800
300
Example – Finding Directions Using Bearings
• A hiker travels at 4 miles per hour at a
heading of S 35° E from a ranger station.
After 3 hours how far south and how far east
is the hiker from the station?
Precalculus
4.8 Applications and Models
11
20sin(36o)
Example – Finding Directions Using Bearings
A ship leaves port at noon and heads due west at 20 knots, or 20
nautical miles (nm) per hour. At 2 P.M. the ship changes course to N
54o W. Find the ship’s bearing and distance from the port of
departure at 3 P.M.
d
a
θ
b
o)
20cos(36
a
20
o
20sin(36 )  a
b
cos(36o ) 
20
o
20cos(36 )  b
sin(36o ) 
40 nmfor 2
20 nmph
hrso )  40
20cos(36
20sin(36o )
tan( ) 
20cos(36o )  40
o

20sin(36
) 
1
  tan 

o
20cos(36
)

40


  11.819o
Bearing: N
78.181o W
20sin(36o)
A ship leaves port at noon and heads due west at 20
knots, or 20 nautical miles (nm) per hour. At 2 P.M. the
ship changes course to N 54o W. Find the ship’s bearing
and distance from the port of departure at 3 P.M.
d
a
θ
b
o)
20cos(36
a
20
o
20sin(36 )  a
b
cos(36o ) 
20
o
20cos(36 )  b
sin(36o ) 
40 nmfor 2
20 nmph
hrso )  40 Bearing: N 78.181o W
20cos(36
 20sin(36 ) 
 20sin(36 ) 
o
o


 20 cos(36 )  40 
2
 20 cos(36 )  40  d 2
2

o
o
2
2
d
57.397 nm  d
Two lookout towers are 50 kilometers apart. Tower A is due west of
tower B. A roadway connects the two towers. A dinosaur is spotted
from each of the towers. The bearing of the dinosaur from A is N 43o
E. The bearing of the dinosaur from tower B is N 58o W. Find the
distance of the dinosaur to the roadway that connects the two towers.
h
43o
58o
47o
A
h
tan(47 ) 
x
x  tan(47o )  h
o
32o
x
50– x
h
tan(32 ) 
50  x
50  x  tan(32o )  h
o
B
h
47o
A
32o
x
x  tan(47o )  h
50 tan(32o )
o

tan(47
)h
o
o
tan(47 )  tan(32 )
19.741  h
19.741
50– x
B
50  x  tan(32o )  h
x  tan(47o )   50  x  tan(32o )
x  tan(47o )  50 tan(32o )  x tan(32o )
x  tan(47o )  x tan(32o )  50 tan(32o )
x  tan(47o )  tan(32o )   50 tan(32o )
50 tan(32o )
x
km
tan(47o )  tan(32o )
Two lookout towers spot a fire at the same time. Tower B is Northeast of
Tower A. The bearing of the fire from tower A is N 33o E and is calculated to
be 45 km from the tower. The bearing of the fire from tower B is N 63o W and
is calculated to be 72 km from the tower. Find the distance between the two
towers and the bearing from tower A to tower B.
a
72
b
33o
A
c
45
s
63o
d
B
45cos(33
b – d 0) – 72sin(630)
a +0)c+ 72sin(630)
45sin(33
a
c
d
b
o
o
o
sin(33 ) 
sin(63 ) 
cos(63 ) 
cos(33 ) 
45
72
72
45
45sin(33o )  a 45cos(33o )  b 72sin(63o )  c 72cos(63o )  d
o
a
c
63o
72
b
33o
A
45
s
0) – 72sin(630)
b45cos(33
–d
o
2
 45cos(33 )  72cos(63 )
o
B
a +0)c+ 72sin(630)
45sin(33
 45cos(33 )  72 cos(63 ) 
o
d
o
2

o

o
 45sin(33 )  72sin(63 )
o
 45sin(33 )  72sin(63 )
88.805km  s
o

2

2
 s2
 s
a
72
b
33o
A
c
45
s
63o
d
B
0) – 72sin(630)
b45cos(33
–d
θ
a +0)c+ 72sin(630)
45sin(33
45cos(33o )  72cos(63o )
tan( ) 
45sin(33o )  72sin(63o )


 
 
 45cos(33o )  72 cos(63o )
  tan 1 
 45sin(33o )  72sin(63o )

  3.262o
o
o
o
N
86.738
E
90    86.738
Homework
4.8 p 326
1, 5, 9, 17-37 Odd
Quiz tomorrow on sections 4.5,4.6, and 4.7
Precalculus
4.8 Applications and Models
19
4.8 Trigonometric Applications
and Models Day 2
Objectives:
•Use harmonic motion to solve real-life problems.
HWQ 11/14
• A plane is 160 miles north and 85 miles east of
an airport. The pilot wants to fly directly to
the airport. What bearing should be taken?
Precalculus
4.8 Applications and Models
21
Terminology
• Harmonic Motion – Simple vibration,
oscillation, rotation, or wave motion. It can
be described using the sine and cosine
functions.
• Displacement – Distance from equilibrium.
Precalculus
4.8 Applications and Models
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Simple Harmonic Motion
• A point that moves on a coordinate line is in
simple harmonic motion if its distance d from the
origin at time t is given by
d  a sin t
a  amplitude
or
d  a cos t
2
 period


 frequency
2
where a and ω are real numbers (ω>0)
and frequency is number of cycles per unit of time.
Precalculus
4.8 Applications and Models
23
Simple Harmonic Motion
10 cm
10 cm
0 cm
0 cm
10 cm
10 cm
Precalculus
4.8 Applications and Models
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Example – Simple Harmonic Motion
Given this equation for simple harmonic motion
3
d  6 cos
t
4
Find:
a) Maximum displacement 6
b) Frequency 83 cycle per unit of time
c) Value of d at t=4 6
d) The least positive value of t when d=0
Precalculus
4.8 Applications and Models
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t
3
25
You Try – Simple Harmonic Motion
A mass attached to a spring vibrates up and down in
simple harmonic motion according to the equation
d  4sin

2
t
Find:
a) Maximum displacement 4
1
b) Frequency 4 cycle per unit of time
c) Value of d at t  2 0
d) 2 lvalues of t for which d=0 0 and 2
Example – Simple Harmonic Motion
A weight attached to the end of a spring is pulled down
5 cm below its equilibrium point and released. It takes
4 seconds to complete one cycle of moving from 5 cm
below the equilibrium point to 5 cm above the
equilibrium point and then returning to its low point.
• Find the sinusoidal function that best represents
the motion of the moving weight. f  t   5cos   t 


2


• Find the position of the weight 9 seconds after it is
released. 0
You Try – Simple Harmonic Motion
A buoy oscillates in simple harmonic motion as waves
go past. At a given time it is noted that the buoy
moves a total of 6 feet from its low point to its high
point, returning to its high point every 15 seconds.
• Write a sinusoidal function that describes the
motion of the buoy if it is at the high point at t=0.
 2 
f  t   3cos 
t
 15 
• Find the position of the buoy 10 seconds after it is
released.  3
2
Homework
4.8 Applications and Models Worksheet
(Bearings and Harmonic Motion)
Test next Tuesday.