Transcript 4.8 Applications and Models using Trigonometry

4.8 Applications and Models
using Trigonometry
Given one side and an acute angle
of a right triangle
Find the remaining parts of the triangle.
a  ____
mA  45.9
mB  ___
mC  90
B
b  12.8
A
c  ___
C
Given one side and an acute angle
of a right triangle
Find the remaining parts of the triangle.
12.8
Cos 45.9 
c
12.8
c
Cos 45.9
c  18.4
B
A
49.5
12.8
C
Given one side and an acute angle
of a right triangle
Find the remaining parts of the triangle.
B
a
Tan45.9 
12.8
12.8  Tan45.9  a
a  13.2
18.4
A
49.5
12.8
C
Given one side and an acute angle
of a right triangle
Find the remaining parts of the triangle.
For Angle B
90  45.9  44.1
B
18.4
13.2
A
49.5
12.8
C
Find of an object in the distance
Finding the height of a tree on a mountain.
T
M
58
1000 ft
22
Find of an object in the distance
Finding the height of a tree on a mountain.
Tan22 
T
M
1000
1000  Tan22  M
404.03  M
M
58
1000 ft
22
Find of an object in the distance
Finding the height of a tree on a mountain.
Tan22 
T
M
1000
1000  Tan22  M
404.03  M
M
58
22
Tan58 
M T
1000
1000 ft
1000  Tan58  404.03  T
1196.3 ft  T
Trigonometry and Bearings
Bearing is an acute angle based off the
North - South line.
N 38º W
N
38
W
E
S
A nautical problem
A yacht is going 14 knots East for 3 hours,
then turns N 42º E for an hour. How far
from port is the yacht.
42
A nautical problem
Need to find a hypotenuse of a larger
triangle. To find the distance.
42
A nautical problem
The extension helps us find the hypotenuse.
We have a few angles and the distance to
add.
42
A nautical problem
The extension helps us find the hypotenuse.
We have a few angles and the distance to
add.
14
42
48
14(3)  42
b
a
A nautical problem
The extension helps us find the hypotenuse.
We have a few angles and the distance to
add.
a
Sin 48 
14
14
42
a
14  Sin 48  a
48
14(3)  42
b
10.4  a
A nautical problem
The extension helps us find the hypotenuse.
We have a few angles and the distance to
add.
b
Cos 48 
14
42
10.4
14  Cos 48  b
48
14(3)  42
14
b
9.36  b
A nautical problem
The extension helps us find the hypotenuse.
We have a few angles and the distance to
add.
b
Cos 48 
14
42
10.4
14  Cos 48  b
48
14(3)  42
14
b
9.36  b
A nautical problem
The extension helps us find the hypotenuse.
We have a few angles and thedistance to
add.
51.36  10.4  52.4
2
2
42
42
14
48
9.36
42  9.36  51.36
10.4
Harmonic Motion (Doing the Wave)
Way of writing the Sine function or Cosine
function with time.
d = a Sin wt or d = a Cos wt
d is the distance from the origin or
Equilibrium
a is for amplitude; w is like b in the normal
function (changes period)
w
2
Period =
Frequency = 2
w
Homework
Page 338 – 342
# 3, 11, 16, 21,
26, 31, 36, 41,
47, 53
Homework
Page 338 – 342
#
9, 14, 19, 24,
27, 33, 40, 43,
52, 55