Download 2. Right Triangle Trigonometry

2. Right Triangle Trigonometry
2.1 Definition II: Right Triangle Trigonometry
2.2 Calculators and Trigonometric Functions of
an Acute Angle
2.3 Solving Right Triangles
2.4 Applications
2.5 Vectors: A Geometric Approaches
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2.1 Definition II: Right Triangle Trigonometry
hypotenuse
c
A
b
Side adjacent A
a
C
Side opposite A
B
If triangle ABC is a right triangle
with C = 90°, then the six
trigonometric functions for A are
defined as follows:
side opposite A a
sinA = hypotenuse = c
side adjacent A
b
=
cosA = hypotenuse
c
side opposite A
a
=
tanA = side adjacent A b
side adjacent A
b
=
cotA = side opposite A a
hypotenuse
c
=
secA = side adjacent A b
hypotenuse
c
=
cscA = side opposite A a
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2.1 Definition II: Right Triangle Trigonometry
1) Three pairs of cofunctions
Sine and Cosine are cofunctions, so are tangent and
cotangent, and secant and cosecant.
2) Cofunction Theorem
A trigonometric function of an angle is equal to the
cofunction of the complement of the angle. i.e.
cosθ = sin(90 − θ )
cotθ = tan(90 − θ )
cscθ = sec(90 − θ )
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2.1 Definition II: Right Triangle Trigonometry
θ
sinθ
cosθ
tanθ
30°
45°
60°
1
2
3
2
1
3
=
3
3
1
2
=
2
2
1
1
2
3
1
2
=
2
2
3
2
TABLE 1. EXACT VALUES (MEMORIZE)
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2.1 Definition II: Right Triangle Trigonometry
(1) ΔABC is a right triangle with C = 90°. Find the six
trigonometric functions of A if a = 3 and b = 7 [6]
(2) Find the following
[12]
sin A
A
cos A
5
tan A
C
sin B
1
cos B
B
tan B
5
2.1 Definition II: Right Triangle Trigonometry
(3) Use the Cofunction theorem to fill in the blanks so that
each expression becomes a true statement
•
•
sin y = cos
cot(12°) = tan
[26]
[24]
(4) Simplify.
•
•
(sin30° + cos30°)2
sin230° + cos230°
[35 modified]
[36 modified]
(5) Find exact value:
•
•
csc30°
cot30°
[50]
[54]
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2.2 Calculators and Trig Functions of an Acute Angle
1)
2)
3)
4)
5)
Degree, minute and second (DMS) and decimal degree
Convert between DMS and decimal degree
Add or subtract angles
Using calculator
Using inverse trig function
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2.2 Calculators and Trig Functions of an Acute Angle
1) Degree, minute and second (DMS) and decimal degree
Expression Read as
52°10’
52 degrees, 10 minutes
13° 24′ 15′′ 13 degrees, 24 minutes, 15 seconds
DMS
Expression Read as
27.36°
27 point 36 degrees
Decimal Degree
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2.2 Calculators and Trig Functions of an Acute Angle
2) Convert between DMS and decimal degree
1° = 60′
1′ = 60′′
or
or
⎛ 1 ⎞
1′ = ⎜ ⎟
⎝ 60 ⎠
⎛ 1 ⎞
1′′ = ⎜ ⎟
⎝ 60 ⎠
(1) Change to degree and minutes:
o
′
18.75°
[18]
21°48′
[26]
ans. 18°45′
(2) Change to decimal points:
ans. 21.8°
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2.2 Calculators and Trig Functions of an Acute Angle
(3) Add or subtract
•
•
63°38′ + 24°52′
90° − (62°25′)
[4]
[8]
ans. 88°30′
ans. 27°35′
(4) Use calculate to find. Round the answer to four
decimal places. [32, 36]
•
•
cos 82.9° =
tan 81.43° =
ans. 0.1236
ans. 0.6357
(5) Use calculate to find (convert to degree decimal).
Round the answer to four decimal places. [44, 50]
•
•
sin 35°10′
Sec 84°48′
ans. 0.4760
ans. 11.0336
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2.2 Calculators and Trig Functions of an Acute Angle
(6) Inverse trig function (more later, similar for sine,
cosine)
•
tanθ = 3.152. Find θ.
[e.g.10]
ans. 72.4°
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2.3 Solving Right Triangles
1) Number of significant digits
2) Accuracy of side and accuracy of angle.
3) Solving right triangles
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2.3 Solving Right Triangles
1) Number of significant digits
Definition. The number of significant digits in a number is
found by counting all the digits from left to right with the first
non-zero digit on the left. When no decimal point is present,
trailing zeros are not considered significant.
0.042 has two significant digits
0.005 has one significant digit
20.5 has three significant digits
6.000 has four significant digits
9,200. has four significant digits
700 has one significant digit
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2.3 Solving Right Triangles
2) Accuracy of side and accuracy of angle.
The relationship between the accuracy of the sides of a
triangle and the accuracy of the angles in the same
triangle.
Accuracy of Sides
Two significant digits
Three significant digits
Four significant digits
Accuracy of angles
Nearest degree
Nearest tenth of a degree
Nearest hundredth of a degree
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2.3 Solving Right Triangles
(1) ΔABC is a right triangle with C = 90°
•
If B = 16.9° and c = 7.55 cm, find b [6]
ans. 2.19 cm
•
If a = 42.3 in and b = 32.4 in, find B [10]
ans. 37.5°
(2) If C = 26° and r = 19, find x
[36]
ans. 2.1
C
26°
r
B
x
r
D
A
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2.3 Solving Right Triangles
(3) ΔABC is a right triangle with C = 90° [46]
If A = 32°, ∠BDC = 48° and AB = 56. Find h, then x.
A
y
D
x
B
ans. h = 56sin(32°) = 30
h
ans. x = h/tan(48°) = 27
C
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2.4 Applications
1) Geometry – isosceles triangle
2) Angle of elevation and angle of depression
3) Distance and bearing
17
2.4 Applications
1) Geometry – isosceles triangle
e.g.1 Two equal sides of an isosceles triangle are each 12
centimeters. If each of the two equal angles measures
52°, find the length of the base and the altitude.
C
24
A
h
52°
b
24
ans. h = 24sin(52°) = 19 cm
52°
ans. b = 2·24cos(52°) = 30 cm
B
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2.4 Applications
2) Angle of elevation and angle of depression
An angle measured from the horizontal up is called
angle of elevation; an angle measured from the
horizontal down is called angle of depression.
horizontal
Angle of
elevation
Angle of
elevation
horizontal
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2.4 Applications
2) Angle of elevation and angle of depression
Angle of elevation. If the angle of elevation of the Sun is
63.4° when a building casts a shadow of 37.5 ft, what is
the height of the building?
[10]
ans. h = 37.5tan(63.4°) = 74.9 ft
h
63.4°
37.5 ft
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2.4 Applications
3) Distance and bearing
Definition. The bearing of a line l is the acute angle formed by
the north-south line and the line l. The notation used to
designate the bearing of a line begins with N or S (North or
South), followed by number of degrees in the angle, and ends
with E or W (East or West)
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2.4 Applications
3) Distance and bearing – four scenarios
N
N
B
W
W
The bearing
Of B from A
Is N65°W
B
The bearing
Of B from A 40°
Is N40°E
65°
E
A
W
E
A
S
S
N
N
The bearing A
70°
Of B from A
Is S70°E
E
A
W
20°
B
The bearing
Of B from A
Is S20°W
E
B
S
S
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2.4 Applications
3) Distance and bearing
A man wandering in the desert walks 2.3 miles in the direction of
S31°W. He then turns 90° and walks 3.5 mile in the direction of
N59°W. At the same time, how far is he from his starting point, and
what is his bearing from his starting point? [18]
x
θ
31°
59°
ans. x = 4.2 mi
ans. θ = 88°
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2.5 Vectors: A Geometric Approach
1)
2)
3)
4)
5)
6)
Vector
Notation of vector
Equality for vectors
Addition and subtraction of vectors
Horizontal and vertical components of a vector
applications
24
2.5 Vectors: A Geometric Approach
1) Vector
•
•
Magnitude
direction
2) A zero vector is a vector with its magnitude equals to
zero
3) Two vectors are equal if they have the same
magnitude and the same direction.
25
2.5 Vectors: A Geometric Approach
Notation The quantity is
v
a vector (boldface)
v
a vector (arrow above the variable)
AB
a vector
x
a scalar
|v|
the magnitude of vector v, a scalar
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2.5 Vectors: A Geometric Approach
4) Addition and Subtraction of Vectors: given u and v
find u + v and u – v (Parallelogram Rule)
v v
u+v
u
v
−v
−v
u
u–v
27
2.5 Vectors: A Geometric Approach
5) Horizontal and vertical components of a vector
Vector v is in standard position if the tail of the vector
is at the origin.
y
Vector v in standard position
vy is the vertical component of v
vy
v
vx
x
vx is the horizontal component of v
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2.5 Vectors: A Geometric Approach
6) Application
Draw a vector representing a velocity of 50 cm/sec at
N30°E
[6]
N
v
30°
W
A
E
S
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2.5 Vectors: A Geometric Approach
6) Application
A person is riding in a hot air balloon. For the first hour, the wind
current is a constant 9.50 mph at N37.5°E. Then the wind changes to
8.00 mph and heads the balloon in the direction S52.5°E. This
continues for 1.5 hours. How far is the balloon from its starting
position?
[9]
52.5°
37.5°
θ
d
ans. d = 15.3 mi
ans. θ = arctan(12/9.5) = 51.6°
ans. bearing = 51.6° + 37.5° = 89.1°
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2.5 Vectors: A Geometric Approach
6) Application
A bullet is fired into the air with an initial velocity of
1,800 ft/sec at 60° from the horizontal. Find the
magnitudes of the horizontal and vertical components of
the velocity.
[24]
N
ans. |vx|= 1800cos(60°) = 900 ft/sec
v
vy
ans. |vy|= 1800sin(60°) = 1600 ft/sec
60°
W
A
S
vx
E
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2.5 Vectors: A Geometric Approach
6) application
A ship travels in direction S12°E, for 68 miles and then
changes its course to S60°E and travels another 110
miles. Find the total distance south and the total
distance east that the ship traveled.
[32]
distance south:
68cos(12°) + 110cos(60°) = 120 mi
12°
distance east:
60°
68sin(12°) + 110sin(60°) = 110 mi
32