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Chapter 2
Trigonometry
§ 2.1
The Tangent Ratio
Hypotenuse
Opposite
Adjacent
The Tangent Ratio
Hypotenuse
(h)
Opposite
(o)
Hypotenuse
(h)
x
Opposite
(o)
Adjacent
(a)
opp
tan X 
adj
TOA
x
Adjacent
(a)
Example #1
Determine the tangent ratio for the following.
Adjacent
(a)
Hypotenuse
(h)
Opposite
(o)
opp
tan V 
adj
3
tan V 
4
opp
tan Q 
adj
4
tan Q 
3
Example #2
Determine the measures of G and J to the nearest tenth of
a degree.
Adjacent
(a)
Opposite
(o)
Hypotenuse
(h)
opp 5

tan J 
4
adj
5
tan 1    J
4
51.3  J
4
opp

tan G 
5
adj
4
tan 1    G
5
38.7  G
Example #3
The latitude of Fort Smith, NWT, is approximately 60o. Determine
whether this design for a solar panel is the best for Fort Smith. Justify
your answer.
The best angle of inclination for the solar
panel is the same as the latitude (60°)
h
x
o
a
9
Opp

tan x 
Adj 12
9
tan    x
 12 
1
 x = 36.8699°
This is not the best design for the solar panel
because it is not equal to the latitude of Fort
Smith (60 °)
Example #4
A 10-ft. ladder leans against the side of a building that is 7 ft. tall.
What angle, to the nearest degree, does the ladder make with the
ground?
a 2 + b 2 = c2
a 2 + o 2 = h2
a2 + 72 = 102
a2 + 49 = 100
a2 = 51
a = √51
7
Opp

tan x 
51
Adj
h
10
on your calculator:
x = tan-1(7÷√(51))
<x = 44.43
<x = 44°
x
√51 a
7
O
§ 2.2
Using the Tangent Ratio to Calculate Lengths
Hypotenuse
Opposite
Adjacent
The Tangent Ratio
Hypotenuse
(h)
Opposite
(o)
Hypotenuse
(h)
x
Opposite
(o)
Adjacent
(a)
opp
tan X 
adj
TOA
x
Adjacent
(a)
Example #1
Determine the length of AB to the nearest tenth of a
centimeter
Adjacent
(a)
Hypotenuse
(h)
Opposite
(o)
opp
tan 30 
adj
AB
10
10  tan 30 
10
10  tan 30  AB
5.8 cm  AB
Example #2 (first way)
Determine the length of EF to the nearest tenth of a
centimeter.
opp
tan 20 
a
O
h
adj
3.5
 EF
EF  tan 20 
EF
EF  tan 20  3.5
tan 20 tan 20
3. 5
EF 
tan 20
EF  9.6 cm
Example #2 (second way)
Determine the length of EF to the nearest tenth of a
centimeter.
opp
Angles in a triangle add up to 90°
O
a
70°
h
tan 70 
adj
EF
 3.5
3.5  tan 70 
3.5
3.5  tan 70  EF
9.6 cm  EF
Example #3
A searchlight beam shines vertically on a cloud. At a horizontal distance
of 250m from the searchlight, the angle between the ground and the
line of sight to the cloud is 75o. Determine the height of the cloud to
the nearest metre.
opp
tan 75 
h
O
a
250 m
adj
opp
 250
250  tan 75 
250
250  tan 75  opp
933 m  opp
The cloud is 933 metres above the ground
§ 2.4
The Sine and Cosine Ratios
Hypotenuse
Opposite
Adjacent
The Sine & Cosine Ratios
Hypotenuse
(h)
x
Opposite
(o)
Hypotenuse
(h)
Opposite
(o)
x
Adjacent
(a)
Adjacent
(a)
opp
sin 
hyp
adj
cos 
hyp
SOH
CAH
Example #1
a) In DEF identify the side opposite D, the side adjacent to D
and the hypotenuse.
Adjacent
Opposite
Hypotenuse
b) Determine D using sin and cos to the nearest hundredth
opp
sin D 
hyp
adj
cos D 
hyp
5
sin D 
13
12
cos D 
13
5
sin    D
 13 
1
D  22.62
 12 
cos    D
 13 
1
D  22.62
Example #2
Determine the measures of G and H to the nearest tenth of a
degree.
adj
cos G 
hyp
6
cos G 
14
6
cos    G
 14 
1
64.6  G
Adjacent
Opposite
Hypotenuse
opp
sin H 
hyp
6
sin H 
14
6
sin    H
 14 
1
25.4  H
Example #3
A water bomber is flying at an altitude of 5000ft. The plane’s radar shows that it is 8000 ft. from
the target site. What is the angle of elevation of the plane measured from the target site to the
nearest degree?
opp
sin x 
hyp
5000
sin x 
8000
 5000 
sin 
  x
 80000 
1
39  x
h
8000 ft
O
5000 ft
x
a
§ 2.5
Using Sine and Cosine Ratios To Calculate
Lengths
Example #1
Determine the length of BC to the nearest tenth of a centimetre.
a
O
h
What trig ratio uses adjacent
and hypotenuse?
BC
Adj

cos(50) =
Hyp 5.2
BC
 5.2
5.2  cos50  
5. 2
5.2 x cos(50) = BC
3.34249557 = BC
BC = 3.3 cm
Example #2
Determine the length of DE to the nearest tenth of a centimetre.
a
h
O
What trig ratio uses opposite
and hypotenuse?
Opp 6.8

sin(55) =
Hyp DE
6.8
 DE
DE  sin 55 
DE
DE  sin( 55)  6.8
sin( 55) sin( 55)
6.8
DE 
sin( 55)
DE  8.3 cm
Example #3
A surveyor made the measurements shown in the diagram. How could
the surveyor determine the distance from the transit to the survey pole
to the nearest hundredth of a meter?
h
O
a
What trig ratio uses adjacent
and hypotenuse?
Adj 20.86

cos(67.3) =
H
Hyp
20.86
H
H  cos67.3 
H
H  cos(67.3)  20.86
cos(67.3) cos(67.3)
20.86
H
cos(67.3)
H  54.05 m
§ 2.6
Applying the Trigonometric Ratios
Example #1
Solve . Give the measures to the nearest tenth.
Solve means find the measures of
all the sides and angles.
a 2 + b 2 = c2
6.02 + 10.02 = c2
36 + 100 = c2
136 = c2
√136 = c
11.7 = c
a
Opp 10.0

tan x 
6 .0
Adj
o
59.0°
h
11.7 cm
31.0°
z
on your calculator:
x = tan-1(10÷6)
<x = 59.0°
Angles in a triangle add up to
59.0° + 90° + z = 180°
149.0° + z = 180°
-149.0°
-149.0°
z = 31.0°
90°
Example #2
Solve this triangle. Give the measures to the nearest tenth
where necessary.
a
Solve means find the measures of all the
z
65°
Opp Opp

tan(
65
)

a 2 + b 2 = c2
Adj 5.0
sides and angles.
11.8
h cm
10.72 + 5.02 = c2
Opp
 5.0
5.0 tan( 65) 
2
114.49 + 25 = c
5.0
139.49 = c2 5.0 x tan(65) = Opp
√139.49 = c
Opp = 10.7 cm
11.8 = c
10.7ocm
Angles in a triangle add up to
25° + 90° + z = 180°
115° + z = 180°
-115°
-115°
z = 65°
90°
Example #3
A helicopter leaves its base, and flies 35km due west to pick up a sick person. It then
flies 58km due north to a hospital.
a) When the helicopter is at the hospital, how far is it from its base to the nearest
kilometre?
58 Km
x
35 Km
a 2 + b 2 = c2
582 + 352 = x2
3364 + 1225 = x2
4589 = x2
√4589 = x
67.7421= x
68 = x
Example #3
A helicopter leaves its base, and flies 35km due west to pick up a sick person. It then
flies 58km due north to a hospital.
b) When the helicopter is at the hospital, what is the measure of the angle between
the path it took due north and the path it will take to return directly to its base? Write
the angel to the nearest degree.
x
58 Km
a
h
68 Km
O
35 Km
on your calculator:
sin x 
Opp 35

Hyp 68
cos x 
58
Adj

68
Hyp
x = cos-1(58÷68)
<x = 31°
tan x 
Opp 35

Adj
58
x = tan-1(35÷58)
<x = 31°
x = sin-1(35÷68)
<x = 31°
§ 2.7
Solving Problems Involving More than One
Right Triangle
Example #1
Calculate the length of CD to the nearest tenth of a centimetre.
What trig ratio uses adjacent and hypotenuse?
What trig ratio uses opposite and hypotenuse?
sin(47) =
Opp 4.2

Hyp
h
4.2
h  sin( 47)   h
h
h  sin( 47)  4.2
sin( 47) sin( 47)
4.2
h
sin( 47)
h  5.7 cm
x
Adj

cos(26) =
Hyp 5.7
5.7  cos( 26) 
x  5.7
5.7
5.7 x cos(26) = x
x = 5.1 cm
O
a
h
5.7 cm
h
O
a
x
Example #2
From the top of a 20-m high building, a surveyor measured the angle of
elevation of the top of another building and the angle of depression of
the base of the building. The surveyor sketched this plan of her
measurements. Determine the height of the taller building to the
Opp 20
nearest tenth of a metre.

tan(15) =
Adj a
What trig ratio uses opposite and adjacent?
20
tan(15) 
a
h
o
74.6 cm
20 o
a
h
Opp Opp

tan(30) =
Adj 74.6
tan( 30) 
Opp
74.6
74.6  tan( 30)  Opp
Opp  43.1 m
Height = 20 + 43.1 = 63.1 m
a
20
tan(15)
a  74.6 m
Example #3
In the given diagram find HJK. Round your answer to the nearest
tenth.
2
2
2
a +b =c
102 + 32 = c2
100 + 9 = c2
109 = c2
√109 = c
o
√109
h
What trig ratio uses opposite and adjacent?
tan(x) =
Opp
109

5
Adj
tan( x) 
109
5
 109 
  x
tan 

 5 
1
HJK  64.4
a
x
Example #4
From the top of a 90-ft. observation tower, a fire ranger observes one fire due west of
the tower at an angle of depression of 5o, and another fire due south of the tower at
an angle of depression of 2o. How far apart are the fires to the nearest foot?
What trig ratio uses opposite and adjacent?
a
h
o
h
ft
1028.7 ft 2577.3
o
Opp Opp

tan(85) =
Adj 90
tan( 85) 
Opp
90
Opp Opp

tan(88) =
Adj 90
tan( 88) 
Opp
90
90  tan( 85)  Opp
90  tan( 88)  Opp
1028.7  Opp
2577.3  Opp
a2 + b2 = c 2
1028.72 + 2577.32 = c2
1058223.69 + 6642475.29= c2
7700698.98 = c2
√ 7700698.98 = c
2775 ft = c