Download 11.3 Primary Trigonometric Ratios

400
11.3 Primary Trigonometric Ratios
Introduction
In the previous section, the relationship between the side lengths of the three sides of a right
triangle was examined. In this section, we will study the relationship between the side lengths of a
right triangle and its acute angle measures. The core concept behind this relationship is based on
the fact that if one of the two acute angles of a right triangle is known, then all right triangles with
that one angle measure will be similar; therefore, their side lengths will be in proportion.
F
D
∆ABC, ∆ADE, ∆AFG are similar and
their side lengths are in proportion.
B
A
θ
C
E
G
A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. Mathematicians have
given special names to the six ratios of the three side lengths, relative to one of the acute angles in the
right triangle, known as θ. They are (1) the sine (sin) ratio, (2) the cosine (cos) ratio, (3) the tangent
(tan) ratio, (4) the cosecant (csc) ratio, (5) the secant (sec) ratio, and (6) the cotangent (cot) ratio.
The first three ratios are known as the primary trigonometric ratios, and will be the focus of this section.
The other three ratios, known as the secondary or reciprocal trigonometric ratios, are the reciprocal
ratios of the three primary trigonometric ratios, respectively. They are not covered in this textbook.
Sine, Cosine, and Tangent Ratios of Angles in a Right Triangle
In a right triangle, recall that the hypotenuse, the longest side, is the side across from the right angle.
In a right triangle with an acute angle θ (i.e., 0°< θ < 90°), the leg that forms the angle θ with the
hypotenuse is known as the adjacent leg and the third side, across (opposite) from angle θ, is known
as the opposite leg. The three primary trigonometric ratios of θ are shown in Exhibit 11.3-a:
Opposite
O
sinθ =
=
Hypotenuse H
Hypotenuse (H)
Opposite (O)
θ
Adjacent (A)
Adjacent
A
=
Hypotenuse H
Opposite
O
tanθ =
=
Adjacent
A
cosθ =
Exhibit 11.3-a Primary Trigonometric Ratios
You may find it helpful to use the acronym SOH-CAH-TOA to remember the three primary
trigonometric ratios:
In any acute angle,
0 < Sin θ < 1
0 < Cos θ < 1
Tan θ > 0
Chapter 11 | Basic Trigonometry
SOH
Sin θ = Opposite/Hypotenuse
CAH
Cos θ = Adjacent/Hypotenuse
TOA
Tan θ = Opposite/Adjacent
Note: Since the lengths of the legs of a right triangle will be greater than 0 but always less than the
hypotenuse, the sine and cosine ratios of any acute angle must be between 0 and 1. However, since
there is no relationship between the opposite leg and the adjacent leg, except that they must both
be greater than 0, the tangent ratio can be any positive number.
401
Example 11.3-a
Calculating Side Lengths Using the Sine Ratio
Calculate the unknown length in the following diagrams:
(i)
(ii)
10 m
h
25°
Solution
Using sin θ =
(i)
sin 25° =
Example 11.3-b
8m
O
H
h
10
(ii)
h = 10 (sin 25°)
50°
x
sin 50° =
8
x
8
sin 50°
= 10.443258... = 10.44 m
x=
= 4.226182... = 4.23 m
Calculating Side Lengths Using the Cosine Ratio
Calculate the unknown length in the following diagrams:
(i)
(ii)
12 m
40°
x
6m
20°
a
Solution
A
H
Using cos θ =
(i)
cos 20 ° =
a
12
(ii)
a = 12 (cos 20°)
= 11.276311... = 11.28 m
Example 11.3-c
cos 40 ° =
6
x
x=
6
cos 40°
= 7.832443... = 7.83 m
Calculating Side Lengths Using the Tangent Ratio
Calculate the unknown length in the following diagrams:
(i)
(ii)
35°
h
y
30°
5m
7m
Solution
Using tan θ =
(i)
tan 30° =
O
A
y
7
y = 7 (tan 30°)
= 4.041451... = 4.04 m
(ii)
tan 35° =
5
h
5
tan 35°
= 7.140740... = 7.14 m
h =
11.3 Primary Trigonometric Ratios
402
Exact Trigonometric Ratios of Special Common Angles
There are trigonometric ratios of special common angles (30°, 45°, and 60°) that can be computed
from special right triangles: the 30-60-90 triangle and 45-45-90 triangle.
30 - 60 - 90 Triangle
A
2
30° 30°
2
60°
60°
B
D
1
To calculate the trigonometric ratios of 30° and 60°, draw an
equilateral triangle, ABC, with side lengths of 2 units each and
draw AD = BC, as shown.
1
C
In the right triangle ABD (30 - 60 - 90), AB = 2 units, BD = 1
unit, and ADh = 3 units (AD is calculated using the Pythagorean
Theorem: AD2 = 22 – 12 = 4 – 1 = 3, which gives AD
h = 3 ).
A
30°
2
Using the above measures in the right triangle ABD, the sine,
cosine, and tangent ratios of 30° and 60° can be calculated exactly:
3
60°
1
B
Since the angle measures are all 60°, the angle at the vertex A is
bisected into 30° each and the base length ‘BC’ is bisected into 1
unit each.
D
sin 30° = BD = 1 , 2
AB
cos 30° =
1
tan 30° = BD =
,
AB
3
3
AD
=
,
2
AB
sin 60° =
3
AD
=
2
AB
cos 60° = BD = 1
2
AB
3
AD
tan 60° =
= h= 3
1
BD
45 - 45 - 90 Triangle
A
45°
2
B
1
45°
1
C
To calculate the trigonometric ratios of 45°, draw a right isosceles
triangle, ABC, with leg side lengths of 1 unit each as shown.
Since the triangle is a right isosceles, the two acute angles must be
equal. Therefore, the two acute angles are each 45°.
In the right isosceles triangle, ABC, (45 - 45 - 90), (AB is calculated using
the Pythagorean Theorem: AB2 = 12 + 12 = 1 + 1 = 2 which gives AB
h = 2)
Using the above measures in the right isosceles triangle ABC, the
sine, cosine, and tangent ratios of 45° can be calculated exactly:
1
1
BC
sin 45° = AC =
,
cos 45° =
=
AB
AB
2
2
Chapter 11 | Basic Trigonometry
tan 45° =
AC 1
=
=1
1
BC
403
Sin 30° = Cos 60° = 1
2
The primary trigonometric ratios of the special common angles are summarized in Exhibit 11.3-b.
These are referred to as the special trig ratios.
3
2
Sin 45° = Cos 45° = 1
2
Tan 30° = 1
3
Sin 60° = Cos 30° =
trigratio
30°
2
3
Tan 45° = 1
60°
Tan 60° = 3
1
ANGLE
30°
45°
60°
Sin
1
2
1
2
3
2
Cos
3
2
1
2
1
2
Tan
1
3
1
45°
2
1
45°
3
1
Exhibit 11.3-b Special Trig Ratios - Trigonometric Ratios of Special Common Angles
Using Calculators to Determine Trigonometric Ratios and Angles
A scientific calculator can be used to determine the trigonometric ratios of any other acute angle.
When using a calculator to determine the trigonometric ratios, ensure that it is in "degree" mode.
Note: Some calculators may yield trigonometric ratios as a decimal number. E.g. sin (60°) = 0.866025...
3
(which represents an equivalent ratio to 2 ).
Example 11.3-d
Using a Calculator to Calculate sin, cos, and tan of Acute Angles
Using a calculator, calculate the sine, cosine, and tangent of the following acute angles, rounded to
four decimal places as required:
Solution
(i) θ = 15°
(ii) θ = 72°
(iii) θ = 36.87°
(i) s in 15° = 0.258819...
= 0.2588
cos 15° = 0.965925...
= 0.9659
tan 15° = 0.267949....
= 0.2679
(ii) sin 72° = 0.951056... (iii) sin 36.87° = 0.600001...
= 0.9511
= 0.6000
cos 72° = 0.309016...
cos 36.87° = 0.799998...
= 0.3090
= 0.8000
tan 72° = 3.077683...
tan 36.87° = 0.750002...
= 3.0777
= 0.7500
Using Calculators to Calculate Angles
If we know the ratio of the lengths two sides of a right triangle, we can determine the angle related to
that ratio using the inverse trigonometric functions of sine, cosine, and tangent, known as arcsine,
arccosine, and arctangent, respectively. These functions often appear on scientific calculators as sin–1,
cos–1, and tan–1.
Example 11.3-e
Using a Calculator to Calculate the Angle Given a Trig Ratio
Using a calculator, calculate the angle measure in degrees (rounded to the nearest tenth of a degree)
for each of the following trigonometric ratios:
Solution
(i)sin θ = 0.9063
(ii) cos θ = 0.6
(iii) tan θ = 0.1467
(i) θ = sin-1(0.9063)
θ = 64.998944... = 65.0°
(ii) θ = cos-1(0.6)
(iii) θ = tan-1(0.1467)
θ = 53.130102... = 53.1°
θ = 8.345761... = 8.3°
11.3 Primary Trigonometric Ratios
404
Solving Right Triangles using Trigonometry
If one side length and one acute angle measure in a right triangle, are provided, trigonometric ratios
may be used to solve for the lenght of the remaining side lengths.
Conversely, if any two side lengths of a right triangle are given, the inverse trigonometric functions
and complementary angles may be used to solve for the two acute angles.
When solving right triangles, there are often several ways to proceed. However, calculations should be
performed using the method that requires as fewer steps as possible, i.e., whenever convenient, measurements
provided in the question, rather than measurements obtained from secondary calculations should be used.
Example 11.3-f
Solving a Right Triangle Given One Side Length and One Acute Angle
Determine the unknown side lengths and missing angle of the following right triangles. Round all
side lengths to the nearest hundredth and all angle measures to the nearest tenth.
(i)
b
Solution
(ii)
c
8.5 cm θ
24°
θ
b
1.64 m
72°
a
(i) First, calculate the length of one of the unknown sides using one of the primary trig ratios:
O
8.5
8.5
tan 24°tan
= 24°==
bA
b
b tan 24° = 8.5
8.5
b=
=
tan 24°
c
8.5 cm θ
b
8.5
≈ 19.1 cm
0.445228...
24°
= 19.091312... ≈ 19.1 cm
Then, calculate the other unknown side length using another trig ratio or the Pythagorean Theorem.
Using trig ratio,
Using Pythagorean Theorem,
O 8.5
8.5
sin 24°sin
= 24° =
H
c
c
c2 = 8.52 + 19.12
c sin 24° = 8.5
or
= 72.25 + 364.81 = 437.06
Using the Pythagorean
Theorem to calculate the
length of the hypotenuse,
c, could have resulted
in a rounding error or a
compound calculation
error had there been an
error in calculating b.
c == 437.06 ≈=20.9
20.905980...
≈ 20.9 cm
cm
8.5
8.5
=
≈ 20.9 cm
sin 24°
0.406736...
= 20.898043...≈ 20.9 cm
Finally, since the acute angles in a right triangle are complimentary: θ = 90° − 24° = 66°.
c=
(ii) First, calculate the length of one of the unknown sides using one of the primary trig ratios:
A
a
a
cos 72 °cos
= 72 °==
H
1.64 1.64
a = 1.64(cos 72°)
= 1.64(0.309016...)
= 0.506787... ≈ 0.51 m
θ
or
b
1.64 m
72°
a
Then, calculate the other unknown side length using another trig ratio or the Pythagorean Theorem:
Chapter 11 | Basic Trigonometry
405
Solution
continued
Using trig ratio,
Using Pythagorean Theorem,
Ob
b
sin 72°sin
= 72° ==
H 1.64
1.64
b = 1.64(sin 72°)
= 1.64(0.951056...)
=1.559732... ≈ 1.56 m
a +b =c
2
2
2
b2 = 1.642 − 0.512
= 2.6986 − 0.260 = 2.4295
or
b = 2.4295 ≈ 1.56 m
Using the Pythagorean
Theorem to calculate the
length of the unknown
leg, b, could have resulted
in a rounding error or a
compound calculation
error had there been an
error in calculating a.
2.4295 ≈ 1.56 m
b= =
1.558685...
Finally, since the acute angles in a right triangle are complimentary: θ = 90° − 72° = 18°.
Example 11.3-g
Solving a Right Triangle Given Two Side Lengths
Determine the unknown side length and acute angles of the following right triangles (round all
answers to the nearest tenth as required):
(i)
(ii)
5 cm
x
y
12 cm
x
6m
a
y
3m
Solution
(i) First, calculate the length of the hypotenuse using the Pythagorean Theorem,
c2 = 52 + 122 = 25 + 144 = 169
c = 169 =13
c
x
5 cm
y
12 cm
As the opposite and adjacent lengths are provided, use the inverse trig function for tan
(i.e., tan–1) to determine the angle, x.
-1
X
S 12 X =
x = tan -1 S 12
67.4°
(2.4) . 67.4°
tan-1-1(2.4)
tan
=67.380135...
5 x ==tan
5 .
Finally, since acute angles in a right triangle are complimentary: y = 90° − 67.4° = 22.6°.
(ii) First, calculate the length of the unknown leg using the Pythagorean Theorem,
a2 = 62 − 32 = 36 − 9 =27
am= 27 ≈ 5.2 m
=5.2
5.196152...
a = 27 ≈
x
6m
a
y
3m
As the opposite and hypotenuse lengths are provided, use the inverse trig function for sin
(i.e., sin–1) to determine the angle, x.
x = sin -1 S 36 X = sin -1 (0.5) = 30°
Finally, since acute angles in a right triangle are complimentary: y = 90° − 30° = 60°.
11.3 Primary Trigonometric Ratios
406
Slopes of Lines and Angles of Elevation and Depression
Slope of Lines
Recall from Section 9.2, that the definition of the slope of a line is the ratio of Rise to Run, or the ratio
of the changes in the y-value to the changes in the x-value between two points, P and Q, on the line:
m=
Change in y value ∆y Rise y2 – y1
=
=
=
Change in x value ∆ x Run x 2 – x1
Now, consider the right triangle that is created when the rise and run of a line are drawn between
points P and Q on that line, as illustrated in Exhibits 11.3-c and 11.3-d
Y
Y
y = mx + b
Q
(x1, y1)
P
(x2, y2)
b P(x1, y1)
(y2 – y1) = ∆y
(x2 – x1) = ∆x
b
0
m=
( y2 – y1)
( x 2 – x1 )
∆y
y = mx + b
∆x
X
0
Q (x2, y2)
X
(y – y ) (y – y )
m= 1 2 = 2 1
( x 2 – x1 ) ( x 2 – x1 )
Exhibit 11.3-d Rise and Run Between Points P and Q
Exhibit 11.3-c Rise and Run Between Points P and Q
Notice that the tangent ratio of the angle θ is almost exactly the same as the slope of the line. The only
difference between the two is that the tangent ratio is irrespective of the direction of the line (i.e., the
tangent ratio is positive, regardless of whether the slope of the line is positive or negative). This yields
the following equation:
y −y
Absolute Rise
= 2 1
tanθ = m =
Absolute Run
x2 − x1
Therefore, we can determine the angle, θ, of any line, by calculating the arc tangent of the slope of the line:
 ∆y
 ∆x
θ = tan −1 ( m ) = tan −1 

 y2 − y1
−1
 = tan 

 x2 − x1



Angles of elevation and Depression
Angle of Elevation: It is the angle above the horizontal line from
the observer's eye to the object, known as the angle of sight.
Angle of Depression: It is the angle below the horizontal line
from the observer's eye to the object known as the line of sight.
Object
e of
Angle of elevation
sigh
t
Horizontal line
Angle of depression
ght
Observer
f si
ne o
Tangents are used to solve problems using angles of elevation and
depression.
Example 11.3-h
Lin
Li
Object
Calculating the Angle of Elevation or Depression
Determine the angle of elevation/depression of the following, rounded to the nearest tenth of a
degree as required:
(i) A ramp with a rise of 1.2 m and a run of 15 m.
(ii) A road with a decline of 500 m over 8 km.
Solution
(i) The slope of the ramp is:
1.2
==0.08
m tan m
θ ==
0.08
15
tan θ = 0.08
Line of sight
θ
Angle of Elevation
Therefore, the angle of elevation is θ = tan-1(0.08) = 4.573921... ≈ 4.6°
Chapter 11 | Basic Trigonometry
1.2 m
15 m
407
(ii) Convert 8 km to 8000 m in order to compare.
The slope of the road (called the grade) is:
500 m
8,000 m
θ
500
m tan m
θ ==
0.0625
== –0.0625
8,000
Therefore, the angle of depression is θ = tan–1(0.0625) = 3.576334... ≈ 3.6°.
Example 11.3-i
Angle of depression
Line of sight
Determining the Slope Given the Angle of Elevation/Depression
Determine the slope of the following, rounded to four decimal places as required:
(i) A skateboard ramp with an angle of elevation of 20°.
(ii) A ski hill with an angle of depression of 33.6°.
Solution
(i) The slope of the ramp is m = tan 20° = 0.363970... ≈ 0.3640.
(ii) The slope of the ski hill is m = tan 33.6° = 0.664398... ≈ 0.6644.
20°
Angle of depression
Angle of elevation
33.6°
Pythagorean Theorem and Trigonometric Ratios
The Pythagorean Theorem and primary trigonometric ratios can be combined to yield an important
result in trigonometry, sin2 θ + cos2 θ = 1. This result is known as the Fundamental Pythagorean
Trigonometric Identity.
Consider the following right triangle:
We know,
c
sin θ =
a
a
b
, cos θ = , and a2 + b2 = c2 (Pythagorean Theorem)
c
c
θ
b
2
2
2
2
a2 + b2 c 2
a b a b
Then sin 2 θ + cos 2 θ =   +   = 2 + 2 =
= 2 =1
c
c2
c
 c  c  c
Therefore, sin2 θ + cos2 θ = 1, regardless of what the value of θ is.
Example 11.3-j
Using the Fundamental Pythagorean Trigonometric Identity to Calculate Exact Trigonometric Ratios
Solution
Using the Fundamental Pythagorean Trigonometric Identity, determine the exact value of cos θ,
1
given that sin θ = (without using a calculator).
3
1
sin θ = , and solving for θ,
Substituting sin
Using sin2 θ + cos2 θ = 1,
3
2
1
2
  + cos θ =1
3
2
 1  = cos2 2 θ = 1 − 1 = 8
–1 
2
=
cos
1
+
=
θ
  + cos θ =1 9 9
 
3
9
cos θ =
8 2 42× 28 2 2
cos
== θ =
==
9
3 9 9
3
8 28 2 2 2
cos θ cos
= θ ==
=.
Therefore,
9
93
3
11.3 Primary Trigonometric Ratios
408
Applications of the Trigonometric Ratios
Example 11.3-k
Determining the Height of the CN Tower
From a point 30 m away from the base of the CN Tower, the angle of elevation to the top of the tower
is 83.5°. If the radius of the base is 33 m, determine the height of the tower, rounded to the nearest
metre.
Solution
First, draw a picture of this to better understand how to solve the question:
Object
tan 83.5° =
Angle of elevation is
the angle between the
horizontal line and the
line from the observer's
eye to the object (when
the observer is below
the level of the object).
h
63
h = 63 (tan 83.5°)
Line of sight
= 63 (8.776887...)
= 552.943903...
≈ 553 m
Therefore, the height of the tower is approximately 553 m.
Angle
of elevation
h
83.5 °
33m
30m
Observer
63m
Example 11.3-l
Determining the Distance Across a Lake
From a point 520 m above an elliptical (oval) lake, the angle of depression to one end of the lake is
40.6° and the angle of depression to the other end of the lake is 33.5°. Determine the length of the
lake, to the nearest ten metres.
Solution
First draw a picture of this to better understand how to solve the question:
Observer
Angle of depression
is the angle between
the horizontal line
and the line from the
observer's eye to the
object (when the object
is below the level of the
observer).
Angle of depression
Line of sight
40.6 ° Angle of depression
Line of sight
520 m
33.5 °
Object
33.5 °
d2
d1
40.6 °
Object
d
tan 33.5° =
520
d1
520
520520
520
= d1 =
≈==785.6
m
≈ 785.6 m
785.634300...
tan 33.5°
0.661885...
tan 33.5°
0.661885...
520
tan 40.6° =
d2
520
520520
520
d2 =
= d2 =
≈= 606.7
m
≈ 606.7 m
=
606.694410...
tan 40.6°
0.875103...
tan 40.6°
0.875103...
d1 =
d = d1 + d2 = 785.6 + 606.7 = 1392.3 m.
Therefore, the length of the lake is approximately 1,392.3 m.
Chapter 11 | Basic Trigonometry
409
Example 11.3-m
Calculating the Heading and Groundspeed of a Plane
An airplane is flying at a groundspeed of 880 km/h. The wind is blowing from due east at a speed of
132 km/h. If the airplane needs to travel due south, find the angle of its trajectory (the “heading”)
rounded to the nearest hundredth of a degree, and its resultant speed (the groundspeed), rounded to
the nearest km/h.
If the plane were to head due south, the wind from the east would push it off course to the west. As
such, the plane needs to fly into the wind (i.e., slightly east of south) in order to fly due south. First,
draw a picture of this to better understand how to solve the question:
Solution
132
= 0.15
880
θ = sin–1(0.15) = 8.626926... ≈ 8.63°
sin θ =
θ
880 km/h
x
Using Pythagorean Theorem,
x 2 = 8802 – 1322 = 774, 400 – 17, 424 = 756,976
x = 756,976 ≈= 870
x870.043677...
= km/h
756,976 ≈ 870 km/h
132 km/h
Therefore, the heading of the plane is 8.63°E and the groundspeed of the plane is 870 km/h.
11.3 Exercises
Answers to odd-numbered problems are available at the end of the textbook.
Given the values of θ in Problems 1 and 2, determine the three, primary trigonometric ratios of θ, rounded to four decimal places.
1.
a.
b.
c.
θ
65°
12.5°
53.13°
sin θ
?
?
?
cos θ
?
?
?
tan θ
?
?
?
2.
a.
b.
c.
θ
24°
82.8°
73.74°
sin θ
?
?
?
cos θ
?
?
?
tan θ
?
?
?
Given one trigonometric ratio, in Problems 3 and 4, determine the corresponding angle θ, rounded to the nearest degree, and
the other two, primary trigonometric ratios, rounded to four decimal places.
3.
a.
b.
c.
θ
?
?
?
sin θ
0.4540
?
?
cos θ
?
0.2924
?
tan θ
?
?
0.3639
4.
a.
b.
c.
θ
?
?
?
sin θ
0.5591
?
?
cos θ
?
0.9743
?
tan θ
?
?
8.1443
Given one trigonometric ratio, in Problems 5 and 6, determine the remaining trigonometric ratios exactly using the Pythagorean
Theorem, and determine the corresponding angle, rounded to the nearest degree.
sin θ
5.
cos θ
tan θ
θ
sin θ
?
?
?
a.
5
13
6.
cos θ
tan θ
θ
?
?
?
a.
3
5
b.
?
24
25
?
?
b.
?
8
17
?
?
c.
?
?
20
21
?
c.
?
?
12
35
?
. tan
. tan
. tan
. tan 60°
sin 60°sin
· cos
60°45°
· cos
– sin
45° 45°
– sin
· cos
45°30°
· cos 30°
sin 60°sin
60°30°
– sin
30° 30°
– sin
30°60°
. tan
special
trig
ratios in
Exhibit
11.3-b to determine
the .exact
value
for Problems
760°
to
12.
sinUse
60°the
· cos
45°
– sin
· cos
30°
sin 60°
tan
30°
– sin
. tan
. tan
sin
60°
· cos
45°45°
– sin
45°
· cos 30°
sin
60°
30°30°
– sin
30°
60°
.
.
.
.
sin7.60°sin
· cos
– sin
· cos
– sin
60°45°
· cos
45°45°
– sin
45°30°
· cos 30° 8. sin 60°sintan
60°30°tan
30°30°
– sintan
30°60°tan 60°
sin 45°sin
· cos
45°45°
· cos 45°
sin 60°sin
– sin
60° 30°
– sin 30°
sin 45°sin
· cos
45°
sin 60°sin
– sin
30°
45°
· 45°
cos 45°
2
tan
45°
2
tan
60°
sin 30°
10. cos 60°
cos
– cos
60° 30°
–– cos
30°
sin9.245°
· cos
45° 45° tan
45°
sin
sin 60°sin
– sin
30°
cos
cos
30°
60°
–
sin
245°
tan· cos
45°
cos 60° – cos30°
30°
2 tan 45°
2 tan 45°
cos
60°
–
cos
30°
cos
60°
–
cos
30°
11.2 60°
12. sin22 45°
sin+2 45°
tan
tan–2 sin
60°22 –
60°
sin+2 60°
cos22+60°
cos2 60°
cos22+30°
cos–2 tan
30°22–30°
tan2 30°
sin 45°sin
tan2 60°tan
– sin
+ 2cos
+ 2cos
– tan
2
2
60°60°
– sin
60° 60°
+ cos2 60°
45° 30°
+ cos
30°30°
– tan2 30°
2
sin2 45°sin
tan2 60°tan
– 2sin
60°
+ 2cos
+ 2cos
– 2tan
60°
– sin
60°2 60°
+ cos2 60°
45°2 30°
+ cos
30°2 30°
– tan2 30°
11.3 Primary Trigonometric Ratios
410
For Problems 13 to 18, determine the length of the unknown side, rounded to the nearest hundredths, for the given right triangles.
13.
16.
14.
x
x x
8 cm
x 8 cm
x
8
cm
8 cm y
8 cm
y cm 13 cm
25° 25° 25°
13 cm
y y 13
13 cm
cm20°
y
25° x
25°
20°13
20°
20°
8 cm20°
13 cm
y
25°
20°
17.
17 cm 17 cm
17 cm
17 cm 17 cm
a
a
a17 cm 63°
63°
63°63°
a 63°
a
a
63°
b
15.
48°
48°
48° 48°48°
z 21 cm 21 cm
z z
z 48°
z
21 cm
21 cm 21 cm
z
21 cm
b
18.
b b
4.5 cm
72° 72° 72°
4.5 cm cm
72°
72°
cm c c c c c
4.5 cm4.54.5
b
55°
55°
72° 27 cm
27 cm
cm55°
55°4.555°
c
27
cm
27 cm 27 cm
b
55°
27 cm
For Problems 19 to 24 determine the value of θ, rounded to the nearest hundredth of a degree, for the given right triangles.
19.
20.
13 cm
13 cm
10 cm 13 cm 13 cm
10 cm
13θcm
13 cm
10 cm 10 cm
θ
10 cm
10 cm θ
θ
θ
22.
21 cm
21 cm
θ
21 cm 21 cm
θ
θ cm θ
21 cm
21
θ θ
25 cm
25 cm
25 cm 25 cm
21.
25 cm
25 cm
θ
23.
16 cm
16 cm
15 cm
16 cm 16 cm
θ
15 cm
16
16 cm15 cm 15 cm
θ cm
θ
θ
15 cm
15 cm
θ θ
30 cm
30 cm
30 cm 30 cm
θ 29 cm
30
30 cm
θ cm
29
θcm29 cm
θ 29 cm
θ 29
θ cm
29 cm
24.
7 cm
7 cm
θ
7 cm 7 cm
θ
θ cm 7 cm7 cm
θ
12
12θcm θ
12 cm
12 cm
6
cm
12 cm
12 cm
6 cm
6 cm 6 cm
6 cm6 cm
8 cmθ θ
8 cm
θ
8 cm 8 θcm
8 cm8 cmθ θ
For Problems 25 to 30, solve the given right triangles fully (i.e., identify all missing side lengths and angle measures).
25.
θ
17 cm
θ θ 17 17
cm
cm
x
θ
17 cm
x x
Φ
θ
17
cm
xθ
ΦΦ
15
cm
17 cm
x
Φ
15 cm
x 15 cm Φ
15 cm
Φ
15 cm
6 cm
Φ
6 cm
cm
Φ15Φ6cm
28.
10 cm
Φ 6 cm
10 10
cmcm Φ 6 cm
10 cm Φ 6 cm
x
θ
10 cm
θ x x
θ cm
10
x
θ
x
θ
x
θ
Chapter 11 | Basic Trigonometry
26.
12 cm
cm
12 12
cm
θ
30°
12
30°cm
θ θ y 30°
x
30°
y y 12 θcm
xcmx
12
θ
30°
y
θ
30° x
y
x
x
y
x
x x
θ
θ θ x
x
29.
14 cm x θ y
14 14
cmcm θ y 16°
y
16°16° θ
14 cm
y
16°
14 cm
y
1416°
cm
y
16°
27.
11 cm
11 cm
11 11
cmcm
11 11
cmcm
11 cm
11
θ
Φ cm
11 cm
11
θ θ
ΦΦ
xcm
11 cmxθ x
11 Φ
cm
θ
Φx
θx
Φ
x
30. x 65°
y
x 65° y y
x 65°
x 65°
y
65°
θ
x
y
x 65° θ29θycm
29 29
cmcm
θ
θ29 cm
θ
29 cm
29 cm
411
For Problems 31 to 34, determine the angle of elevation/depression, rounded to the nearest degree.
31. A wheelchair ramp that rises 1 inch for every 12 inches along the ground.
32. A bicycle ramp that rises 3 feet for every 8 feet along the ground.
33. A road that descends 500 m vertically for every 6 km horizontally.
34. A ski jump ramp that descends 7 m vertically for every 10 m horizontally.
For Problems 35 to 38, determine the slope, rounded to the nearest thousandth, that corresponds to the angles of elevation/descent.
35. A snowboard ramp with a 12.5° angle of elevation.
36. A mountain railroad with a steep 16.8° angle of elevation.
37. A skiing ramp with a 12.5° angle of descent.
38. A water slide with a 75° angle of descent.
Solve the following application problems:
39. The Burj Khalifa is the tallest building in the world, soaring over Dubai at a pinnacle height of approximately 830 m.
From a point on the ground, an observer measures the angle of elevation to the aircraft beacon at the very top of the
building to be 77.5°. He then measures the angle of elevation to the observation deck to be 67.85°. Determine the
height of the observation deck, rounded to the nearest metre.
40. The distance between the CN Tower, the tallest free-standing building in Canada, and First Canadian Place, the tallest
skyscraper in Canada, is 818 m (horizontally). From the top of First Canadian Place, the angle of depression to the
bottom of the CN Tower is 20.0°, and the angle of elevation to the space-deck of the CN Tower is 10.4°.
a. Determine the height of First Canadian Place, rounded to the nearest metre.
b. Determine the height of the CN Tower space-deck, rounded to the nearest metre.
41. From the cockpit of a light aircraft 1,980 m above the ground, the angle of depression to the closer bank of a small lake
is 52.5°, and the angle of depression to the farther bank of the same lake is 31.6°. Determine the distance across the
lake, rounded to the nearest ten metres.
42. To measure the height of a hill, a surveyor records a 32.5° angle of elevation from the ground to the top of the hill. The
surveyor moves 12 m closer on the flat ground and records a 43.5° angle of elevation to the top of the hill. Determine
the height of the hill, rounded to the nearest tenth of a metre.
43. A goose is flying north at a groundspeed of 65 km/h. There is a cross wind coming from the west, blowing at a speed of
30 km/h. Determine the goose’s resulting trajectory and groundspeed.
44. A swimmer is attempting to swim across a river. She wishes to land at a point on the opposite shore, directly across from
the point she is starting. She swims at a speed of 4 km/h and the current is flowing at a speed of 2.4 km/h downstream.
a. At what angle will she need to swim upstream in order to reach her desired point on the opposite shore?
b. If the river is 400 m wide, how long will it take her to reach the other shore?
45. Determine the area of an isosceles triangle with base length of 32 cm and an opposite angle measuring 32°. Round the
answer to the nearest tenth of a cm2.
46. A segment of a circle is the area bounded between a chord and the boundary of the circle. Determine the area of a
segment bounded by a chord of length 24 cm in a circle of radius 14 cm. Round the answer to the nearest hundredth
of a cm2.
11.4 Laws of Sine and Cosine
Introduction
In the previous section, methods for solving right triangles using primary trigonometric ratios,
sine, cosine, and tangent were outlined. However, if the triangle is oblique (acute or obtuse), which
does not include a right angle, the primary trigonometric ratios do not apply.
In this section, the sine and cosine ratios will be used to develop two laws, namely the Sine Law
and the Cosine Law for solving oblique triangles. One of the benefits of these laws is that they
apply not only to oblique triangles, but also to right triangles.
As well, applications of these laws involving oblique triangles, will be discussed.
11.4 Laws of Sine and Cosine