259

5.7 | Primary Trigonometric Ratios

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.

B ΔABC, ΔADE, ΔAFG are similar;

their side lengths are in proportion.
θ
A
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 4. the cosecant (csc) ratio
2. the cosine (cos) ratio 5. the secant (sec) ratio
3. the tangent (tan) ratio 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 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 5.7-a:

Opposite O
sinθ = =
Hypotenuse H
Hypotenuse (H)
Opposite (O) Adjacent A
cosθ = =
Hypotenuse H
θ Opposite O
Adjacent (A) tanθ = =
Adjacent A
Exhibit 5.7-a Primary Trigonometric Ratios

You may find it helpful to use the acronym SOH-CAH-TOA to remember the three primary
trigonometric ratios:
SOH Sin θ = Opposite/Hypotenuse
CAH Cos θ = Adjacent/Hypotenuse
TOA Tan θ = Opposite/Adjacent

In any acute angle, Note: Since the lengths of the legs of a right triangle will be greater than 0 but always less than the
0 < sin θ < 1 hypotenuse, the sine and cosine ratios of any acute angle must be between 0 and 1. However, since
0 < cos θ < 1 there is no relationship between the opposite leg and the adjacent leg, except that they must both
tan θ > 0 be greater than 0, the tangent ratio can be any positive number.

Chapter 5 | Basic Geometry and Trigonometry

260

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
To calculate the trigonometric ratios of 30° and 60°, draw an
equilateral triangle, ABC, with side lengths of 2 units each and
2 30° 30° 2 draw AD = BC, as shown.
Since the angle measures are all 60°, the angle at the vertex A is
60° 60° bisected into two 30° angles and the base length ‘BC’ is bisected
B D C into 1 unit each.
1 1

A In the right triangle ABD (30 - 60 - 90), AB = 2 units, BD = 1 unit,

and AD = √3 units (calculated using the Pythagorean Theorem:
30° AD2 = 22 – 12 = 4 – 1 = 3, which gives AD = √3).
2
3 Using the above measures in the right triangle ABD, the sine, cosine,
and tangent ratios of 30° and 60° can be calculated exactly:
60°
B O BD 1 AD 3
1 D Using sin θ = , sin 30° = = , sin 60° = =
H AB 2 AB 2
A AD 3 , cos 60° = BD = 1
Using cos θ = , cos 30° = =
H AB 2 AB 2
O BD AD 3
Using tan θ = , tan 30° = = 1 , tan 60° = = = √3
A AD 3 BD 1

45 - 45 - 90 Triangle

A To calculate the trigonometric ratios of 45°, draw a right isosceles

triangle, ABC, with leg side lengths of 1 unit each, as shown.
45° Since the triangle is a right isosceles, the two acute angles must be
2
1 equal. Therefore, the two acute angles are each 45°.
45° In the right isosceles triangle ABC (45 - 45 - 90), AC = 1 unit,
B 1 C BC = 1 unit, and AB = √2 units (calculated using the Pythagorean
Theorem: AB2 = 12 + 12 = 1 + 1 = 2, which gives AB = √2).
Using the above measures in the right isosceles triangle ABC, the
sine, cosine, and tangent ratios of 45° can be calculated exactly:
O AC BC
Using sin θ = , sin 45° = = = 1
H AB AB 2
A BC AC
Using cos θ = , cos 45° = = = 1
H AB AB 2
O AC BC 1
Using tan θ = , tan 45° = = = =1
A BC AC 1

Chapter 5 | Basic Geometry and Trigonometry

 261

The primary trigonometric ratios of the special common angles are summarized in Exhibit 5.7-b.
1
sin 30° = cos 60° =
2
We refer to these as the special trig ratios.
3
sin 60° = cos 30° =
2
ANGLE
1 TRIG- 0° 30° 45° 60° 90°
sin 45° = cos 45° =
2 30° RATIO
1 2
tan 30° =
3
3 Sin 0
1 1 3 1
tan 45° = 11 2 2 2
tan 60° = 3 60°

1
Cos 1 3 1 1 0
45° 2 2 2
2 1
1
1
45° Tan 0 1 3 undefined
3
1

Exhibit 5.7-b Special Trig Ratios

Using Calculators to Determine Trigonometric Ratios

A scientific calculator can be used to determine the trigonometric ratios of any 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.
3
e.g., sin 60° = 0.866025... (which is equivalent to to ).
2

Example 5.7-a 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:

(i) θ = 15° (ii) θ = 72° (iii) θ = 36.87°

Solution (i) sin 15° = 0.258819... (ii) sin 72° = 0.951056... (iii) sin 36.87° = 0.600001...
= 0.2588 = 0.9511 = 0.6000
cos 15° = 0.965925... cos 72° = 0.309016... cos 36.87° = 0.799998...
= 0.9659 = 0.3090 = 0.8000
tan 15° = 0.267949.... tan 72° = 3.077683... tan 36.87° = 0.750002...
= 0.2679 = 3.0777 = 0.7500

Example 5.7-b Calculating Side Lengths Using the Sine Ratio

Calculate the unknown side lengths in the following diagrams. Round to the nearest hundredth,
wherever required.
(i) (ii)
50°
10 m x
h
25°
8m

Solution O
Using sin θ = ,
H

Chapter 5 | Basic Geometry and Trigonometry

262

h 8
(i) sin 25° = (ii) sin 50° =
10 x
h = 10 (sin 25°) 8
x=
= 4.226182... = 4.23 m sin 50°
= 10.443258... = 10.44 m

Example 5.7-c Calculating Side Lengths Using the Cosine Ratio

Calculate the unknown side lengths in the following diagrams. Round to the nearest hundredth,
wherever required.
(i) (ii)
12 m 40°
x
6m
20°
a

Solution A
Using cos θ = ,
H

a 6
(i) cos 20° = (ii) cos 40° =
12 x
a = 12 (cos 20°) 6
x=
= 11.276311... = 11.28 m cos 40°
= 7.832443... = 7.83 m

Example 5.7-d Calculating Side Lengths Using the Tangent Ratio

Calculate the unknown side lengths in the following diagrams. Round to the nearest hundredth,
wherever required.
(i) (ii)
35°
h
y
30°
7m 5m

Solution O
Using tan θ = ,
A
y 5
(i) tan 30° = (ii) tan 35° =
7 h
5
y = 7 (tan 30°) h=
tan 35°
= 4.041451... = 4.04 m
= 7.140740... = 7.14 m

Chapter 5 | Basic Geometry and Trigonometry

 263

Using Calculators to Calculate Angles

If we know the ratio of the lengths of 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 5.7-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:

(i) sin θ = 0.9063 (ii) cos θ = 0.6 (iii) tan θ = 0.1467

Solution (i) θ = sin-1(0.9063) (ii) θ = cos-1(0.6) (iii) θ = tan-1(0.1467)

θ = 64.998944... = 65.0° θ = 53.130102... = 53.1° θ = 8.345761... = 8.3°

Example 5.7-f Calculating Angles using the Inverse Trigonometric Ratios

Calculate the unknown angle in the following diagrams. Round to the nearest degree, wherever
required.

(i) (ii) 10 (iii)

5.6 p
3.2 24.5
x 20
A
9.8

Solution O A O
(i) Using sin θ = , (ii) Using cos θ = , (iii) Using tan θ = ,
H H A
3.2 10 1 24.5
sin x = cos p = = tan A =
5.6 20 2 9.8
3.2 1 24.5
x = sin–1 e o p = cos–1 e o A = tan–1 e o
5.6 2 9.8
This is one is the
x = 34.849904... = 35° p = 60° special angles,
A = 68.198590... = 68°
since the ratio
1
cos p =
2

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 lengths of the remaining sides.
Conversely, if any two side lengths of a right triangle are given, the inverse trigonometric functions
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 the fewest number of steps. Whenever possible,
measurements provided in the question, rather than measurements obtained from secondary
calculations, should be used.

Chapter 5 | Basic Geometry and Trigonometry

264

Example 5.7-g 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) (ii)
θ
θ c
8.5 cm
1.64 m
b
24°
b 72°
a

Solution (i) Calculate the length of one of the unknown sides using a primary trig ratio. Note that either
b or c can be calculated first.
O 8.5 θ c
tan 24° = = 8.5 cm
A b
24°
b
8.5 8.5
b= =
tan 24° 0.445228...
= 19.091312... ≈ 19.09 cm
Then, calculate the other unknown side length using another trig ratio. Make sure to only use
the information that was given in the original question.
O 8.5 Using the Pythagorean Theorem
sin 24° = = to calculate the length of the
H c hypotenuse, c, could have resulted
in a rounding error, or a compound
8.5 8.5 calculation error had there been
c= = an error in calculating b. It is for
sin 24° 0.406736... this reason that it is important to
use measurements provided in
the question whenever possible.
= 20.898043... ≈ 20.90 cm
Finally, since the acute angles in a right-triangle are complementary: θ = 90° – 24° = 66°
(ii) Calculate the length of one of the unknown sides using a primary trig ratio. Note that either
a or b can be calculated first.
A a
cos 72° = =
H 1.64 θ
1.64 m
b
a = 1.64(cos 72°) = 1.64(0.309016...)
= 0.506787... ≈ 0.51 m 72°
a

Then, calculate the other unknown side length using another trig ratio. Make sure to only use the
information that was given in the original question.
O b
sin 72° = =
H 1.64
b=
 1.64(sin 72°) = 1.64(0.951056...)
=1.559732... ≈ 1.56 m
Finally, since the acute angles in a right-triangle are complementary: θ = 90° − 72° = 18°

Chapter 5 | Basic Geometry and Trigonometry

 265

Example 5.7-h Solving a Right-Triangle Given Two Side Lengths

Determine the unknown side length and acute angles of the following right triangles. Round all
side lengths to the nearest hundredth and all angle measures to the nearest tenth.
(i) (ii)
x
x c
5 cm a 6m
y
12 cm y
3m

Solution (i) Calculate the length of the hypotenuse using the Pythagorean Theorem,
2 2 2
c = 5 + 12 = 25 + 144 = 169
c
c = √169 = 13 cm x
5 cm
y
12 cm
Then, solve for one of the acute angles using the inverse trig functions. Note that either x or y
can be calculated first.

Notice every calculation in As the opposite and adjacent lengths are Similarly, determining the angle y,
this example only involves provided, we use the inverse trig function
those values that were given –1
in the original question.
for tan (i.e., tan ) to determine the angle x.
12 5
tan x = tan y =
5 12
–1 12 5
x = tan e n y = tan e n
–1
5 12
= 67.380135… = 67.4° = 22.619864... = 22.6°

(ii) Calculate the length of the unknown leg using the Pythagorean Theorem,
2 2 2
a = 6 − 3 = 36 − 9 = 27
a = √27 = 5.196152... ≈ 5.20 m x
a 6m

y
3m

Then, solve for one of the acute angles using the inverse trig functions. Note that either x or y
can be calculated first.

As the hypotenuse length and the length As the hypotenuse length and the length
opposite to x are provided, we use the adjacent to y are provided, we use the inverse
inverse trig function for sin (i.e., sin–1) to trig function for cos (i.e., cos–1) to determine
determine the angle x. the angle y.
3 3
sin x = cos y =
6 6
–1 3 –1 1 –1 3 –1 1
x = sin f p = sin f p y = cos f p = cos f p
6 2 6 2
= 30° = 60°

Chapter 5 | Basic Geometry and Trigonometry

 266

Angles of Elevation and Depression

The angle of elevation is the angle above the
Object
horizontal line from the observer’s eye to an object, Lin
known as the line of sight. The angle of depression Rise
e of
sigh
t
is the angle below the horizontal line from the Angle of elevation

Observer
observer’s eye to an object (also known as the line Horizontal line
Run
of sight).
In problems involving angles of elevation and Run
Recall the definition of slope: depression, we are usually given a slope. That is, Horizontal line
Observer
m=
Rise in the right triangle formed by the horizontal line Angle of depression
Rise t
Run and the line of sight, we are given the side measure of sigh
Line
opposite to the angle of elevation/depression (rise) Object
and the side measure adjacent to the angle of
elevation/depression (run). Therefore, tangents are Exhibit 5.7-c Angles of Elevation and Depression
used to solve problems involving angles of elevation
and depression.

Example 5.7-i 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 Rise 1.2 Line of sight

(i) The slope of the ramp is m = =
Run 15 1.2 m
θ
15 m
Therefore, Angle of Elevation
O 1.2
tan θ = = Solving for θ, the angle of elevation,
A 15
1.2
θ = tan f p
–1
15
= 4.573921... = 4.6°
Therefore, the angle of elevation is 4.6°.

(ii) First, convert 8 km to 8,000 m so that the units of 8,000 m Angle of depression
measurement are the same. θ
Regardless of whether you are 500 m Line of sight
calculating an angle of elevation or
Rise 500
depression, use positive values for
Then, the slope of the road is m = =
rise and run in your calculations.
Run 8,000
Therefore,
O 500
tan θ = = Solving for θ, the angle of depression,
A 8,000
500
θ = tan f p
–1
8,000
= 3.576334... = 3.6°
Therefore, the angle of depression is 3.6°.

Chapter 5 | Basic Geometry and Trigonometry

 267

Applications of the Trigonometric Ratios

Example 5.7-j 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:
O h
Angle of elevation is Object tan 83.5° = =
the angle between the A 63
horizontal line and the
line from the observer’s h = 63(tan 83.5°)
eye to the object (when Line of sight
= 63(8.776887...)
the object is above the
level of the observer). = 552.943903...
h Angle
≈ 553 m
of elevation Therefore, the height of the tower is approximately 553 m.
83.5 °

33m 30m Observer

63m

Example 5.7-k Determining the Distance Across a Lake

From a point 520 m above a lake in the shape of an ellipse (oval), 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 metre.

Solution First draw a picture of this to better understand how to solve the question:

Angle of depression is Observer

the angle between the Angle of depression 33.5 ° 40.6 ° Angle of depression
horizontal line and the
line from the observer’s Line of sight 520 m Line of sight
eye to the object (when
the object is below the 33.5 ° 40.6 °
level of the observer). Object Object
d1 d2

Chapter 5 | Basic Geometry and Trigonometry

268

Calculating the horizontal distance from the observer to one side of the lake:
O 520
tan 33.5° = =
A d1
520 520
d1 = = = 785.634300 ≈ 785.6 m
tan 33.5° 0.661885...

Calculating the horizontal distance from the observer to the other side of the lake:
O 520
tan 40.6° = =
A d2
520 520
d2 = = = 606.694410... ≈ 606.7 m
tan 40.6° 0.857103...
Calculating the total length of the lake:
d = d1 + d2 = 785.6 + 606.7 ≈ 1,392 m
Therefore, the length of the lake is approximately 1,392 m.

Example 5.7-l Calculating the Heading and Groundspeed of a Plane

An airplane is flying at a speed 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.

Solution 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:

Calculating the angle of trajectory:

O 132 θ
sin θ = = = 0.15 880 km/h
H 880
–1
x
θ = sin (0.15) = 8.626926... ≈ 8.63°
Calculating the resultant speed, using the Pythagorean Theorem:
x2 = 8802 – 1322 = 774,400 – 17,424 = 756,976 132 km/h
x = √756,976 = 870.043677... ≈ 870 km/h

Therefore, the heading of the plane is S 8.63° E and the groundspeed of the plane is 870 km/h.

5.7 | Exercises Answers to odd-numbered problems are available at the end of the textbook.

For Problems 1 and 2, given the values of θ, determine the three primary trigonometric ratios of θ, rounded to four decimal
places.
1. θ sin θ cos θ tan θ 2. θ sin θ cos θ tan θ
a. 65° ? ? ? a. 24° ? ? ?
b. 12.5° ? ? ? b. 82.8° ? ? ?
c. 53.13° ? ? ? c. 73.74° ? ? ?

Chapter 5 | Basic Geometry and Trigonometry

 269

For Problems 3 and 4, given one trigonometric ratio, determine the corresponding angle θ, rounded to the nearest degree, and
the other two, primary trigonometric ratios, rounded to four decimal places.
3. θ sin θ cos θ tan θ 4. θ sin θ cos θ tan θ
a. ? 0.4540 ? ? a. ? 0.5591 ? ?
b. ? ? 0.2924 ? b. ? ? 0.9743 ?
c. ? ? ? 0.3639 c. ? ? ? 8.1443
For Problems 5 and 6, given one trigonometric ratio, determine the remaining trigonometric ratios exactly using the
Pythagorean Theorem, and determine the corresponding angle, rounded to the nearest degree.
5. sin θ cos θ tan θ θ 6. sin θ cos θ tan θ θ
3 5
a. ? ? ? a. ? ? ?
5 13
24 8
b. ? ? ? b. ? ? ?
25 17
20 12
c. ? ? ? c. ? ? ?
21 35

For Problems 7 to 12, use the special trig ratios in Exhibit 5.7-b to determine the exact values.
7. sin 60° ∙ cos 45° – sin 45° ∙ cos 30° 8. sin 60° ∙ tan 30° – sin 30° ∙ tan 60°
x
sin 45° . cos 45° sin 60° – sin 30° 8 cm
9. 10. y
2 tan 45° cos 60° – cos 30° 25°
20°
2 2 2
11. tan 60° – sin 60° + cos 60° 12. sin2 45° – cos2 30° + tan2 30°

For Problems 13 to 18, determine the length of the unknown side, rounded to the nearest hundredth, for the given right triangles.
13. 14. 15. 16. 17 cm b
xxx 48°
88cm 48°
48°
8cm
cm 7
25°
25°
25° yyy 1313cm
13 cm
cm z zz 2121cm
21 cm
cm
20°
20°
20° a 63°

21 cm
17. In a right triangle with angle θ = 72°
bbband a hypotenuse of 27 cm, determine the length of the side adjacent to θ.
13 cm
1717cm
17 cm
cm
18. In a right triangle with angle θ = 55° 72°72°
and 4.5tocm
4.5 cm 10 cm the length of the side
θ equal to 4.5c ccm, determine
72° the side adjacent 4.5 cm c
opposite to θ. θ
aaa 63°
63°
63° 55°
55°
27 55°
27cm
27
For Problems 19 to 24 determine the value of θ, rounded cm
cmto the nearest hundredth of a degree, for the given right triangles.
19. 20.
21 cm 21. 22.
21 cm
θ21 cm 16 cm 30
13 cm θ θ
13 cm 7 cm 15 cm
10 cm 13 cm 7 cm7 cm θ
θ
10 cm
10 cm 25 cm θ θ
θ 12 cm
θ 25 cm
25 cm
θ 12 cm
12 cm
6 cm
6 cm6 cm
23. 16 a right triangle with angle θ, a30
Incm cm
hypotenuse of 30 cm, and the side adjacent to θ equal to 29 cm, determine θ.
16 cm 30 cm
30 cm
16 cm15 cm
24. In aθ right triangle with angle θ, a θ
hypotenuse 8 cm θ to θθequal to 6 cm, determine θ.
15 cm
15 cm 29 cm of 8 cm, and the side opposite
8 cm8 cm θ
θ θ θ 29θ cm29 cm

Chapter 5 | Basic Geometry and Trigonometry

 17 cm
x
270 Φ
15 cm
For Problems 25 to 30, solve the given right triangles fully (i.e., identify all missing sides and angle measures).
12 cm
25. θ 17 cm 26. 12 cm 12 cm 27. 28.
θ θ
17 cm θ 30° 11 cm 11 cm
x 17 cm θ θ30° 30° 11 cm 11 cm11 cm 11 cm Φ 6 cm
x y x
x Φ y y x 10 cm 14 cm
x θ Φ
Φ Φ θ
15 cm θx Φ Φ
15 cm 15 cm x x
x θ x
x x
θ
Φ 6 cm θ
Φ 6 cmΦ 6 cm θ
29. 10Incm
a right triangle with angle θ 14
= 16°
cmand the x65°65°to 14ycm, determine the missing angle measure
y side adjacent to θ equal
10 cmand 10
thecm 14 cm
two missing side lengths. 16° 16°
14 cm y x x y65°
y y
16°
30. In aθright xtriangle with angle θ = 65° and the side opposite to θ equal to 29 cm, θ determine the missing angle measure
θ x x 29 θ
cm θ
θ
and the two missing side lengths.
29 cm 29 cm
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.
35. 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.

36. The distance between the CN Tower, the tallest free-standing building in Canada, and the 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 the First Canadian Place, rounded to the nearest metre.
b. Determine the height of the CN Tower space-deck, rounded to the nearest metre.

37. 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.

38. 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.

39. A goose is flying north at an air-speed 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 ground-speed.

40. 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?
41. 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.

42. 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.

Chapter 5 | Basic Geometry and Trigonometry