Download 4.1 Reviewing the Trigonometry of Right Triangles

4.1 Reviewing the Trigonometry of Right Triangles
I NVESTIGATE & I NQUIRE
In the short story The Musgrave Ritual, Sherlock
Holmes found the solution to a mystery at a
certain point. To find the point, he had to start
near the stump of an elm tree and take 20 paces
north, then 10 paces east, then 4 paces south,
and finally 2 paces west.
Let E be the point where Holmes started
pacing and S be the point where he stopped.
Draw a diagram of his path. Join ES. Draw
another line segment so that ES is the
hypotenuse of a right triangle.
1.
What are the lengths of the perpendicular
sides of the right triangle?
2.
What trigonometric ratio can you use to find
∠E from the lengths of the perpendicular sides?
3.
4.
Find ∠E, to the nearest degree.
What methods could you use to find the
length of ES, in paces?
5.
6.
What is the length of ES, to the nearest pace?
7. In what direction, and for how many paces,
could Holmes have walked in order to go
directly from E to S?
After Holmes arrived at S, he looked back at
the location of point E on the ground. What
angle did his line of sight make with the ground,
to the nearest degree? Assume that the ground
was level and that the height of his eyes above
the ground was approximately equal to the
length of 2 paces.
8.
266 MHR • Chapter 4
The primary trigonometric ratios are
opposite
sine θ = hypotenuse
hypotenuse
opposite
adjacent
cosine θ = hypotenuse
θ
opposite
tangent θ = adjacent
adjacent
To solve a right triangle means to find all the unknown sides and the
unknown angles.
EXAMPLE 1 Solving a Right Triangle, Given a Side and an Angle
In ABC, ∠B = 90°, ∠A = 18.6°, and b = 11.3 cm. Solve the triangle
by finding
a) the unknown angle
b) the unknown sides, to the nearest tenth of a centimetre
SOLUTION
Using the given information,
∠C = 90° − 18.6°
= 71.4°
b) From the diagram,
a
= sin 18.6°
11.3
a = 11.3 × sin 18.6°
=⋅ 3.6
c
= cos 18.6°
11.3
c = 11.3 × cos 18.6°
=⋅ 10.7
a)
C
A
b =11.3 cm
a
18.6°
c
B
Using the mode settings, ensure that
the calculator is in degree mode.
In ABC, ∠C = 71.4°, a = 3.6 cm, and c = 10.7 cm.
4.1 Reviewing the Trigonometry of Right Triangles • MHR 267
EXAMPLE 2 Solving a Right Triangle, Given Two Sides
In DEF, ∠E = 90°, d = 7.4 m, and f = 6.5 m. Solve the triangle by
finding
a) the unknown angles, to the nearest tenth of a degree
b) the unknown side, to the nearest tenth of a metre
SOLUTION
From the diagram,
7.4
tan D = 6.5
⋅
∠D = 48.7°
∠F = 90° − 48.7°
= 41.3°
b) From the diagram,
7.4
sin 48.7° = e
e × sin 48.7° = 7.4
7.4
e=
sin 48.7°
⋅= 9.9
a)
F
e
D
f = 6.5 m
d = 7.4 m
E
In DEF, ∠D = 48.7°, ∠F = 41.3°, and e = 9.9 m.
If you are standing on a cliff beside a river,
and you look down at a boat, the angle that
your line of sight makes with the horizontal
is called the angle of depression. If you look
up at a helicopter, the angle that your line
of sight makes with the horizontal is called
the angle of elevation.
268 MHR • Chapter 4
Angle of elevation
Angle of depression
Horizontal
EXAMPLE 3 Western Red Cedars
Cathedral Grove, on Vancouver Island, is a rain forest of firs and western red
cedars. From a point 40 m from the foot of one cedar, the angle of elevation
of the top is 65°. Find the height of the cedar, to the nearest metre.
SOLUTION
Draw and label a diagram.
Let h represent the height of the cedar.
h
= tan 65°
40
h = 40tan 65°
=⋅ 86
h
The cedar is 86 m tall, to the nearest metre.
65°
40 m
On the Earth, a parallel of latitude is a circle parallel
to the equator.
EXAMPLE 4 Parallel of Latitude
Find the length of the 35° parallel of latitude, to the
nearest 10 km. Assume that the radius of the Earth is
6380 km.
35° parallel
35°
6380 km B
SOLUTION
In the diagram, B is the centre of the Earth, and A is a point
on the equator. D is the centre of the circle defined by the
35° parallel, and E is a point on its circumference. DE is the
radius, r, of the 35° parallel.
E
6380 km
A
r
35°
D
35°
6380 km B
∠BDE is a right angle.
BA = BE, because both are radii of the Earth.
∠DEB = ∠ABE (alternate angles)
4.1 Reviewing the Trigonometry of Right Triangles • MHR 269
In DEB,
r
= cos 35°
6380
r = 6380cos 35°
=⋅ 5226
The length of the 35° parallel of latitude is its circumference, C.
C = 2πr
Estimate
=⋅ 2π(5226)
2 × 3 × 5000 = 30 000
≈ 32 840
To copy the previous answer to
the cursor location, enter Ans by
pressing 2nd (–).
The length of the 35° parallel of latitude is 32 840 km, to the nearest 10 km.
EXAMPLE 5 Rock Pillars
Rock pillars are interesting geological features found
in several national parks in Ontario. Rock pillars,
found in rivers and lakes, have been sculpted by
wind and water. A geologist wanted to determine
the height of a rock pillar in a river. The geologist
set up a theodolite at C and measured ∠ACB to be
28.5°. A baseline CD was marked off, perpendicular
to BC. The length of CD is 10 m, and
∠CDB = 56.4°. If the height of the theodolite is
1.6 m, what is the height of the rock pillar, to the
nearest tenth of a metre?
A
B
SOLUTION
Calculate the length of AB, and then add the height
of the theodolite to determine the height of the rock pillar.
In BCD,
In ABC,
BC
AB
= tan ∠BCD
= tan ∠ACB
CD
BC
BC
AB
= tan 56.4°
= tan 28.5°
10
15.1
BC = 10 tan 56.4°
AB = 15.1tan 28.5°
BC =⋅ 15.1
AB =⋅ 8.2
8.2 + 1.6 = 9.8
So, the height of the rock pillar is 9.8 m, to the nearest tenth of a metre.
270 MHR • Chapter 4
28
.5
°
C
56.4°
D
Key
Concepts
• For any acute angle θ in a right triangle,
opposite
adjacent
sin θ = cos θ = hypotenuse
hypotenuse
opposite
tan θ = adjacent
hypotenuse
opposite
θ
adjacent
• To use trigonometry to solve a right triangle, given the measure of one acute
angle and the length of one side, find
a) the measure of the third angle using the angle sum in the triangle
b) the measure of the other two sides using the sine, cosine, or tangent ratios
• To use trigonometry to solve a right triangle, given the lengths of two sides,
find
a) the measure of one angle using its sine, cosine, or tangent ratio
b) the measure of the third angle using the angle sum in the triangle
c) the measure of the third side using a sine, cosine, or tangent ratio
• To use trigonometry to solve a problem involving two right triangles,
a) use a diagram showing the given information and the unknown side
length(s) or angle measure(s)
b) identify the two triangles that can be used to solve the problem, and plan
how to use each triangle
c) carry out the plan
Communicate
Yo u r
Understanding
Describe how you would solve ABC, given ∠B = 90°, ∠A = 36°, and
c = 12 cm.
2. Describe how you would solve RST, given ∠S = 90°, s = 22 cm, and
t = 15 cm.
3. Describe the difference between an angle of elevation and an angle of
depression.
D
4. Describe how you would find the
measure of ∠A.
1.
49°
A
8.5 m
6.5 m
C
B
4.1 Reviewing the Trigonometry of Right Triangles • MHR 271
Practise
A
1. Solve each triangle. Round each side
length to the nearest unit and each angle to
the nearest degree.
a)
c = 56 cm
B
b) E
A
f = 60 m
41°
D
b
e
d
33°
a
C
F
c)
d)
U
s = 15 m
t = 10 m
u
S
Q
r = 13 cm
p = 8 cm
P
Find the measure of ∠θ, to the nearest
tenth of a degree.
4.
q
R
T
Solve each triangle. Round answers to the
nearest tenth, if necessary.
a) In XYZ, ∠X = 90°, x = 9.5 cm,
z = 4.2 cm
b) In KLM, ∠M = 90°, ∠K = 37°,
m = 12.3 cm
c) In ABC, ∠A = 90°, ∠B = 55.1°,
b = 4.8 m
d) In DEF, ∠E = 90°, d = 18.2 cm,
f = 14.9 cm
3.
a)
A
Solve each triangle. Round each side
length to the nearest tenth of a unit and each
angle to the nearest tenth of a degree.
2.
y
91.7°
C
B 19.4 cm
D
x
X
b)
63.5°
9.6 cm
b)
θ
28.8 cm
W
a)
E
Y
A
5 cm D 7 cm
C
L
18 cm
n
57.4°
θ
M
m
20.3 m
N
B
c)
G
d)
c)
K
R
θ
i
H
72.3 cm
j
24.5 m
12.8 m
I
272 MHR • Chapter 4
S
J
68.8 cm
L
39.4°
33.5 m
T 27.2 m U
D
d)
Find RT, to the
nearest centimetre.
8.
θ
E
R
38°
14.6 m
11.7 m
F
43 cm
S
52.3°
U
G
Find AB, to the nearest
tenth of a metre.
5.
52°
A
T
53.5°
D
B
38.7°
4.6 m
Find MN, to the
nearest tenth of a
centimetre.
9.
C
N
M
Find RS, to the nearest tenth of a
P
centimetre.
22.7°
6.
L
57.8°
Q 9.8 cm R
7.
28°
Find AB, to
the nearest metre.
10.
29.6°
S
K
25.1°
5.8 cm
D
44°
48 m
Find FH, to the nearest tenth of a metre.
D
10.4 m
F
A
G
B
C
81.4°
8.7 m
11.5 m
E
H
Apply, Solve, Communicate
The highest dam in Canada is the Mica Dam, one of three
dams on the Columbia River in British Columbia. From a point 600 m from
the foot of the dam, the angle of elevation of the top of the dam is 22°. What
is the height of the dam, to the nearest metre?
11. Mica Dam
12. Arctic Circle Find the length of the Arctic Circle, which is 66.55° north,
to the nearest 10 km. Assume that the radius of the Earth is 6380 km.
4.1 Reviewing the Trigonometry of Right Triangles • MHR 273
B
A
A surveyor measured the
height of a vertical rock face by determining
the measurements shown. If the surveyor’s
theodolite had a height of 1.7 m, find the
height of the rock face, AB, to the nearest
tenth of a metre.
13. Surveying
2°
60.
B
C
44.7°
28
.3
m
D
Find the area
of DEF, to the nearest tenth of a
square metre.
b) What other minimum sets of
conditions (sides, angles) would allow
you to calculate the area of DEF?
14. Communication a)
D
14.8 m
E
57.4°
F
Find the length of the 20° parallel of latitude, to the nearest
10 km. Assume that the radius of the Earth is 6380 km.
b) Find the length of the parallel of latitude where you live, to the nearest 10 km.
15. Latitude a)
From 1857 to 1860, Great Britain financed the construction
of ten lighthouses in British North America. They were built because obsolete
navigational aids were hindering economic growth. The lighthouses are
called the Imperial Lights. Four of them were built along the approaches to
the Saint Lawrence, and six were built on the eastern shore of Lake Huron.
The Point Clark lighthouse, on Lake Huron, is 28.3 m tall. From the top of the
lighthouse, the angle of depression of a ship is 3.3°. How far is the ship from
the lighthouse, to the nearest metre?
16. Application
17. Inquiry/Problem Solving Show that the length of any parallel of latitude
is equal to the length of the equator times the cosine of the latitude angle.
18. Measurement Find the volume of the triangular
prism, to the nearest cubic centimetre.
56 cm
52°
44 cm
274 MHR • Chapter 4
C
The Great Pyramid of Khufu has a
square base with a side length of about 230 m. The four
triangular faces of the pyramid are congruent and
isosceles. The altitude of each triangular face makes an
angle of 52° with the base. Find the measure of each base
angle of the triangular faces, to the nearest degree.
19. Great Pyramid
52°
230 m
base angle
ABC is an acute triangle. Show that the area, A,
A
of ABC can be found using the formula A = 0.5acsin B.
c
b) Show that the area, A, of ABC can also be found using the
formula A = 0.5absin C.
B
c) Find the other formula, similar to those in parts a) and b),
a
for the area of ABC.
d) Describe the pattern in the formulas.
e) What given information is sufficient to find the area of an acute triangle?
Explain.
X
f) Find the area of the triangle to the right, to the nearest tenth of a
square metre.
20. Geometry a)
b
C
8.5 m
63.4°
7.6 m
Z
Y
Find the area of the triangle to the right, to the nearest tenth of
a square centimetre.
h) Do the formulas from parts a), b), and c) apply to obtuse
triangles? Explain and justify your reasoning using diagrams.
g)
D
72.5° 56.8°
19.5 cm
E
22.3 cm
F
A C H I E V E M E N T Check
Knowledge/Understanding Thinking/Inquiry/Problem Solving Communication Application
On a wall, a spider is 100 cm above a fly. The fly starts moving horizontally
at the speed of 10 cm/s. After 1 s, the spider begins moving at twice the
speed of the fly, in such a way as to intercept the fly by taking a straight line
path. In what direction does the spider move, and how far has the fly moved
when they meet?
4.1 Reviewing the Trigonometry of Right Triangles • MHR 275