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Trigonometry
Trigonometry is concerned with the relationship between the angles
and sides of triangles. An understanding of these relationships enables
unknown angles and sides to be calculated without recourse to direct
measurement. Applications include finding heights/distances of objects.
Discuss how the following
sequence of diagrams allows
us to determine the height
of the Eiffel Tower without
actually having to climb it.
?
Trigonometry
Trigonometry is concerned with the relationship between the angles
and sides of triangles. An understanding of these relationships enables
unknown angles and sides to be calculated without recourse to direct
measurement. Applications include finding heights/distances of objects.
Discuss how the following
sequence of diagrams allows
us to determine the height
of the Eiffel Tower without
actually having to climb it.
30o
Trigonometry
Trigonometry is concerned with the relationship between the angles
and sides of triangles. An understanding of these relationships enables
unknown angles and sides to be calculated without recourse to direct
measurement. Applications include finding heights/distances of objects.
Discuss how the following
sequence of diagrams allows
us to determine the height
of the Eiffel Tower without
actually having to climb it.
35o
Trigonometry
Trigonometry is concerned with the relationship between the angles
and sides of triangles. An understanding of these relationships enables
unknown angles and sides to be calculated without recourse to direct
measurement. Applications include finding heights/distances of objects.
Discuss how the following
sequence of diagrams allows
us to determine the height
of the Eiffel Tower without
actually having to climb it.
40o
Trigonometry
Trigonometry is concerned with the relationship between the angles
and sides of triangles. An understanding of these relationships enables
unknown angles and sides to be calculated without recourse to direct
measurement. Applications include finding heights/distances of objects.
Discuss how the following
sequence of diagrams allows
us to determine the height
of the Eiffel Tower without
actually having to climb it.
?
What’s he going
to do next?
45o
Trigonometry
Trigonometry is concerned with the relationship between the angles
and sides of triangles. An understanding of these relationships enables
unknown angles and sides to be calculated without recourse to direct
measurement. Applications include finding heights/distances of objects.
Discuss how the following
sequence of diagrams allows
us to determine the height
of the Eiffel Tower without
actually having to climb it.
?
What’s he going
to do next?
45o
324 m
Trigonometry
Trigonometry is concerned with the relationship between the angles
and sides of triangles. An understanding of these relationships enables
unknown angles and sides to be calculated without recourse to direct
measurement. Applications include finding heights/distances of objects.
324 m
45o
324 m
Trigonometry
Eiffel Tower Facts:
•Designed by Gustave Eiffel.
•Completed in 1889 to celebrate the centenary of
the French Revolution.
•Intended to have been dismantled after the 1900
Paris Expo.
•Took 26 months to build.
•The structure is very light and only weighs 7 300
tonnes.
•18 000 pieces, 2½ million rivets.
•1665 steps.
•Some tricky equations had to be solved for its
design. 1 H 2
H
2 x
f (x )  cons tantx (H  x ) 

x
xw (x )f (x )dx
324 m
Early
Beginnings
Trigonometry (Three-angle-measure)
The Great Pyramid (Cheops) at Giza, near Cairo, one of the 7 wonders
of the ancient word. (The only one still surviving).
Trigonometry
Thales of Miletus
640 – 546 B.C. The
first Greek
Mathematician. He
predicted the Solar
Eclipse of 585 BC.
An early application of trigonometry was made by Thales on a
visit to Egypt. He was surprised that no one could tell him the
height of the 2000 year old Cheops pyramid. He used his
knowledge of the relationship between the heights of objects
and the length of their shadows to calculate the height for
them. (This will later become the Tangent ratio.) Can you see what
this relationship is, based on the drawings below?
h
480 ft
720 ft
Similar
Triangles
Sun’s rays casting shadows mid-afternoon
Similar 6 ft
Triangles
9 ft
Sun’s rays casting shadows late afternoon
Thales may not have used similar triangles directly to solve the problem but
h that6the ratio of the6vertical
x 720to horizontal sides of each triangle was
he knew

 hfor different heights
 480
ft(Egyptian
course)
constant and unchanging
of the
sun. Can feet
you of
use
the
720
9
9
measurements shown above to find the height of Cheops?
Trigonometry
Thales of Miletus
640 – 546 B.C. The
first Mathematician.
He predicted the
Solar eclipse of 585
BC.
An early application of trigonometry was made by Thales, on a
visit to Egypt. He was surprised that no one could tell him the
height of the 2000 year old Cheops pyramid. He used his
knowledge of the relationship between the heights of objects
and the length of their shadows to calculate the height for
them. (This will later become the Tangent ratio.) Can you see what
this relationship is, based on the drawings below?
h
480 ft
720 ft
Similar
Triangles
Sun’s rays casting shadows mid-afternoon
Similar 6 ft
Triangles
9 ft
Sun’s rays casting shadows late afternoon
This relationship is a consequence of the triangles being similar.
h
6
h 720
6x 720


Since 
h
720 9
6
9
9
Later, during the Golden Age of Athens (5C BC.), the philosophers and
mathematicians were not particularly interested in the practical side of
mathematics so trigonometry was not further developed. It was another 250
years or so, when the centre of learning had switched to Alexandria (current
day Egypt) that the ideas behind trigonometry were more fully explored. The
astronomer and mathematician, Hipparchus was the first person to construct
tables of trigonometric ratios. Amongst his many notable achievements was
his determination of the distance to the moon with an error of only 5%. He
used the diameter of the Earth (previously calculated by Eratosthenes)
together with angular measurements that had been taken during the total
solar eclipse of March 190 BC.
Eratosthenes Hipparchus of Rhodes
190-120 BC
275 – 194 BC
The library of Alexandria was the
foremost seat of learning in the world and
functioned like a university. The library
contained 600 000 manuscripts.
Early Applications of Trigonometry
Finding the height of a
mountain/hill.
h
x
Constructing sundials to
estimate the time from
the sun’s shadow.
25o
20o
d
Finding the distance to
the moon.
Historically trigonometry was developed for work in
Astronomy and Geography. Today it is used
extensively in mathematics and many other areas of
the sciences.
•Surveying
•Navigation
•Physics
•Engineering
Trigonometry
The ideas behind trigonometry are based firmly on the previous
work on similar triangles. In particular we are interested in similar
right-angled triangles.
Explain why triangles A and B are similar.
Explain why triangles C and D are similar.
A
C A
B
50o
50o
40o
6 cm
D
10 cm
3 cm
40o
8 cm
5 cm
4 cm
Because they are equiangular.
Because corresponding sides are in
proportion: 5/10 = 4/8 = 3/6 = ½
This means of course that A is
an enlargement of B
C is enlargement of D by scale
factor x 2
Trigonometry
C A
6 cm
D
Corresponding sides are in proportion
10 cm
5 cm
3 cm
8 cm
5/10 = 4/8 = 3/6 = ½
4 cm
Compare the ratio of any two sides in triangle C to the corresponding
pair in triangle D. What do you notice?
C A
6 cm
The ratio of any two sides in one
triangle is equal to the ratio of the
corresponding pair in the other.
D
10 cm
3 cm
8 cm
5 cm
4 cm
6/10 = 3/5 (= 0.6)
6/8 = 3/4 (= 0 .75)
8/10 = 4/5 (= 0.8)
This relationship is always true for similar right-angled triangles.
In similar right-angled triangles the ratios of any pair of sides in one
triangle is equal to the ratio of the corresponding pair in all others. It
is this idea that forms the basis for trigonometry.
C A
D
6/10 = 3/5 (= 0.6)
10 cm
6 cm
3 cm
8 cm
6/8 = 3/4 (= 0 .75)
5 cm
8/10 = 4/5 (= 0.8)
4 cm
Convention for labelling triangles.
•Angles denoted by CAPITAL letters.
A
b
C
c
a
Side a opposite angle A
•Sides opposite a given angle use the
same letter but in lower case.
B
Side b opposite angle B
Side c opposite angle C
In similar right-angled triangles the ratios of any pair of sides in one
triangle is equal to the ratio of the corresponding pair in all others. It
is this idea forms the basis for trigonometry.
C A
6 cm
D
6/10 = 3/5 (= 0.6)
10 cm
5 cm
3 cm
8 cm
6/8 = 3/4 (= 0 .75)
8/10 = 4/5 (= 0.8)
4 cm
Convention for labelling triangles.
P
•Sides opposite a given angle use the
same letter but in lower case.
r
q
R
•Angles denoted by CAPITAL letters.
p
Side p opposite angle P
Q
Side q opposite angle Q
Side r opposite angle R
In similar right-angled triangles the ratios of any pair of sides in one
triangle is equal to the ratio of the corresponding pair in all others. It
is this idea forms the basis for trigonometry.
C A
D
6/10 = 3/5 (= 0.6)
10 cm
6 cm
3 cm
8 cm
6/8 = 3/4 (= 0 .75)
5 cm
8/10 = 4/5 (= 0.8)
4 cm
Convention for naming sides.
•The side opposite the right-angle is
called the hypotenuse.
adjacent
A
C
65o
•The side opposite a given angle is
called the opposite side.
hypotenuse
opposite
B
•The side next to (or adjacent to) a
given angle is called the adjacent side.
In similar right-angled triangles the ratios of any pair of sides in one
triangle is equal to the ratio of the corresponding pair in all others. It
is this idea forms the basis for trigonometry.
C A
6 cm
D
6/10 = 3/5 (= 0.6)
10 cm
3 cm
8 cm
6/8 = 3/4 (= 0 .75)
5 cm
8/10 = 4/5 (= 0.8)
4 cm
Convention for naming sides.
•The side opposite the right-angle is
called the hypotenuse
opposite
A
C
hypotenuse
25o
adjacent
•The side opposite a given angle is
called the opposite side.
B
•The sides next to (or adjacent to) a
given angle is called adjacent side.
In similar right-angled triangles the ratios of any pair of sides in one
triangle is equal to the ratio of the corresponding pair in all others. It
is this idea forms the basis for trigonometry.
C A
D
6/10 = 3/5 (= 0.6)
10 cm
6 cm
3 cm
8 cm
5 cm
6/8 = 3/4 (= 0 .75)
8/10 = 4/5 (= 0.8)
4 cm
Convention for naming sides.
A
adjacent
C
38o
hypotenuse
opposite
B
•The side opposite the right-angle is
called the hypotenuse
•The side opposite a given angle is
called the opposite side.
•The sides next to (or adjacent to) a
given angle is called adjacent side.
The Trigonometric Ratios
A
adjacent
C
B
hypotenuse
Sine A 
B
opposite
opposite
hypotenuse
C
Opposite
Hypotenuse
SinA 
O
H
Adjacent
Hypotenuse
CosA 
A
H
Opposite
Adjacent
TanA 
O
A
Cosine A 
Tangent A 
adjacent
Make up a Mnemonic!
A
S O H C A H T O A
Make up a Mnemonic!
S O H C A H T O A
U A
W P
T I
R R
E S
N P
R
L
Y
E
Harry
The Trigonometric Ratios (Finding an unknown side).
SinA 
O
H
30o
CosA 
A
H
30o
TanA 
O
A
30o
Use your worksheet and a ruler to measure as accurately as
possible the sides of each triangle above. Find approximate
values for the sine, cosine and tangent ratios for an angle of 30o.
True Values (2 dp)
Sin 30o = 0.50
Cos 30o = 0.87
Tan 30o = 0.58
To Trig Tables >
The fixed values of these ratios were
calculated for every angle and stored in a
table of sines, cosines and tangents.
Nowadays a calculator computes each value
instead. Because these values are constant
we can use them to find unknown sides.
The Trigonometric Ratios (Finding an unknown side).
SinA 
O
H
30o
True Values (2 dp)
Sin 30o = 0.50
Cos 30o = 0.87
CosA 
A
H
TanA 
30o
O
A
30o
For example, anytime we come across a
right-angled triangle containing an angle
of 30o we can find an unknown side if we
are given the value of one other.
Tan 30o = 0.58
Tan 300 
h
75
 h  75Tan 300
 h  75x 0.58  43.5m
30o
75 m
h
The Trigonometric Ratios (Finding an unknown side).
Example 1. In triangle ABC find side CB. S O H C A H T O A
A
CB
Diagrams
Sin 700 
70o
12 cm
12
not to
scale.
12Sin 700  CB  11.3 cm (1dp )
C
B
Opp
Example 2. In triangle PQR find side PQ. S O H C A H T O A
P
7.2
7.2
Cos 220 
 PQ 
PQ
Cos 220
22o
Q
PQ  7.8 cm (1dp )
R
7.2 cm
Example 3. In triangle LMN find side MN. S O H C A H T O A
L
4.3 m
4.3
4.3
M
 MN 
Tan 750 
Tan 750
MN
75o
MN  1.2 m (1dp )
N
The Trigonometric Ratios (Finding an unknown angle).
True Values (2 dp)
Sin 30o = 0.50
Cos 30o = 0.87
Tan 30o = 0.58
Anytime we come across a right-angled
triangle containing 2 given sides we can
calculate the ratio of the sides then
look up (or calculate) the angle that
corresponds to this ratio.
S O H C A H T O A
Tanx 0 
xoo
30
75 m
43.5
 0.58
75
43.5 m
The Trigonometric Ratios (Finding an unknown angle).
Example 1. In triangle ABC find angle A. S O H C A H T O A
A
12 cm
C
11.3 cm
11.3
Sin A 
12
Key Sequence
Sin-1(11.3  12) =
0

Angle
A

70
(nearest deg ree )
B
Example 2. In triangle LMN find angle N. S O H C A H T O A
L
4.3 m
Key Sequence
M
4.3
Tan N 
Tan-1(4.3  1.2) =
1.2
1.2 m
Diagrams not
o

Angle
N

7
4
(nearest degree)
N
to scale.
Example 3. In triangle PQR find angle Q. S O H C A H T O A
P
7.8 cm
Key Sequence
7.2
Cos Q 
-1(7.2  7.8) =
Cos
7.8
Q
R
7.2 cm
 Angle Q  23o (nearest degree)
Applications of Trigonometry
A boat sails due East from a Harbour (H), to a marker buoy (B), 15 miles away.
At B the boat turns due South and sails for 6.4 miles to a Lighthouse (L). It then
returns to harbour. Make a sketch of the trip and calculate the bearing of the
harbour from the lighthouse to the nearest degree.
H
15 miles
B
15
Tan L 
6.4
 Angle L  66.90
6.4 miles
Bearing  360  66.9  293o
L
SOH CAH TOA
Applications of Trigonometry
A 12 ft ladder rests against the side of a house. The top of
the ladder is 9.5 ft from the floor. Calculate the angle that
the foot of ladder makes with the ground.
9.5
Sin L 
12
o
 Angle L  52
12 ft
9.5 ft
Lo
SOH CAH TOA
Applications of Trigonometry
An AWACS aircraft takes off from RAF
Waddington (W) on a navigation
exercise. It flies 430 miles North to a
point P before turning left and flying
for 570 miles to a second point Q,
West of W. It then returns to base.
Not to Scale
P
(a) Make a sketch of the flight.
(b) Find the bearing of Q from P.
570 miles
430
Cos P 
570
430 miles
 Angle P  41o
Bearing  180  41  221
0
Q
W
SOH CAH TOA
Angles of Elevation and Depression.
An angle of elevation is the angle measured upwards from a
horizontal to a fixed point. The angle of depression is the angle
measured downwards from a horizontal to a fixed point.
Horizontal
Angle of depression
Explain why the angles of
elevation and depression are
always equal.
25o
Angle of elevation
Horizontal
25o
Applications of Trigonometry
A man stands at a point P, 45 m from the base of a building
that is 20 m high. Find the angle of elevation of the top of the
building from the man.
Tan P 
20
45
Angle P  240 (nearest deg ree )
20 m
45 m
P
SOH CAH TOA
A 25 m tall lighthouse sits on a cliff top, 30 m above sea level. A fishing
boat is seen 100m from the base of the cliff, (vertically below the
lighthouse). Find the angle of depression from the top of the lighthouse to
the boat.
100
Tan C 
55
Angle C  61.2o
Angle D  90  61.20  290 (nearest deg ree )
C
D
55 m
100 m
D
Or more directly since the angles of elevation
and depression are equal.
SOH CAH TOA
Tan D 
55
 Angle D  29o
100
A 22 m tall lighthouse sits on a cliff top, 35 m above sea level. The angle
of depression of a fishing boat is measured from the top of the lighthouse
as 30o. How far is the fishing boat from the base of the cliff?
x
Tan 60 
57
x  57Tan 60
=99m (nearest m)
30o
60o
57 m
30o
xm
SOH CAH TOA
Or more directly since the angles of elevation
and depression are equal. Tan 30  57
x
x 
57
 99m
Tan 30
The origins of trigonometry are closely tied up with problems involving
circles. One particular problem is that of finding the lengths of
chords subtended by different angles at the centre of a circle. The
Arabs called the half chord (“ardha-jya”). This became misinterpreted and mis-translated over the centuries and eventually
ended up as “sinus” in Latin, meaning cove or bay. Other derivations
include: bulge, bosom, sinus, cavity, nose and skull. The cosinus simply
means the compliment of the sinus, since SinA = Cos (90 – A)
(Sin 60 = Cos 30, Sin 70 = Cos 20 etc)
Angular Bisector
O
M
Half Chord PM (Sinus)
Chord
P
OM (Cosinus)
Q
The following diagrams show the relationships between the 3 trigonometric
ratios for a circle of radius 1 unit. Tangent means “To touch”
P
Half Chord PM
Sin  = O/H = PM/1 = PM
Cos  = A/H = OM/1 = OM
O
o
M
Tan  = PT/1 = PT
Chord
Angular Bisector
P
P
O
O
P
1
1


M
M

1
O
O
T
o
Sinus
Q
Cosinus M
To show that Tan =
Sin
(O / A)
Cos 
Since triangles OMP and OPT are similar we have Tan =
1
P
T
PT PM Sin
=
=
1
OM Cos 
The Trigonometric Ratios (Finding an unknown side).
SinA 
O
H
30o
CosA 
A
H
30o
TanA 
O
A
30o
Use your worksheet and a ruler to measure as accurately as
possible the sides of each triangle above. Find approximate
values for the sin, cos and tan ratios for an angle of 30o.
Estimated Values
Sin 30o =
Cos 30o =
Tan 30o =
Worksheet 1
Trig Tables
Click to
go back
The Trigonometric Ratios (Finding an unknown side).
Example 1. In triangle ABC find side CB.
A
70o
12 cm
C
B
Opp
Example 2. In triangle PQR find side PQ.
P
R
22o
Q
7.2 cm
Example 3. In triangle LMN find side MN.
L
4.3 m
M
75o
N
Worksheet 2
The Trigonometric Ratios (Finding an unknown angle).
Example 1. In triangle ABC find angle A.
A
12 cm
C
11.3 cm
B
Example 2. In triangle LMN find angle N.
L
4.3 m
M
1.2 m
N
Example 3. In triangle PQR find angle Q.
P
7.8 cm
R
7.2 cm
Q
Worksheet 3