Download sin(A) = cos(A) = x tan(A) = csc(A) = 1 = sec(A

9.0 Trigonometry CH 13
9.1 Right Angle Triangles Trigonometric Ratios
It is observed that the sides and angles of a right angle triangle are related. By knowing
the length of two sides or an angle and a side length, the length of the remaining sides and
angles can be found. Given the following right angle triangle:
Hypotenuse
R
Y
Opposite
A
X
Adjacent
the hypotenuse is the side opposite the right angle
- the following three fundamental trig ratios can be defined
sin(A ) =
csc(A ) =
y
r
=
1
sin(A )
opp
hyp
=
hyp
opp
cos(A ) =
sec(A ) =
x
r
=
1
sec(A )
adj
hyp
=
hyp
adj
tan(A ) =
cot(A ) =
y
r
=
1
tan(A )
opp
adj
=
adj
opp
The Pythagorean Theorem permits the calculation of the length of the hypotenuse given the
lengths of the remaining sides.
r2 =
x2 + y2
r =
x2 + y2
Pythagoras and the Pythagoreans: an interesting diversion
Pythagoras was born about 570 BC on the Greek island of Samos. Due to political
differences with the Samian tyrant Polycrates, Pythagoras left his homeland and travelled
throught the eastern Mediterrean. He spent many years in Eygpt and acquired knowledge
in mathematics and mysticism. Eventually he immigrated to the southern Italian town of
Cortona. Here Pythagoras gained power through a religious and political cult. Philosophy
for the Pythagoreans was the basis of life, leading to the salvation of the soul. Purity was to
be sought by silence, self-examination, and abstention from flesh and beans. The
prohibition of meat was the result of the belief in the transmigration of the soul: a sheep may
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contain the soul of an ancestor. New initiates into the brotherhood were only permitted to
hear the voice of Pythagorus and not see him. After several years of purification they were
allowed to see him.
It was believed that an understanding of the kosmos can be achieved through reason
and observation. All phenomena in the universe could be reduced to natural numbers or
their ratios. The Pythagoreans chanted prayers based on natural numbers. They
associated one with reason, two stood for woman, three for man, five represented marriage,
since the union was composed of woman plus man. Justice is represented by four since it
was a product of equals. All even numbers were feminine and the odd numbers are
masculine.
Pythagoreans discovered the basic relation between musical harmony and
mathematics. A single stretched string produces the ground note when plucked. The notes
that sound harmonious with the ground note are produced by dividing the string into an
exact number of parts: into exactly two parts or into 3 parts or into 4 parts and so on. If the
still point on the string , a node, that does not fall on one of these exact points is plucked the
sound is discordant. As the node is shifted along the string, we recognise the notes that are
harmonious when we reach a prescribed point. When the node is at the midpoint the note is
an octive above the ground note. The next octive or fifth is heard when the node is at one
third of the way along the string. The next octive or fourth is heard when the node is on
fourth of the way. It is true that the most harmonious chords correspond to exact divisions
of the string in whole numbers. The brotherhood embraced the mystic force of whole
numbers where not only the sounds of nature but the whole universe operated with whole or
ratios of whole numbers. They believed that the orbits of heavenly bodies ( which the
Greeks pictured as carried round the earth on crystal spheres ) must be related to musical
intervals. All regularities in nature were considered musical and for them the movement of
the heavens was the music of the spheres.
By tradition we attribute the famous theorem to Pythagoras but the concept was
understood by the Babylonians, Egyptians and Chinese. However the theorem was proved
by Pythagoras. The following is a geometric proof:
( a + b )2 = c2 + 2 ab
a2 + 2 ab + b2 = c2 + 2 ab
a2 + b2 = c2
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The discovery of incommensurables ( irrational numbers ) proved to be scandalous to
the Pythagoreans. No ratio of natural numbers exists that describes the relation between
the diagonal of a square and its sides. Given a square with a length of 1, Pythagoras was
unable to solve for the length of the diagonal as a rational number. The Greek term logos
(word,speech) was used for a rational number. The newly discovered irrational was
described as alogon, this term had a double meaning -"not a ratio" and "not to be spoken".
Legend has it that the Pythagoreans tried to cover up the existence of irrationals. One
Hippasus revealed the secret and he was executed by the brotherhood.
It is claimed that Pythagoras fell in love with one of his students, Theano. They would
have lived happily however the brotherhood became aristocratic and the populace of
Crotona did not agree with their politics. The Crotonians banished the Pythagoreans.
Enemies followed Pythagoras into exile and legend has it that Pythagoras was slain in the
presence of his young wife.
Using the SHARP EL546 Calculator to evaluate angles and trig ratios
- the following are for inclass student practise ( calculator must be in degree mode )
eg) Use your calculator to evaluate and roundoff to four decimal places
a)
cos 35.50 o = 0.8141
b)
tan 89.55 0 = 127.3213
c)
csc 45 o = 1.4142
- when using the SHARP EL546 calculator, note that csc =
1 ( / ) ( sin ) 45 ( = )
1
cos 123.5
d)
sec 123.5 o =
e)
cot 85.6 o =
f)
sin −1 0.235 = arcsin 0.235 = 13.5916 o
1
tan 85.6
1
sin 1.4142
= −1.8118
= 0.0769
−1
note: sin 0.235 ! (sin 0.235 )
−1
=
1
sin 0.235
- when using the SHARP EL546 calculator
( 2ndF ) ( sin ) 0.235 ( = ) 13.5916
g)
tan −1 1.511 = arctan 1.511 = 56.5028 o
cos = 0.876
h)
= arc cos 0.876 = 28.8365 o
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csc = 1.75
i)
1 = 1.75
sin sin =
1
1.75
= arcsin
1 = 34.8499 o
1.75
( 2ndF ) ( sin ) ( 1/1.75 ) ( = ) 34.8499o
Homework pg 414 11, 17, 21, 25, 31, 41, 47, 49, 53, 57
Using Trig Ratios, Inverse trig and the Pythagorean Theorem
1. Given the following right angle triangle, find the remaining side lengths and angles.
- to minimize round off error use the values given in the problem
a) find r
B
r
A
y
50
C
3
y
tan(A ) = x
y
tan(50 o ) =
3
y = 3 tan 50 = 3.575
b) find r
cos(A) = xr
cos(50 o ) = 3r
r=
3
= 4.667
cos 50
c) find B
remember: that given any triangle the angles sum to 180o
A + B + C = 180o
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B = 180 o − (a + c )
= 180 o − (50 o + 90 o ) = 40 o
2. Find the remaining sides and angles
B
60
5
A
y
x
C
a) find A
A = 180o - ( 60o + 90o ) = 30o
b) find y
y
sin(A) = r
y
sin(30 o ) =
5
y = 5 sin(30 o ) = 2.5
c) find x
cos(A ) = xr
x
cos(30 o ) = 5
x = 5 cos(30 o ) = 4.33
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3. Find the length of the remaining side
B
5
A
4
C
- use the Pythagorean Theorem
r2 = x2 + y2
x2 = 52 − 42 = 9
x=3
4. From question 3, find angle A
y
sin(A ) = r = 4
5 = 0.8
A = arcsin(0.8 ) = 53.1 o
Homework pg 417 1, 5, 11, 15, 19, 23
Homework pg 419 1, 5, 9, 15, 23
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In-Class Practice problems
1.
Solve the right triangle ABC if A = 18.6o and c= 135.
B = 180 - ( 90 + 18.6 ) = 71.4o
B
sin 18.6 o = a
135
a = 135 sin 18.6 = 43.1 o
135
18.6
A
o
C
cos 18.6 = b
135
b = 135 cos 18.6 = 128 o
2.
Solve the right triangle ABC if B = 1.25 rad and a = 350 ( use radians )
B
A = 2 − ( + 1.25 ) = 1.89rad
1.25 rad
tan 1.25 = b
350
b = 350 tan 1.25 = 1053.3
350
A
C
cos 1.25 = 350
c
c = 350 = 1110.0
cos 1.25
3.
Solve the right triangle ABC if a = 1.48 and b = 2.25.
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c = 1.48 2 + 2.25 2 = 2.69
tan A = 1.48 = 0.6578
2.25
A = tan −1 0.6578 = 33.3 o
B = 180 − (90 + 33.3 ) = 56.7 o
Eg)
B
1
1
Find the height of a unit equilateral triangle
A
0.5 2.25 0.5
1.48
C
1 2 = 0.5 2 + h 2
h = 1 2 − 0.5 2 = 0.866
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