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COURSE OUTLINE:
I. ARITHMETIC
• Basic Operation Involving Whole Numbers• Basic Operation Involving Fractions
• Basic Operation Involving Decimals
• Ratio, Proportion and Percent
• Powers and Roots
II. SYSTEMS OF MEASUREMENT
• Metric System
• Unit Conversion and Conversion Factors
III. GEOMETRICAL APPLICATION
• Points, Lines and Planes
• Angles and Triangles
• Polygons and Circles
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IV. BASIC MENSURATION
• Areas and Volumes
V. LINES AND THE CARTESIAN PLANE
VI. BASIC TRIGONOMETRICAL FUNCTION
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Operations with Whole Numbers
Addition represents the idea of finding a total count, or
summing up, of values. Since we use only ten digits in our
system (remember base 10), it is often necessary to use
place value to “carry” digits.
Properties of Addition
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EXERCISES:
I. Perform the following additions and subtractions.
1. 2,456 + 8,946 2. 534 – 276
II. Perform the following multiplications.
1. 1,859 • 68 2. 2,695 • 465
III. Perform the following divisions.
1. 14,846 ÷ 124 2. 33,429 ÷ 132
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TYPES OF FRACTIONS
1. Proper Fractions – are fractions in which the numerator is
smaller than or less than ( < ) the denominator.
Example:
2
3
,4
5
,3
8
,7
12
2. Improper Fractions – are fractions in which the
numerator is larger than or greater than ( > ) the
denominator. ( The value of an improper fraction is
always greater than one.)
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Example:
5
3
,7
6
,5
4
,12
7
3. Mixed Fractions
- A simplified improper fraction.- Numbers used to express amounts that are
greater than a whole and have a fractional part.
- A combination of whole number and a proper
fraction.Example:
7
1= ;
1
61
3
54
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Simplification of Fractions:
Rule: To reduce fraction to its simplest form, divide both the
numerator and denominator by a common number. If both
numerator and denominator are no longer divisible by a
common number except one ( 1 ), then it is in simplest form.
Example:
6
15=
2
5
( divide both numerator and
denominator by 3 )
Least Common Denominator:
1. In addition and subtraction of fractions, the numerator are
added or subtracted directly for common denominator.
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Example:
1
3+
1
3+
2
3= =
1 + 1 + 2
3
4
3or
1
31
1
5 –
2
5+
4
5=
1 – 2 + 4
5=
3
5
Note: The numerator of fractions having the same
denominator can be added or subtracted directly.
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2. In addition or subtraction of fractions whose denominatorare not the same, we find the least common denominator.
LCD – the lowest number that is exactly divisible by all the
denominators of the fractions to be added or subtracted.
To find the least common denominator ( LCD ), factor each
denominator of the given fractions and multiply all factorsto the highest exponent as it appear in any of the
denominators.
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Example:
1
5+ 1
10+ 1
25
Denominators: 5 = 5 x 1
10 = 5 x 2 x 1
25 = ( 5 )2 x 1
The factors that are present ( not necessarily common ) are 5,
2 & 1
The highest exponent of 5 is square ( 2 )
The highest exponent of 2 is one ( 1 )
The highest exponent of 1 is one ( 1 )
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Therefore the LCD is ( 52 x 2 x 1 ) = 50
1
5+
1
10+
1
25
=10 + 5 + 2
50
17
50
Example:
1
2+ 1
3 – 1
9
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2 = 2 * 1
3 = 3 * 1
9 = 32 * 1
LCD = 2 * 32 * 1 = 18
9 + 6 – 2
18=
13
18
Denominators:
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Operations of Fractions:
1. Addition of Fractions:
To add fractions of the same denominator, add the
numerator and copy the common denominator andreduce answer to its simplest form.
Example:
1
3+ 2
3= 3
3=a. 1
b.1
4+
1
4=
2
4=
1
2
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To add fractions of different denominators, find the least
common denominator and add. Simplify answer to its
simplest form.
Example:
34
+a. 78
= =6 + 78
or138
58
1
2. Subtraction of Fractions:
To subtract fractions having the same denominator. Subtract
the numerators and copy the common denominator and
reduce answer to simplest form.
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7
5 – a.
2
5= =
7 – 2
5
5
5
Example:
1=
3
4 – b.
1
4=
3 – 1
4=
2
4=
1
2
To subtract fractions having different denominators, find the
least denominator (LCD) and subtract the numerators, and
reduce final answer to simplest form.
Example:3
5
– a.1
4
= =12 – 5
20
7
20
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4. Division of Fractions:
To reduce fraction by another fraction, take the reciprocal
of the divisor and then proceed to the multiplications
process. Simplify final fraction to simplest form.
Example:1
2 – a.
1
4=
1
2*
4
1= or
4
22
3
4 – b. 2 =
3
4*
1
2=
3
8
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ROUNDING OFF NUMBERS
1. Round to the nearest Tenths
How do round to the nearest tenths? Follow the steps outlined below.
1. First find the tenths digit. This the first digit after the decimal place.
2.241 (Ex.1) | 4.567 (Ex.2)The tenths digit is shown using the red arrow.
2. Now, find the next digit to the right.
2.241 (Ex.1) | 4.567 (Ex.2)
The next digit is shown using the green arrow.
3. We round up or down depending on the value of the next digit.
(a) If the next digit is less than 5, round down. To round down,
remove all the digits beginning at the next digit.
In Ex 1., the number to the nearest tenths is 2.2.
BASIC OPERATIONS INVOLVING DECIMALS
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ROUNDING OFF NUMBERS
(b) If the next digit is greater than or equal to 5, round up. Toround up, remove all digits beginning at the next digit, and add
1 to the tenths digit. In Ex 2., the number 4.567 rounded to the
nearest tenths is 4.6.
Here are two more examples.
Example 1. Round each of the two numbers to the nearest tenths.
(a) 235.233
Solution: First find the tenths digit, 235.233. Now find the next digit to
the right , 235.233, the next digit is a 3, which is less than 5, so we
round down. The answer
(b) 45.581
Solution: First find the tenths digit, 45.581. Now find the next digit to
the right, 45.581, the next digit is an 8, which is greater than 5, so we
round up. The answer is
235.2
45.6
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ROUNDING OFF NUMBERS
(b) If the next digit is greater than or equal to 5, round up. To round up,remove all digits beginning at the next digit, and add 1 to the
hundredths digit. In Ex 2, the number 4.567 rounded to the nearest
hundredths is 4.56.
Example 2. Round each of the two numbers to the nearest hundredths.
(a) 17.2363
Solution: First find the hundredths digit, 17.2363. Now find the next digit to
the right, 17.2363, the next digit is a 6, which is greater than 5, so we
rounded up. Remove all digits beginning with the next digit and increase
the hundredths digit by one. The answer
(b) 54.581
Solution: Find the hundredths digit, 54.581. Now find the next digit to the
right, 54.581, a 1, which is less than 5, so we round down.
The answer
17.24
54.58
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ROUNDING OFF NUMBERS
3. Round to nearest Whole Number
How do you round to the nearest whole number? This is also called rounding to
the nearest units. Follow the steps outlined below.
1. First find the units digit, the first digit before the decimal place.
2.241 (Ex.1) | 4.567 (Ex.2)The units digit is shown using the red arrow.
2. Now, find the next digit to the right .
2.241 (Ex.1) | 4.567 (Ex.2)
The next digit is shown using green arrow.
3. We round up or down depending on the value of the next digit.
(a) If the next digit is less than 5, round down. To round down,
remove all digits beginning at the next digit. In Ex 1, the number 2.241
rounded to the nearest units (or whole number) is 2.
(b) If the next digit is greater than or equal to 5, round up.
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ROUNDING OFF NUMBERS
To round up, remove all digits beginning at the next digit, and add 1 tothe units digit. In Ex 2, the number 4.567 rounded to the nearest units ( whole
number ) is 5.
Example 3. Round each of the two numbers to the nearest whole number ( units ).
(a) 17.2363
Solution: First find the units digit, 17.2363. Now find the next digit to the right,
17.2363, the next digit is a 2, which is less than 5, so we round down. The
answer .
(b) 54.581
Solution: First find the units digit, 54.581. Now find the next digit to the right,
54.581, the next digit is 5, which is greater than 5, so we round up. The
answer .
17
55
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ROUNDING OFF NUMBERS
4. Round to nearest Tens
How do you round to the nearest tens? Follow the steps outlined below.
1. First find the tens digit, the second digit before the decimal place.
62.64 (Ex 1) | 27.56 (Ex 2)
The tens digit is shown using the red arrow.
2. Now, find the next digit to the right .
62.64 (Ex 1) | 27.56 (Ex 2)
The next digit is shown using the green arrow.
3. We round up or down depending on the value of the next digit.
(a) If the next digit is less than 5, round down. To round down, remove
all digits beginning at the next digit. In Ex 1, the number 62.24
rounded to the nearest tens place is 60.
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ROUNDING OFF NUMBERS
(b) If the next digit is greater than or equal to 5, round up. To round up,remove all digits beginning at the next digit, and add 1 to the tens digit. In Ex
2, the number 27.65 rounded to the nearest tens place is 30.
Example 4. Round each of the two numbers to the nearest tens.
(a) 17.2363
Solution: First find the tens digit, 17.2363. Now find the next digit to the right,
17.2363, the next digit is a 7, which is greater than 5,so we round up. The
answer is
(b) 54.581
Solution: First find the tens digit, 54.581. Now find the next digit to the right,
54.581, the next digit is a 4, which is less than 5, so we round down. The
answer is
20
50
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ROUNDING OFF NUMBERS
5. Round to nearest Hundreds
How do you round to the nearest hundreds? Follow the steps outlined below.
1. First find the number digit, the third digit before the decimal place.
642.24 ( Ex 1 ) | 272.56 ( Ex 2 )
The hundreds digit is shown using the red arrow.
2. Now, find the next digit to the right .
642.24 ( Ex 1 ) | 272.56 ( Ex 2 )
The next digit is shown using the green arrow.
3. We round up or down depending on the value of the next digit.
(a) If the next digit is less than 5, round down. To round down, remove
all the digits beginning at the next digit. In Ex 1, the number 642.24
rounded to the nearest hundreds place is 600.
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ROUNDING OFF NUMBERS
(b) If the next digit is greater than or equal to 5, round up. To round up,remove all the digits beginning at the next digit, and add 1 to the hundreds
digit. In Ex 2, the number 272.56 rounded to the nearest hundreds place is
300.
Example 5. Round each of the two numbers to the nearest hundreds.
(a) 717.23
Solution: First find the hundreds digit, 717.23. Now find the next digit to the
right, 717.23, the next digit is a 1, which is less than 5, so we round down.
The answer is
(b) 254.58
Solution: First find the hundreds digit, 254.58. Now find the next digit to the
right, 254.58, the next digit is a 5, which is greater than or equal to 5, so we
round up. The answer is
700
300
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ROUNDING OFF NUMBERS
6. Round to nearest Thousands
How do you round to the nearest thousands? Follow the steps outlined below.
1. First find the thousands place, the fourth digit before the decimal place.
6242.24 ( Ex 1 ) | 2721.56 ( Ex 2 )
The thousands digit is shown using the red arrow.
2. Now, find the next digit to the right .
6242.24 ( Ex 1 ) | 2721.56 ( Ex 2 )
The next digit is shown using the green arrow.
3. We round up or down depending on the value of the next digit.
(a) If the next digit is less than 5, round down. To round down,
remove all digits beginning at the next digit. In Ex 1, the number
6242.24 rounded to the nearest thousands place is 6000.
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ROUNDING OFF NUMBERS
(b) If the next digit is greater than or equal to 5, round up. Toround up, remove all digits beginning at the next digit, and add
1 to the thousands digit. In Ex 2, the number 2721.56 rounded
to the nearest thousands place is 3000.
Example 6. Round each of the two numbers to the nearest thousands.
(a) 1717.23
Solution: First find the thousands digit, 1717.23. Now find the next digit to the
right, 1717.23, the next digit is a 7, which is greater than 5, so we round up.
The answer is
(b) 8254.58
Solution: First find the thousands digit, 8254.58. Now find the next digit to the
right, 8254.58, the next digit is a 2, which is less than 5, so we round down.
The answer is
2000
8000
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ROUNDING OFF NUMBERS
(b) If the next digit is greater than or equal to 5, round up. To round up,remove all digits beginning at the next digit, and add 1 to the ten thousands
digit. In Ex 2, the number 27,213, rounded to the nearest thousands place is
30,000.
Example 7. Round each of the two numbers to the nearest ten thousands.
(a) 51,717.2
Solution: First find the ten thousands digit, 51,717.2. Now find the next digit to
the right, 51,717.2, the next digit is a 1, which is less than 5, so we round
down. The answer is
(b) 38,254
Solution: First find the ten thousands digit, 38,254. Now find the next digit to
the right, 38,254, the next digit is a 8, which is greater than 5, so we round
up. The answer is
50,000
40,000
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EXERCISES: ROUNDING OFF NUMBERS
Exercises 1: Round each to the nearest tenths.
1. 2.67 3. 9.457 5. 17.232
2. 5.71 4. 12.59
Exercises 2: Round each to the nearest hundredths
1. 5.347 3. 0.136 5. 1.345
2. 12.423 4. 39.126
Exercises 3: Round each to the nearest whole numbers
1. 2.67 3. 9.457 5. 7.23
2. 13.59 4. 57.23
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Ratio
A ratio is a pair of numbers that compares two quantities
or describes a rate. A ratio can be written in three ways:
Using the word to 3 to 4
Using a colon 3:4Writing a fraction 3/4
When a ratio is used to compare two different kinds of
quantities, such as miles to gallons, it is called a rate.Equal ratios make the same comparison. To find equal
ratios, multiply or divide each term of the given ratio by
the same number.
These are all read 3 to 4.
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EXAMPLE:
In the shop, there are 7 MS plates and 12 Aluminum
plates. In three different ways, write the ratio of:
a. MS plates to Aluminum plates
b. Aluminum plates to MS plates
c. MS plates to pieces of plates
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Proportion
A proportion is a statement that two ratios are equal. A
proportion shows that the numbers in two different ratios
compare to each other in the same way.
In a proportion, the cross products are equal. 3 and
10 are the means, or the terms in the middle of the
proportion. 2 and 15 are the extremes, or the terms atthe beginning and end of the proportion.
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EXAMPLE:
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Use the figure below. The small gear makes 8 turns for
every 5 turns of the large gear.
EXERCISE:
How many turns will the large gear
make if the small gear makes 80 turns?
How many turns will the small gear make
if the large gear makes 62.5 turns?
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Percentage:
In machine shop and in daily life, we often need to work with
percentage. Like decimals, percents are closely related to
fractions. Decimals are a simple way of writing fractions of
tenths, hundredths, thousandths, and ten thousandths.
Similarly, percents are a simple means of writing fractions of
hundredths, Percent are also a means of expressing ratios.
In working with percents as ratio, we make a comparison
against a base of 100.
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Interpretation of Percents:
Percent means “ per one hundred “. Percent written in
fractional form have 100 as the base of comparison
(denominator).
Example: 3
100
, 25
100
, 63
100
, 145
100
Fraction of hundredths can be written in decimal form:
0.03 0.25 0.63 1.45
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We also can write fractions of hundredths as percents by
using symbol ( % ). Thus, in percent notation, these fractions
are written:
3% 25% 63% 145%
As ratio expressions, percents compare an amount to a base
of 100. For example 3% ( 3 per 100 ) is the same as the ratio
3 to 100.
3% =3
100= 0.03 = 3 : 100
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Since percents are ratio expressions, they can be used in
stating proportions. In a proportion expression, the percentratio (amount per 100) is equated with another ratio. The two
value that form the other ratio are called the percentage and
the base.
Percentage is the number expressed as a percent of another
number. The base is the value to which the percentage is
being compared. For example, if we say that 20 is 50% of
40, 20 is the percentage, 40 is the base to which it is being
compared, and 50 is the percent.
Thus, in percent proportion, the ratio of the percent to 100 is
equal to the ratio of percentage to the base.
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Percent
100
=
percentage
base
Example: What percent of 75 is 15
x
100=
15
75
x =100 ( 15 )
75= 20 %
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RULE:
To find the percent, multiply the percentage by 100 and
divide by the base.
Percent =100 x percentage
Base
To find the percentage, multiply the percent by the baseand divide by 100.
Percentage =percent x base
100
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Operations with powers
Multiplication and division of powers.
Power of product of some factors.
Power of a quotient (fraction).
Raising of power to a power.
Negative, zero and fractional exponents of a power.
Operations with roots.
Arithmetical root.
Root of product of some factors.
Root of quotient (fraction).Raising of root to a power.
Proportional change of degrees of a root and its radicand.
About meaningless expressions.
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34
The base
figure
The figure „4‟ is
the power or
„replicator‟
A power describes or states by how many times
the base figure should be multiplied by itself.
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34
= 3x3x3x3 Typical economics and business examples where
powers can be used include:
- calculating compounded figures such asinterest payments on a mortgage;
- working out areas or volumes;
- estimating the expected return from an
investment project (net present values);and
- calculating the value of an individual‟s
savings or pension in times of high and
sustained inflation.
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1. At multiplying of powers with the same base their exponentsare added:
a m · a n = a m + n .
2. At dividing
of powers with the same base their exponents are
subtracted:
3. A power of product of two or some factors is equal to a
product of powers of these factors:
( abc… ) n = a n · b n · c n …
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4. A power of a quotient (fraction) is equal to a quotient of
powers of a dividend (numerator) and a divisor (denominator):
( a / b ) n = a n / b n .
5. At raising of a power to a power their exponents are
multiplied:
( a
m
)
n
= a
m
n
.
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Work out the following expressions using simple powers:
33 =
53 =
66 =
113 =
4.53 =
N ti t f A f b
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Similarly, powers can be used in a „symmetrical way‟. A
negative power can be expressed as a positive power when
it is the reciprocal.
Example:
3-2 = 1/32 = 1/9
4-3 = 1/43 = 1/(4x4x4) = 1/64
5
-5
= 1/5
5
= 1/(5x5x5x5x5) = 1/3125
Negative exponent of a power. A power of some number
with a negative (integer) exponent is defined as unit divided
by the power of the same number with the exponent equal
to an absolute value of the negative exponent:
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In all below mentioned formulas a symbolmeans an arithmetical root
( all radicands are considered here only positive ).
1. A root of product of some factors is equal to a product of roots of these factors:
2. A root of a quotient is equal to a quotient of roots of a dividend and a divisor:
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3. At raising a root to a power it is sufficient to raise a radicand to this power:
4. If to increase a degree of a root by n times and to raise simultaneously its radicand to the n-
th power, the root value doesn’t
change:
5. If to decrease a degree of a root by n times and to extract simultaneously the n-th degree root
of the radicand, the root valuedoesn’t change:
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Zero exponent o f a power . A power of any non-zero number
with zero exponent is equal to 1.
E x a m p l e s: 2 0 = 1, ( – 5 ) 0 = 1, ( – 3 / 5 ) 0 = 1.
Fract ional exponent of a power .
To raise a real number a to a power with an exponentm / n it is necessary to extract the n-th degree root from
the m-th power of this number a:
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ENGLISH AND METRIC SYSTEMOF MEASUREMENT
Wh M t i ?
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Why Metric?
A. The metric system is much easier. All metric units are
related by factors of 10.B. Nearly the entire world (95%), except the United States,
now uses the metric system. U.S. economic competitiveness
would be strengthened by converting to the metric system.
C. Metric is used exclusively in science -- therefore,understanding of scientific and technical issues by non-
scientists will be enhanced if the metric system is universally
adopted.
D. Because the metric system uses units related by factorsof ten and the types of units (distance, area, volume, mass)
are simply-related, performing calculations with the metric
system is much easier ¾ thus facilitating quantitative
analysis and understanding in science.
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Comparison of simple conversion operations in the English
(customary) and Metric systems
Notice the unusual numbers relating the various units in the English system and the
simplicity of the powers of ten in Metric.
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UNIT CONVERSION AND CONVERSION FACTORS
A unit conversion expresses the same property as a differentunit of measurement.
A conversion factor is a number used to change one set of
units to another, by multiplying or dividing.
When a conversion is necessary, the appropriate conversion
factor to an equal value must be used.
For example, to convert inches to feet, the appropriate
conversion value is 12 inches equal 1 foot.
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A unit cancellation table is developed by using known units,
conversion factors, and the fact that a unit of measure ÷ the
same unit of measure cancels out that unit. The table is set up
so all the units cancel except for the unit desired. To cancel a
unit, the same unit must be in the numerator and in the
denominator. When you multiply across the table, the top
number will be divided by the bottom number, and the result
will be the answer in the desired units.
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GEOMETRICAL APPLICATION
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POINTS, LINES, AND PLANES
A Point has no dimension. It is usually represented by a
small dot.
A Line extends in one dimension. It is usually representedby a straight line with two arrowheads to indicate that the
line extends without end in two directions.
A Plane extends in two dimensions. It is usually representedby a shape that looks like a tabletop or wall.
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A few basic concepts in geometry must also be commonly
understood without being defined. One such concept is the idea
that a point lies on a line or a plane.
are points that lie on the same line.
are points that lie on the same plane.
Collinear points
Coplanar points
Examp le: Nam ing Co l l inear and Coplanar Points
a. Name three points that are collinear.
b. Name four points that are coplanar.
c. Name three points that are not collinear.
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Another undefined concept in geometry is the idea that a point
on a line is between two other points on the line.
Consider the line AB (symbolized by AB ).
The line segment or segment AB (symbolized by AB )
consists of the endpoints A and B, and all points on AB
that are between A and B.
A B
Line
A B
Segment
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The ray AB (symbolized by AB ) consists of the initial point A
and all points on AB that lie on the same side of A as point
B.
Note that AB is the same as BA , and AB is the same as BA
. However, AB and BA are not the same. They have different
initial points and extend in different directions.
A B
Ray
A B
Ray
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If C is between A and B, then CA and CB are opposite
rays
A B
Ray
C
Like points, segments and rays are collinear if they lie on
the same line. So, any two opposite rays are collinear.
Segments, rays, and lines are coplanar if they lie on the
same plane.
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INTERSECTIONS OF LINES AND PLANES
Two or more geometric figures intersect if they have one ormore points in common. The intersection of the figures is the
set of points the figures have
in common.
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ANGLES AND TRIANGLES
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DIFFERENT TYPES OF ANGLES
ACUTE ANGLE OBTUSE ANGLE
RIGHT ANGLE
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REFLEX ANGLE
STRAIGHT ANGLE
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DIFFERENT TYPES OF TRIANGLES
ACUTE TRIANGLE
EQUILATERAL TRIANGLE
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ISOSCELES TRIANGLE
OBTUSE TRIANGLE
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RIGHT TRIANGLE
SCALENE TRIANGLE
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POLYGONS AND CIRCLES
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POLYGONS
A regular polygon is a polygon whose sides are all the
same length and whose interior angles all have the same
measure.
An equilateral triangle is a 3-sided regular polygon;
a square is a 4-sided regular polygon.
CIRCLE
A circle is the locus of all points equidistant from a
central point.
I i ti d Ci i ti
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Inscription and Circumscription
Certain geometric figures are created by combining circles
with other geometric figures, such as polygons. There are
two simple ways to unite a circle with a polygon. One is
inscription, and the other is circumscription.
When a polygon is inscribed in a circle, it means that
each of the vertices of that polygon intersects the circle.
When a polygon is circumscribed about a circle, it
means that each of the sides of the polygon is tangentto the circle. Below these situations are pictured.
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INSCRIBED CIRCUMSCRIBED
Above on the left, the hexagon ABCDEF is inscribed in
the circle G. On the right, the quadrilateral ABCD is
circumscribed about the circle E.
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Concentric circles are circles that have the same center.
Just because a circle is inside another circle doesn'tmean they are concentric; they must have the same
point as their center. Any number of circles can be
concentric to one another, provided that they all share a
center.
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BASIC MENSURATION
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Circumference:
The Circle:
The circle has been called the perfect geometric form. It is
perhaps also the most important of all geometric forms.
Definition:
A circle is a plane curve consisting of all the points that are
the same distance from a fixed point called the center, thecommon distance of the points on the curve from the center
is called the radius. The region bounded by (enclosed
within) the circle is also often referred to as a circle.
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The curve, (that is, the circle as defined)) is referred to as the
perimeter or circumference of the circle. These terms are
used both to denote the curve itself and its length.
Figure 1: Circle
A diameter of a circle is any line drawn from one side of the
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y
circumference to the other passing through the center. A
diameter divides both the area and the circumference into two
equal (congruent) parts called semicircles. Any line drawn fromthe center of the circumference is called radius. (Note: the
plural form of “ radius “ is “ radii “).
Any portion of the circumference is called an arc.
A chord is a line segment connecting any two points in the
circle. In the figure as shown below, DE is the chord. The
chord is said to subtend its arc. The chord DE subtends the arc
DME. The area bounded by an arc and a chord is called a
segment; as in the figure, the area DME is a segment. The
area bounded by two radii and an arc is called a sector. A
sector is an area shaped like a wedge of pie, as in the area
BOC in figure 1.
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Relationship between Diameter, Radius, and Circumference:
Rules:
The radius equals one-half the diameter, or, the diameter
equals twice the radius.
r =d
2
or d = 2r
The circumference equals the diameters times ii.
C = d ii
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This may also be expressed in either of the following ways,
which are usually more convenient to use:
Rule: The area of a circle equals = P1 times the square
of the radius; or, the area of a circle equals one-fourth
times the square of the diameter.
If A = area, C = circumference, d = diameter, and
r = radius, these rules are expressed in the following
formulas:
π
π
C r
2 A =
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d2
4 A =
A = r 2π or r = A
π
π
or A = 0.7854 d2 or A
0.7854d =
Example:
Find the radius of a circle whose area is 320 cm2 .
Using the formula r = A / and substituting the
given numbers,
Radius = 10.09 cm
π )
r =320
3.1416
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Pythagorean Theorem:
The Pythagorean theorem is one of the most important
mathematical principles. Pythagorean theorem states that in
a right triangle, the square of the hypotenuse, which is the
longest side of a right triangle, is equal to the sum of the
squares of the other two sides of the triangle. This theorem is
only applicable to right triangles.
A right triangle is a triangle with one of its angles measures
90o .
a
b
c
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The triangle in the figure is a right triangle. Side c is the
hypotenuse. Sides a and b are called legs.
Using the above figure, Pythagorean theorem can be
represented by the equation
c2 = a2 + b2
NOTE: We always need to know the measurements of two
sides of a triangle in order to use the basic formula.
RULE: To find the length of the hypotenuse when the length of
the altitude and base are known, use the formula;
c
2
= a
2
+ b
2
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RULE: To find the length of one of the legs (altitude or base)
of a right triangle when the length of the hypotenuse and theother length are known, use one of the formulas;
b2 = c2 – a2 or a2 = c2 – b2
Example:
Given a right as shown below, calculate the missing
side.
a = 76 mm c = ?
b = 102 mm
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Required: c = ?
Solution:
Using c2 = a2 + b2
= ( 76 )2 + ( 102 )2
= 5776 + 10.404
c = 127.2 mm
)
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Shape & Perimeter:
1. Square – a kind of polygon having 4 equal sides
S
Area = s2
Perimeter = 4s
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2. Hexagon – a polygon having 6 sides
da
s3
2 A = a2
d = 2 a = 1.155s
s = 0.866d
3. Triangle – a kind of polygon having three sides
A = ½ bh
Where: b = base
h = heightb
h
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Units of time and angle
h = hoursmin = minutes
s = seconds
o = degrees
„ = minutes
“ = seconds
1. Time units
The unit of time is the second. A second is the 24 x 60
x 60th part of a mean solar day.
60 min
1 min = 60 s1 h =
Conclusion: The conversion factor from unit to unit is 60.
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2. Angular units
The derived unit of angle is the degree. A degree isthe 360th part of the circumference of a circle.
60 „
1 „
1o =
= 60 “
Conclusion:
The conversion factor from unit to unit is 60.
Result
The difference between units of time and angle consists
only in the way they are named.
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3. Summary
The conversion factor for units of time and angle is 60.
Conversion
Into the next smallest unit: x 60
Into the next largest unit: : 60
4. Example
A time of 1.48 hours was needed to grind the
clearance angle of a face-cutting miller. Calculate the
exact time ( t ) in h, min, s.
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find h, min, s
given that t = 1.48 h
solution
1.48 h =
0.48 h = 0.48 * 60 = 28.8 min0.8 min = 0.8 * 60 =
preliminary consideration
next smallest unit by 60
1 h +
28 min +
48 s
1 h 28 min 48 s
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Areas of regular quadrilaterals:
A = area
l = length of side
h = height of area
Note:
In the case of rectangle, the opposite sides are parallel.
1. Square
l
l
area = length * height
A = l * l
= l
2
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2. Rhombus
l
l
harea = length * height
A = l * h
3. Rectangle
h
l
area = length * height
A = l * h
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4. Parallelogram
h
l
area = length * height
A = l * h
5. Summary
For calculating the areas of rectangles, the formula is:
area = length x height
Note
The height is taken as being the perpendicular to the
line.
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C l l ti f II
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Calculation of areas II
A = area
l = length of side
h = height of area
1. Triangle
hh
l l
If we add a supplementary
area to triangle to form a
familiar rectangle, we have:length * width = 2 * A
Conclusion: l * h
2
A =
N t
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a
a
h
Note
In the case of equilateral triangles,
Pythagoras‟ Theorem gives us a
height of:
a2 = h2 +a
2( )
2
a
2( )
2
h2 = a2 – =3
4
a2
r =3
4
a2
=1
23 * a
h = 0.866 a
2 T i
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2. Trapezium
Any trapezium can be divided into two triangles. Thus:
A =L * h
2+
l * h
2
=l * h
2* h
A = l m * h
Result
L + l
2
is the length of the rectangle in the diagram.
A2
A1
h
L
l
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h
l m
3. Summary
triangle: A =l * h
2
trapezium: A =L + l
2* h
4 Example
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4. Example
A =l * h
2
A die with a triangular area has a cross section of 1015
mm2 and a height of 35 mm. Calculate the length of thebase in mm.
find l
given that A = 1015 mm
2
h = 35 mm
solution
l =2 * A
h=
2 * 1015
35
mm2
mm*
l = 58 mm
A
l
h
Ci l
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Circular area:
A = aread = diameter
b = arc length
= sector angleα
4
π = 0.785
1.Circle
If we divide a circle into section, we obtain an
approximate parallelogram with
area = base line * height
A =d *
2
π *
2
d=
d 2 *
4
π
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1
2
3 4
5
6
7
8
910
11
12
Note
For the calculation of the area or diameter of a circle the
use of tables is recommended.
2. Sector
If we again divide a sector into sectors, we have
b
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1
2 3
4
A
d
A =b * r
2
For a full circle of 360o A = d 2 * 0.785
For a sector of α A = d 2 * 0.785360 o
*α
3 Circular ring
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3. Circular ring
The difference between two circle areas gives a
circular ring area.
A = A1 – A2
D2 *
4
π = –
d 2 *
4
π
=
4
π * ( D2 – d2 )
A = 0.785 * ( D2 – d2 )
D
4 S
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4. Summary
circle A = d 2 * 0.785
sector A = d 2 * 0.785
360 o*
α
A =b * r
2
Circular ring A = A1 – A2 = 0.785 * ( D2
– d 2
)
5 E l
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5. Example
For a grinding machine we require a sheet metalcover in the form of a sector with a diameter of 400 mm
and a sector angle of 220o. Calculate the amount of sheet
metal needed in cm2.
find A in cm2
given that d = 400 mm
α = 200o
solution A = d 2 * 0.785
360 o* α
= ( 40 cm )2 * 0.785 220o
360o*
A = 767.56 cm2
C l l ti th l f i id b di
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Calculating the volume of rigid bodies
A = area
V = volume
H = height of body
Note
In the case of right bodies, the sides or curved surface are
perpendicular to the parallel base and top area.
1 C b
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1. Cubevolume = base area * height
V = A * H
Note
Square base areas can be found in tables.
H
l
2. Prism volume = base area * height
V = A * H
Note
In the case of a prism, the base area can
be of any shape.
H
l
Al
3. Cylinderl b * h i ht
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3. Cylindervolume = base area * height
V = A * H
Note
It is easier to find the base number d if
one has the tabular value for the base area A.
H
4. Summary
For calculating the volume of right bodies, we have
volume = base area * height
V = A * H
5. Example
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5. Example
The capacity in liters is to be calculated for a
cylindrical water container with a diameter of 350 mm and aheight of 750 mm.
find V in l
given that d = 350 mmH = 750 mm
H
preliminary consideration
volume = base area * height
solution
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solution
V = A * H
A = d 2 * 0.785
= 3.52 dm2 * 0.785
A = 9.62 dm2
V = 9.62 dm2 * 7.5 dm = 72.15 dm3
Note
1 dm3 = 1 liter
Calculating the volume of pointed and truncated bodies:
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g p
A = area V = volume
H = vertical height of body
1. Cone / Pyramid
In the case of pointed bodies, the slanting lines
converge in a straight line on a point. One prism equals
three pointed bodies in volume if the base area and heightare the same.
V
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H
t
V
3V
V = V prism
3
V = base area x height
3
V = A . H
3
Note:
The base area can be of any shape
2 Truncated Cone / Pyramid
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2. Truncated Cone / Pyramid
In the case of truncated bodies, the upper part of a
pointed body is cut off parallel to the base. Using the base
and top measurements, we take the mean:
dm = D + d
2
Or Am = A1 + A2
2
Thus we have:
Volume = mean area x height
V = Am x H H
D
d
H
A2
A1m
3 SUMMARY
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3. SUMMARY
V pointed = base area x height
3V = A . H
3
V truncated = mean area x height
V = Am x H
4. Example
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4. Example
A cone with a diameter of 210mm has a volume of
3056cm3. Calculate the height in cm.
Required: H in cm
Given: V = 3056cm3
d = 210mm
Preliminary Consideration:
Volume of cone = 1/3 volume of cylinder
S l ti V A H
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Solution: V = A . H
3
H = 3 . V
A
H = 3 . 3056cm3
346.36cm2
H = 26.47 cm
Calculation of mass
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Calculation of mass
m = mass, result of weighting
V = volume
p = density ( pronounce prho )
1. Mass
The Sl unit for mass is 1 kilogram. In practice we also use
derived units for calculation:
1 t = 1000 kg
1 kg = 1000 g
1 g = 1000 mg
Conclusion
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The conversion factor from unit to unit is 1000.
1 g 1 kg
watersugar
Masses of comparison
1 Mg = 1 t
1m3
t kg g mg
000 000 000 000
2. DensityWe weigh the mass of a material with a
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We weigh the mass of a material with a
volume unit of 1cm3 or 1dm3 or 1m3. We
relate this mass to the volume unit in questionand call the relationship “ DENSITY ”.
g
cm3
kg
dm3
t
m3
m
V
or or = =
Note: These units of mass always belong together
1 2.7
7.85
water
aluminumsteel
ᵨ
3. Summary
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The units of mass obtained give the basic equation
Mass = volume x density
m = V x ᵨ
Note : In commerce and industry, the
result of weighing is known as
“weight”.
“Weights” for semi -finished
materials are given in tables.
mass in kg
Weight-force in N
4. Example
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Calculate the mass in kg of a rectangular plate with the
Dimensions 220 , 330 and 15 mm.
Required: m in kg
Given : A = 2.2 x 3.3 dm2
s = 0.15 dm
p = 7.85 kg/dm3
Preliminary consideration : there is a relationship between kg and dm3
Solution:
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Solution:m = V x ᵨ
V = A x s = 2.2 dm x 3.3 dm x 0.15 dm
= 1.089 dm3
m = 1.089 dm3 x 7.85 kg/dm3 = 8.55 kg
Note : In the case of mass per unit area
(plate weight), it is better to use the
rule of three. For steel, the
statement referring to the unit is :1m2 of 1 mm thickness = 7.85kg
Proof : 1m2 . 1mm = 1dm3, since
100dm2 . 0.01 dm = 1dm3
s
t
Lines and the Cartesian Plane
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Lines and the Cartesian Plane
The cartesian plane provides a pallate on which to graph
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The cartesian plane provides a pallate on which to graph,
and it is built by taking two number lines and making them
cross at a right angle where both of them are zero.
Then, we have one number line going up
and down, and another number line going from left to right.
Notice that there is an x at the
right end of the horizontal
number line and a y at the top of
the vertical number line. We call
these two number lines axes.Specifically, we call the horizontal
number line the x-axis and the
vertical number line the y-axis.
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Each one has its own
number, which we write as a
roman numeral.
That is, quadrants I (1),II (2), III (3) and IV (4).
There are four distinct areas of
the cartesian plane
We name them quadrants
because there are four of them
Notice that in quadrants I and II y is always positive In
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Notice that in quadrants I and II, y is always positive. In
quadrants I and IV, x is always positive.
(+,+)
(-,+)
(-,-)
(+,-)
EXAMPLE:What are the coordinates
of the following:
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Point M
Point Z
Point D
Point A
M
Z
D A
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BASIC TRIGONOMETRICAL FUNCTION
U d it i l t d fi iti t i t i (lit "t i l
TRIGONOMETRY
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Under its simplest definition, a trigonometric (lit. "triangle-
measuring") function, is one of the many functions that relateone non-right angle of a right triangle to the ratio of the
lengths of any two sides of the triangle (or vice versa).
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Any trigonometric function (f), therefore, always satisfieseither of the following equations:
f(Ɵ) = a / b OR f(a / b) = Ɵ,
where Ɵ is the measure of a certain angle in the triangle, and
a and b are the lengths of two specific sides.
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The side lengths (a b c) of a right triangle form a so-
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The side lengths (a , b, c) of a right triangle form a so
called Pythagorean triple. A triangle that is not a right triangle
is sometimes called an oblique triangle. Special cases of theright triangle include the isosceles right triangle (middle
figure) and 30-60-90 triangle (right figure).
Applications of Right Triangle
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Trigonometric ratios (sine, cosine, and tangent) will be used
in real world applications.
Right triangle, one angle is 90º and the side across from
this angle is called the hypotenuse. The two sides which
form the 90º angle are called the legs of the right triangle.
The legs are defined as either “opposite” or “adjacent” (next
to) the angle A.
We shall call the opposite side "opp," the adjacent side "adj" and
the hypotenuse "hyp."
In the following definitions, sine is called "sin," cosine is called
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g
"cos" and tangent is called "tan." The origin of these terms
relates to arcs and tangents to a circle.
i.sin(A) =
ii.cos(A) =
iii.tan(A) =
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In a right triangle, there are actually six possible trigonometric ratios, or
Trigonometric Functions
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In a right triangle, there are actually six possible trigonometric ratios, or
functions.
A Greek letter (such as theta or phi ) will now be used to represent
the angle.
Notice that the three new ratios at the right are reciprocals of
th ti th l ft
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the ratios on the left.
Applying a little algebra shows the connection between these
functions.
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