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MTM 2 Final Exam Review
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Unit 73: Introduction to Computer Numerical Control (CNC)
Big Skill: You should be able to locate points on a Cartesian plane given their coordinates, and state the
coordinates of a given point on a Cartesian plane.
Unit 74: Control Systems: Absolute and Incremental Positioning
absolute positioning: when x and y coordinates are stated relative to a fixed origin
incremental positioning: when positions are stated relative to the previous position… take the new
absolute position minus the old absolute position.
Unit 17: Introduction to Percents
Definition of Percent: A percent (%) indicates the number of hundredths of a whole.
1
 0.01  1%
1 part of 100 parts 
100
To convert a decimal fraction to a percent: move the decimal two places to the right, then add the %
symbol. 0.0987  0.0.9.87  9.87%
To convert common fractions and mixed numbers to a percent: use a calculator to express the
5
number as a decimal fraction, then convert that decimal to a percent.  0.625  0.6.2.5  62.5%
8
To convert percents to decimal fractions: drop the % symbol, then move the decimal two places to the
left. 2.3%  .0.2.3  .023
To convert percents to common fractions: express the percent as a decimal fraction, then find the
62.5 625 5  5  5  5 5



common fraction equivalent of the decimal (and reduce). 62.5% 
100 1000 5  5  5  8 8
MTM 2 Final Exam Review
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Unit 18: Basic Calculations of Percentages, Percents, and Rates
150% of $20
is
$30
150%    $20 


$30
percentage
rate  base
3 out of 4 parts is 75%
3
 75%
4
percentage
 rate
base
20 is the number
such that 10 is 50% of that number...
10
20 
50%
percentage
base 
rate
Finding the Percentage, given the Base and Rate: percentage  rate  base
Example: What is 15% of $60.00? percentage  0.15  $60.00  $9.00
Finding the Rate (Percent), given the Base and Percentage: rate 
Example: 12.2 is what percent of 28.9? rate 
percentage
base
12.2
 0.422  42.2%
28.9
Finding the Base , given the Rate (Percent) and Percentage: base 
Example: 317 is 72% of what number? base 
percentage
rate
317
 440.3
0.72
Unit 19: Percent Practical Applications
Big Idea: The three percentage formulas we practiced in the Unit 18 can be used to solve any realworld problems involving percents.
Big Skill: You should be able to solve practical applications of percents.
150% of $20
is
$30
150%    $20 


$30
percentage
rate  base
3 out of 4 parts is 75%
3
 75%
4
percentage
 rate
base
20 is the number
such that 10 is 50% of that number...
10
20 
50%
percentage
base 
rate
MTM 2 Final Exam Review
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Unit 48: Arcs and Angles of Circles, Tangent Circles
Principle 18:
A central angle is equal to its intercepted arc.
An angle formed by two chords that intersect
inside a circle is equal to one-half the sum of
the two intercepted arcs.
An inscribed angle is equal to one-half of its
intercepted arc.
MTM 2 Final Exam Review
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Principle 19:
An angle formed by a tangent and a chord at the
tangent point is equal to one-half of its
intercepted arc.
MTM 2 Final Exam Review
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Principle 20:
An angle formed at a point outside a circle by
two secants, two tangents, or a secant and a
tangent is equal to one-half the difference of the
intercepted arc.
MTM 2 Final Exam Review
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Tangent circles are tangent to the same line at the same point.
Internally tangent circles are on the same side of a common tangent line.
Externally tangent circles are on opposite sides of a common tangent line.
Principle 21:
A line connecting the centers of tangent circles passes through the point of tangency and is
perpendicular to the tangent line.
Principle 21 is useful for when two or more radii blend to give a smooth curve.
 Hint: It is usually helpful to draw a line through the centers of tangent circles when making
calculations.
MTM 2 Final Exam Review
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Unit 49: Fundamental Geometric Constructions
Construction 1: Construct a Perpendicular Bisector of a Line Segment
1. Strike equal radius arcs from each endpoint.
2. Connect the intersection points of the arcs with a line.
3. The line is the perpendicular bisector.
Construction 2: Construct a Perpendicular to a Line Segment at a given point on the Line
Segment
1. Strike an arc from the point so that it intersects the segment twice.
2. Construct the perpendicular bisector of the new segment formed by the intersection points.
Construction 3: Construct a Line parallel to a Given Line at a Given Distance
1. Strike multiple arcs from points on the line where the radius is equal to the given distance.
2. Connect the arcs with a tangent line.
MTM 2 Final Exam Review
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Construction 4: To Bisect a Given Angle
1. Strike an arc from the vertex of the angle.
2. Bisect the segment that results from connecting the intersection points.
Construction 5: To Construct Tangents to a Circle from an Outside Point
1. Bisect the segment connecting the circle center and the outside point.
2. Strike an arc from the midpoint of that segment through the center of the circle.
3. Connect the outside point to the intersection points.
MTM 2 Final Exam Review
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Construction 6: To Divide a Line Segment into a Given Number of Equal Parts
1. Draw another segment from one of the endpoints to make an angle.
2. Mark off the number of desired segments on the new segment.
3. Construct lines that are parallel to the line through the endpoints on each side of the angle
through each segment’s endpoint.
Construction 7: To Copy a Given Triangle
Note: this can be used to copy any polygon by first dividing the polygon into sub-triangles.
1. Mark off one of the sides on a given line.
2. Strike radii from each endpoint that match the lengths of the other two sides.
Construction 8: To Copy a Given Angle
1. Connect the endpoints of the angle.
2. Copy the resulting triangle.
Construction 9: To Create an Equilateral Triangle
Note: this is useful for constructing 60 and 30 angles.
1. Mark off a segment on a line with the compass.
2. Strike arcs of that length from each endpoint.
3. Connect the intersection point to each endpoint of the segment.
MTM 2 Final Exam Review
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Unit 58: Introduction to Trigonometric Functions
The Ratios of the Six Trigonometric Functions:
Function
Sine of Angle A
Symbol
sin(A)
Cosine of Angle A
cos(A)
Tangent of Angle A
tan(A)
Cotangent of Angle A
cot(A)
Secant of Angle A
sec(A)
Cosecant of Angle A
csc(A)
Function Definition
opp a
sin  A  

hyp c
adj b
cos  A 

hyp c
opp a
tan  A 

adj b
adj b
cot  A 

opp a
hyp c
sec  A 

adj b
hyp c
csc  A 

opp a
To find the Cotangent, Secant, and Cosecant Functions:
These functions are never built into any calculators. To calculate these functions, you have to use the
fact that these three functions are reciprocals of the sine, cosine, and tangent functions.
1
1
1
cot  A 
sec  A 
csc  A 
tan  A
cos  A
sin  A
MTM 2 Final Exam Review
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To find an angle given a ratio of sides:

If sin  A  x , then A  sin 1  x  .

If cos  A  x , then A  cos1  x  .

If tan  A  x , then A  tan 1  x  .
1
If cot  A  x , then A  tan 1   .
 x
1
 If sec  A  x , then A  cos 1   .
 x
1
 If csc  A  x , then A  sin 1   .
 x
Unit 62: Complex Practical Machine Applications

How to solve complex applied right triangle problems
4. Figure out what dimensions would help you relate the given dimensions to the unknown
dimension.
5. Draw extra lines on the diagram to form right triangles that relate one dimension to the next.
a. Always draw lines from the center of a circle to any tangent points.
b. Draw a line from the center of a circle to an exterior point formed by the intersection of
two tangent lines.
c. Draw lines to form a right triangle where an angle is given.
d. Draw lines to form a right triangle when two perpendicular dimensions are given.
MTM 2 Final Exam Review
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6. Solve each right triangle formed until the unknown dimension can be calculated.
x = c + RADIUS – a
MTM 2 Final Exam Review
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Unit 63: The Cartesian Coordinate System
When thinking about angles bigger than 90, you should plot a point on the unit circle equal at that
angle. Form a right triangle by moving straight toward the x axis. The reference angle is the angle
formed by your triangle. The x coordinate of the point is the cosine of the angle, and the y coordinate of
the point is the sine of the angle.
Unit 64: Oblique Triangles: Law of Sines and Law of Cosines
The Law of Sines
a
b
c


sin A sin B sin C
OR
sin A sin B sin C


a
b
c
MTM 2 Final Exam Review
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Sometimes, the Law of Sines can give two possible correct answers. Situations like this are called the
“ambiguous case,” because it is unclear about which triangle to solve unless you have more information.
The two triangles below illustrate the ambiguous case. Notice that to get the obtuse angle in the second
triangle, take the answer from the first application of the law of sines to the first triangle and subtract it
from 180.
The Law of Cosines
cos A 
a 2  b 2  c 2  2bc cos A
OR
OR
b 2  c 2  a 2  2ca cos B
OR
c 2  a 2  b 2  2ab cos C
b2  c 2  a 2
2bc
cos B 
a 2  c2  b2
2ac
OR
a 2  b2  c2
cos C 
2ab
How to solve oblique triangle problems
1. Figure out what dimensions would help you relate the given dimensions to the unknown
dimension.
2. Draw extra lines on the diagram to form triangles that relate one dimension to the next.
a. Always draw lines from the center of a circle to any tangent points.
b. Draw lines to form triangles when you know three of the measurements of the triangle.
3. Solve each right triangle formed until the unknown dimension can be calculated.