Performing Shop Computations (Intermediate)

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Notes Format of Learner’s Guide

Course: Machining NC II
Unit of competency: Perform Shop Computations (Intermediate)
Performing Shop Computations (Intermediate)
Module: At the end of the session, learners should be able to:
Learning outcomes:

1. Perform calculations involving triangles
2. Calculate taper

Duration: 24 Hrs.
Situating Learning: In a fabrication of Plastic Mould Company, one of the customers came and
meets with the manager and supervisor. The customer wants the mould for
the front panel of car stereo, given the product and the mould drawing. After
the meeting, they agreed to finish it in two months. You are a machinist and
your supervisor tasks you to fabricate that mould and then he will give you
an incentive if you finish it in one and a half month. As a machinist you
should know how to interpret drawing, the steps to produce parts of the
mould and to calculate the missing dimensions. Before you start fabricating
the mould, you should:

 Study the drawing first and to figure out the missing dimensions.

 Calculate the missing dimensions so that you will have an idea if the
whole mould fits right with each parts.

Assessment Criteria: After checking the drawing and calculating the missing dimensions, then,
you can start machining the parts considering the safety rules. After finishing
the parts, you can assemble the mould and then test it so that you can
check the product if it meet the required dimensions.
-Performing problems involving right triangles using the trigonometric
functions.
-Performing problems involving non-right triangles using the trigonometric
functions.
-Calculating taper work correctly using appropriate formula.

Learning chunk Performance Criteria Learning Activities Learning documents
(Brief description of (Documents
Sub-task 1:  Two system of strategies, sequence
Determine the system of measurements are of lesson, evaluation) referenced by each
measurements identified learning activity)
1.1.1. Identify the two
 Units use in each system of  Read information
system of measurements and give sheet 1.1 – System
measurements are examples of Measurement
identified - English system
- Metric system  Answer Worksheet
 Units of 1.1.1
measurements are 1.1.2. Identify and
converted discuss the units use to  Assessment:
each system of Oral test
measurements Written test
(questionaire and
answer sheet)

Pedagogical Training in Instructional Design & Delivery for TVET Page 1
© 2010, Institute of Technical Education, Singapore

DLM 03 Developing print-based Learner’s Guide including all related documents
Notes Format of Learner’s Guide

Learning chunk Performance Criteria Learning Activities Learning documents
(Brief description of (Documents
Sub-task 1:  Different geometrical strategies, sequence
Determine unknown shapes are identified of lesson, evaluation) referenced by each
dimensions to calculate learning activity)
 Missing dimensions 1.1.3. Discuss
Sub-task 2: from the drawing are conversion of unit of  Read information
Perform calculations identified measurements and give sheet 1.2 – Different
involving right examples on how to geometrical shapes
triangles convert measurements
from English to Metric  Answer Worksheet
system (vice versa) 1.1.2
- Exercises on
conversion of unit of  Assessment:
measurements. Oral test
1.2.1. Identify and Written test
discuss the different (questionaire and
geometrical shapes answer sheet)

1.2.2. Identify and  Read information
discuss the missing sheet 2.1.1 – Parts
dimensions from the of a triangle
drawing
 Read information
 Parts of a triangle is 2.1.1. Using slides, identify sheet 2.1.2 – History
identified and discuss the parts of a of trigonometry
triangle
 History of  Read information
trigonometry is 2.1.2. Discuss the history sheet 2.1.3 –
explained of trigonometry Trigonometric
function
 The six trigonometric 2.1.3. Discuss the
trigonometric function and  Answer Worksheet
functions are 2.1.1
their relationship
identified - sine  Assessment:
Oral test
- cosine Written test
(questionaire and
- tangent answer sheet)

- cotangent

- secant

- cosecant

 Procedure in solving 2.1.4. Give sample
problem on how to use
a concrete problem is trigonometric function and

listed discuss procedures in

solving a concrete problem

- Discuss how to apply it in

the workplace:

- give drawing with

missing dimension

- sample workpiece to

measure

 Pythagorean theorem 2.1.5. Discuss
Pythagorean theorem in
is used
solving unknown side

- Give example drawing

and apply it

- Give exercises

Pedagogical Training in Instructional Design & Delivery for TVET Page 2
© 2010, Institute of Technical Education, Singapore

DLM 03 Developing print-based Learner’s Guide including all related documents
Notes Format of Learner’s Guide

Learning chunk Performance Criteria Learning Activities Learning documents
(Brief description of (Documents
Sub-task 2: strategies, sequence
Perform calculations of lesson, evaluation) referenced by each
involving non-right learning activity)
triangles
 Read information
Sub-task 3: sheet 2.1.4 –
Calculate Taper Pythagorean
theorem

 Answer Worksheet
2.1.2

 Assessment:

Oral test

Written test

(questionaire and

answer sheet)

 Oblique or non-right 2.2.1. Using slides, discuss  Read information

triangle is defined the definition of oblique or sheet 2.2.1 –
non-right triangle Perform calculation
involving non-right
 The sine rule and - Show examples triangle
2.2.2a. Discuss the general
cosine rule are use to formula for sine rule

calculate the missing

dimension of a 2.2.3aDiscuss the different  Read information

triangle cases of triangles using sheet 2.2.2, 2.2.3,

sine function. and 2.2.4 – Law of
sine
 The different cases of - Give sample problem and
how to apply it
sine and cosine
- written problem
 Answer Worksheet
function are identified - drawing with missing

dimension 2.2.1

2.2.2b. Discuss the general  Read information
formula for cosine rule
sheet 2.2.5, 2.2.6,

2.2.3b. Discuss the and 2.2.7 – Law of
different cases of triangles cosine

using cosine function.

- Give sample problem and  Answer Worksheet

how to apply it 2.2.2

- written problem

- drawing with missing  Assessment:
dimension Oral test

Written test

(questionaire and

answer sheet)

 Taper is defined 3.1.1. Using slides, discuss  Read information

 Taper is calculated about taper sheet 3.1.1 –
correctly using
appropriate formula 3.1.2. Give sample Definition of taper
problem and solve it using
appropriate formula  Read information
sheet 3.1.3 – Taper

calculation

3.1.3. Give sample  Answer Worksheet
workpiece to measure the 3.1.1
taper
 Assessment:

Oral test

Written test

(questionaire and

Pedagogical Training in Instructional Design & Delivery for TVET Page 3
© 2010, Institute of Technical Education, Singapore

DLM 03 Developing print-based Learner’s Guide including all related documents
Notes Format of Learner’s Guide

Learning chunk Performance Criteria Learning Activities Learning documents
(Brief description of (Documents
strategies, sequence
of lesson, evaluation) referenced by each
learning activity)

answer sheet)

:
:

Sub-task N:

Pedagogical Training in Instructional Design & Delivery for TVET Page 4
© 2010, Institute of Technical Education, Singapore

Information Sheet 1.1: Conversion of Measurement
Learning outcomes:
1 Perform calculations involving conversion of unit of measurements
Learning Activity:
1.1 Performing calculations involving conversion of unit of measurements

Objectives:
 Express English and metric equivalent linear units.
 Express English and metric equivalent liquid units.
 Express English and metric equivalent weight units.
 Express English and metric equivalent units of area.
 Express English and metric equivalent units of volume.

The machinist must be able to measure and work accurately with measurement units.
Sometimes the machinist must use equivalent units in the English or metric systems. A good
understanding of the two systems and a set of conversation tables make it easy to express
units in another system.

1.1.1: System of Measurement

1. English system of measurement is commonly used in the United States.

2. Metric system of measurement is much easier to understand than the English system.

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 1

Information Sheet 1.1: Conversion of Measurement

1.1.1.1: English System

1.1.1.1.1: English Linear Units
Linear measurement is the measurement of length. The yard is the standard unit of length in
English system. It is divided into three equal parts, each called a foot, and into thirty-six equal
parts, each called an inch.

English Linear Relationships

Large Unit to Small Unit Small Unit to Large Unit

1 yard = 3 feet 1 foot = 1/3 yard or 0.333 yd.
1 yard = 36 inches
1 foot = 12 inches 1 inch = 1/36 yard or 0.02778 yd.
1 mile = 5,280 feet
1 inch = 1/12 foot or 0.0833 ft.

1 foot = 1/5,280 mile or 0.0001894 mi.

Examples:
1. Express 4 yards in feet.

Since 1 yard = 3 feet, 4 yards = (4 × 3) feet = 12 feet

2. Express 2 feet in yards.

Since 1 foot = 1/3 yard, 2 feet = (2 × 1/3) yard = 2/3 yard or 0.66667 yd.

(Note: Multiply by 1/3 is the same as dividing by 3)

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 2

Information Sheet 1.1: Conversion of Measurement
3. Express 12.5 feet in inches.

Since 1 foot = 12 inches, 12.5 feet = (12.5 × 12) inches = 150 inches

4. Express 24 inches in feet.
Since 1 inch = 1/12 foot, 24 inches = (24 × 1/12) feet = 2 feet
(Note: Multiply by 1/12 is the same as dividing by 12)

5. Express 0.25 feet in inches.
Since 1 foot = 12 inches, 0.25 feet = (0.25 × 12) inches = 4.00 inches

6. Express 0.85 feet in inches.
Since 1 foot = 12 inches, 0.85 feet = (0.85 × 12) inches = 10.20 inches

Notice in all the examples, when large units are changed to small units, the table used is
“Large Unit to Small Unit”. When small units are changed to large units, the table used is
“Small Unit to Large Unit”.

1.1.1.1.2: English Liquid Units
Many liquids are measured by the volume or amount of space they occupy. Certain volumes
are standardized as pints, quarts, and gallons.

English Liquid Relationships
1 gallon (gal.) = 4 quarts (qt.) 1 quart = 1/4 gallon or 0.25 gallon
1 quart (qt.) = 2 pints (pt.) 1 pint = 1/2 quart or 0.5 quart (qt.)
1 gallon (gal.) = 8 pints (pt.) 1 pint = 1/8 gallon or 0.125 gallon

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 3

Information Sheet 1.1: Conversion of Measurement

Examples:

1. Express 3 gallons in quarts.
Since 1 gallon = 4 quarts, 3 gallons = (3 × 4) quarts = 12 quarts

2. Express 3 gallons in pints.
Since 1 gallon = 8 pints, 3 gallons = (3 × 8) pints = 24 pints

3. Express 15 quarts in gallons.
Since 1 quart = 1/4 gallon, 15 quarts = (15 × 1/4) gallon = 15/4 gallon = 3 ¾ gallons

4. Express 0.75 gallon in pints.
Since 1 gallon = 8 pints, 0.75 gallon = (0.75 × 8) pints = 6.00 pints = 6 pints

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 4

Information Sheet 1.1: Conversion of Measurement
1.1.1.1.3: English Weight Units

Certain weights are standardized as ounces, pounds, and tons in the English system of
weights.

1 pound (lb.) = English Weight Relationship 0.0625 lb.
1 ton (T) = 16 ounces (oz.) 1 ounce (oz.) = 1/6 pound or 0.0005 T.
2,000 pounds (lb.) 1 pound (lb.) = 1/2,000 ton or

Examples:

1. Express 68 ounces as pounds.
Since 1 ounces = 1/16 pound, 68 oz. × 1lb./16 oz. = 68/16 oz. = 4 4/16 lb. = 4 ¼ lb. or
4.25 lb.

2. Express 2 tons as pounds.
Since 1 ton = 2,000 pounds, 2 T. × 2,000 lb./T. = 4,000 lb.

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 5

Information Sheet 1.1: Conversion of Measurement

1.1.1.2: Metric System

1.1.1.2.1: Metric Linear Units
The meter is the standard unit of length in the metric system. It is about to 39.37 inches, which
is slightly longer than the Standard English yard. The meter is subdivided into 10 decimeters,
100 centimeters, or 1000 millimeters. It is multiplied by 10 to form a decameter, by 100 to form
a hectometer, or by 1000 to form a kilometer.

Metric Linear Relationships

1 meter (m) = 1000 millimeters (mm) 1 millimeter = 1/1000 meter or 0.001 m

1 meter (m) = 100 centimeters (cm) 1 centimeter = 1/100 meter or 0.01 m

1 meter (m) = 10 decimeters (dm) 1 decimeter = 1/10 meter or 0.1 m

1 kilometer (km) = 1000 meters (m) 1 meter = 1/1000 kilometer or 0.001 km

Examples:
1. Express 2 meters in centimeters.

Since 1 m = 100 cm, 2 m = (2 × 100) cm = 200 cm

2. Express 0.75 kilometer in meters.
Since 1 km = 1000 m, 0.75 km = (0.75 × 1000) = 750 meters

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 6

Information Sheet 1.1: Conversion of Measurement

3. Express 3 decimeters in centimeters.
There is not enough information in the table for this problem. However, the relationship
can be found from the facts given in the table.
If 1 m = 10 dm and 1 m = 100 cm, then 10 dm = 100 cm.
Therefore, 1 dm = 10 cm; 3 dm = (3 × 10) cm = 30 cm

1.1.1.2.2: Metric Liquid Units

Liters and milliliters are common standard units of measurement of liquids in the metric
system.

Metric Liquid Relationships

1 liter (L) = 1000 milliliters (mL) 1 milliliter (mL) = 0.001 liter (L)

Examples:

1. Express 3.5 liters as milliliters.
Since 1 liter = 1000 milliliters, there are 1000 milliliters per liter.
3.5 L = 3.5 L/1 × 1000 mL/L = 3500.0 mL

2. Express 4250 milliliters as liters.
Since 1 milliliter = 0.001 liter, there is 0.001 liter per milliliter.
4250 = 4250 mL/1 × 0.001 L/mL = 4.25 L

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 7

Information Sheet 1.1: Conversion of Measurement

1.1.1.2.3: Metric Weight Units
The units of gram, kilogram, and metric ton are the common standard units of weight in the
metric system.

Metric Weight Relationships

1 kilogram (kg) = 1000 grams (g) 1 gram (g) = 0.001 kilogram (kg)

1 metric ton (t) = 1000 kilograms (kg) 1 kilogram (kg) = 0.001 metric ton (t)

Examples:

1. Express 2.5 kilograms in grams.
Since 1 kilogram = 1000 grams, there are 1000 grams per kilogram.
2.5 kg = 2.5 kg/1 × 1000 g/kg = 2500 g

2. Express 4250 grams in kilograms.
Since 1 gram = 0.001 kilogram, there is 0.001 kilogram per gram.
4250 grams = 4250 g/1 × 0.001 kg/g = 4.25 kg

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 8

Information Sheet 1.1: Conversion of Measurement

1.1.1.2.4: Metric Units Of Area
Units of area describe a surface having
length and width. In the metric system, area
is expressed in square meters, square
decimeters, square centimeters, etc. One
square meters describes a surface one
meter long and one meter wide. One square
decimeter describes a surface one
decimeter long and one decimeter wide. A
square centimeter describes a surface one
centimeter long and one centimeter wide.

One Square Meter

Metric Area Relationships

1 square meter (m²) = 100 square decimeters (dm²)

1 square meter (m²) = 10000 square centimeters (cm²)

1 square meter (m²) = 1000000 square millimeters (mm²)

1 square decimeter (dm²) = 100 square centimeters (cm²)

1 square decimeter (dm²) = 10000 square millimeters (mm²)

1 square centimeter (cm²) = 100 square millimeters (mm²)

1 square decimeter (dm²) = 1/100 square meter or 0.01 m²

1 square centimeter (cm²) = 1/10000 square meter or 0.0001 m²

1 square millimeter (mm²) = 1/1000000 square meter or 0.000001 m²

1 square centimeter (cm²) = 1/100 square decimeter or 0.01 dm²

1 square millimeter (mm²) = 1/10000 square decimeter or 0.0001 dm²

1 square millimeter (mm²) = 1/100 square centimeter or 0.01 cm²

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 9

Information Sheet 1.1: Conversion of Measurement
Examples:

1. Express 3 square meters as square decimeters.
Since 1 m² = 100 dm², 3 m² = (3 × 100) dm² = 300 dm²

2. Express 7.5 square centimeters as square millimeters.
Since 1 cm² = 100 mm², 7.5 cm² = (7.5 × 100) mm² = 750 mm²

3. Express 5000 square millimeters as square decimeters.
Since 1 mm² = 1/10000 dm², 5000 mm² = (5000 × 1/10000) dm² = 0.5 dm²

4. Express 750 square decimeters as square meters.
Since 1 dm² = 1/100 m², 750 dm² = (750 × 1/100) m² = 7.5 m²

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 10

Information Sheet 1.1: Conversion of Measurement
1.1.1.3: Equivalent Units in the English and Metric System

1.1.1.3.1: Equivalent English and Metric Linear Units

The following table shows the relation between some common English and metric linear units.
The correct conversion factor must be used.

Metric to English English to Metric
1 m = 39.37 in. 1 in. = 25.4 mm
1 m = 3.28 ft. 1 in. = 2.54 cm
1 m = 1.09 yd. 1 in. = 0.254 dm
1 mm = 0.039 in. 1 in. = 0.0254 m
1 cm = 0.39 in. 1 ft. = 0.305 m
1 dm = 3.937 in. 1 ft. = 3.05 dm
1 km = 1,093.61 yd. 1 yd. = 0.9144 m
1 km = 0.621 mi. 1 mi. = 1.6093 km

Examples:
1. Express 6 feet in meters.

Since 1 ft. = 0.305 m, 6 ft. = (6 × 0.305) m = 1.830 m

2. Express 12 miles in kilometers.
Since 1 mi. = 1.6093 km, 12 mi. = (12 × 1.6093) km = 19.3116 km

3. Express 2.5 decimeters in inches.
Since 1 dm = 3.937 in., 2.5 dm = (2.5 × 3.937) in. = 9.8425 in.

4. Express 0.5 kilometer in miles.
Since 1 km = 0.621 mi., 0.5 km = (0.5 × 0.621) mi. = 0.3105 mi.

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 11

Information Sheet 1.1: Conversion of Measurement

1.1.1.3.2: Equivalent English and Metric Liquid Units
In the English system, pints, quarts, and gallons are the common units used to measure liquid
volume. The liter is used for liquid volume in the metric system. Gasoline, oil, or any liquid can
be measured by liters.

Metric to English English to Metric

1L = 0.264 gal. 1 gal. = 3.7853 L
1L = 1.06 qt. 1 qt. = 0.9463 L
1L = 2.12 pt. 1 pt. = 0.4732 L

Examples:
1. Express 3 quarts as liters.

Since 1 qt. = 0.9463 L, 3 qt. = (3 × 0.9463) L = 2.8389 L

2. Express 2.5 liters as pints.
Since 1 L = 2.12 pt., 2.5 L = (2.5 × 2.12) pt. = 5.3 pt.

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 12

Information Sheet 1.1: Conversion of Measurement

1.1.1.3.3: Equivalent English and Metric Units of Weight
The following table shows the relation between some common English and metric units of
weight.

Metric to English English to Metric

1g = 0.03527 oz. 1 oz. = 28.34953 g
1 kg = 2.205 lb. 1 lb. = 0.45359 kg
1t = 2,205 lb. 1 lb. = 0.00045 t

Examples:
1. Express 5.2 kilograms in pounds.

Since 1 kg = 2.205 lb., 5.2 kg = 5.2 kg/1 × 2.205 lb./kg = 11.4660 lb.

2. Express 20 pounds in kilograms.
Since 1 lb. = 0.45359 kg, 20 lb. = 20 lb./1 × 0.45359 kg/lb. = 9.071 kg

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 13

Information Sheet 1.1: Conversion of Measurement

1.1.1.3.4: Equivalent English and Metric Units of Area
Area describes a surface having length and width. In the English system area is measured in
square units. This includes square inches, square feet, and square yards. In the metric system
area is measured in square meters, square centimeters, etc. The following table shows the
relation between some English and metric units of area.

Metric to English English to Metric

1 mm² = 0.00155 sq. in. 1 sq. in. = 645.2 mm²

1 cm² = 0.1550 sq. in. 1 sq. in. = 6.452 cm²

1dm² = 0.107639 sq. ft. 1 sq. ft. = 9.29 dm²

1 m² = 10.7639 sq. ft. 1 sq. ft. = 0.0929 m²

1 m² = 1.196 sq. yd. 1 sq. yd. = 0.8361 m²

1 km² = 0.3861 sq. mi. 1 sq. mi. = 2.59 km²

Note: Area units in the metric system are written with an exponent, such as mm². This is not
used in English system.

Examples:
1. Express 45 square meters in square yards.

Since 1 m² = 1.196 sq. yd., 45 m² = (45 × 1.196) sq. yd. = 53.82 sq. yd.

2. Express 2 ½ square feet in square meters.
Since 1 sq. ft. = 0.0929 m², 2 ½ sq. ft. = (2 ½ × 0.0929) m² = 0.23225 m²

3. Express 30.25 square centimeter in square inches.
Since 1 cm² = 0.1550 sq. in., 30.25 cm ² = (30.25 × 0.1550) sq. in. = 4.68875 sq. in.

4. Express 0.75 square mile in square kilometers.

Since 1 sq. mi. = 2.59 km²; 0.75 sq. mi. = (0.75 × 2.59) km² = 1.9425 km²

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 14

Information Sheet 1.1: Conversion of Measurement

1.1.1.3.5: Equivalent English and Metric Units of Volume
Volume describes a space having length, width, and height. In the English system volume is
measured in cubic units. These include cubic inches, cubic feet, and cubic yards. In the metric
system volume is measured by cubic meters, cubic centimeters, liters, etc. The following table
shows the relation between some English and metric units of volume.

Metric to English English to Metric

1 cm³ = 0.0610 cu. in. 1 cu. in. = 16.39 cm³
1 cm³ = 35.31 cu. ft. 1 cu. ft. = 0.0283 m³
1 cm³ = 1.31 cu. yd. 1 cu. yd. = 0.7646 m³
1L = 61.0237 cu. in. 1 cu. in. = 0.0164 L
1L = 0.0353 cu. ft. 1 cu. ft. = 28.32 L

Examples:

1. Express 40 cubic centimeters in cubic inches.
Since 1 cm³ = 0.061 cu. in., 40 cm³ = (40 × 0.061) cu. in. = 2.44 cu. in.

2. Express 2 ¼ cubic feet in cubic meters.
Since 1 cu. ft. = 0.0283 m³, 2 ¼ cu. ft. = (2 ¼ × 0.0283) m³ = (2.25 × 0.0283) m³ =
0.063675 m³

3. Express 7.05 cubic yards in cubic meters.
Since 1 cu. yd. = 0.7646 m³, 7.05 cu. yd. = (7.05 × 0.7646) m³ = 5.39 m³

4. Express 16.5 liters in cubic feet.

Since 1 L = 0.0353 cu. ft., 16.5 L = (16.5 × 0.0353) cu. ft. = 0.58245 cu. ft.

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 15

Worksheet 1.1.1: Conversion of Measurement

Learning outcomes:
1 Perform calculations involving conversion of unit of measurements
Learning Activity:
1.2 Performing calculations involving conversion of unit of measurements

Calculate each of the following as indicated.

1. 48 in. = __________ ft.

2. 9 ft. = __________ yd.

3. 8 quarts = __________ pints

4. 24 pints = __________ gallons

5. 0.5 gallon = __________ quarts

6. 40 oz. = __________ lb.

7. 500 lb. = __________ T.

8. 4000 mm = __________ m

9. 18 cm = __________ mm

10. 10 cm = __________ mm

11. 3500 m = __________ km

12. 0.625 m = __________ dm

13. 2 L = __________ mL

14. 5000 mL = __________ L

Code No. Servicing Starting System Date: Developed Date: Revised Page #
ALT723307
Nov. 28, 2003 Mar 01, 2006 1

Worksheet 1.1.1: Conversion of Measurement

15. 3 kg = __________ g

16. 4000 kg = __________ t

17. 19 cm² = __________ mm²

18. 25 m² = __________ cm²

19. 3400 cm² = __________ dm²

20. 4 m = __________ in.

21. 23.5 m = __________ yd.

22. 18 ft. = __________ mi.

23. 9 pt. = __________ L

24. 4 L = __________qt.

25. 30 L = __________ gal.

26. 75 g = __________ oz.

27. 4 t = __________ lb.

28. 10 lb. = __________ kg
__________ sq. in.
29. 9 cm² = __________ sq. ft.
__________ sq. in.
30. 86.25 m² = __________ km²
__________ sq. yd.
31. 2540 mm² =

32. 20 sq. mi. =

33. 9.1 m² =

Code No. Servicing Starting System Date: Developed Date: Revised Page #
ALT723307
Nov. 28, 2003 Mar 01, 2006 2

Worksheet 1.1.1: Conversion of Measurement

34. 18 cu. in. = __________ cm³
__________ m³
35. 4.25 cu. yd. = __________ cu. ft.
__________ L
36. 280 m³ = __________ L
__________ m²
37. 144 cu. in. = __________ kg

38. 2.5 cu. ft. =

39. 34.5 sq. yd. =

40. 48 lb. =

Code No. Servicing Starting System Date: Developed Date: Revised Page #
ALT723307
Nov. 28, 2003 Mar 01, 2006 3

Information Sheet 1.2: Geometrical Shapes

Learning outcomes:
1 Perform calculations involving different geometrical shapes.
Learning Activity:
1.1 Performing calculations involving different geometrical shapes.

Objectives:
 Identify the units of angular measurements and the kinds of angles.
 Add and subtract angular measurements.
 Identify the different kinds of triangle and calculation of areas.
 Identify quadrilateral and calculation of areas.
 Identify parts of circle and formulas.

The theories of geometry may be approached in one of two ways – from the stand point of
mathematics or from the standpoint of practical application. In this lesson the latter method is
used because of the great importance of geometrical principles in technology and industry.
Hardly a machine is designed or an article constructed without some application of geometrical
theorems.

1.2.1: The Angle

An angle is formed by two lines which meet
at a point called a vertex. The two lines are
the sides of the angle. The size of an angle
is determined by the number of degrees
between the sides of the angle it is not
depend on the length of its sides.

Figure 1

Code No. Performing Shop Computations Date: Developed Date: Revised Page #
MEE722207 (Intermediate) 1

Information Sheet 1.2: Geometrical Shapes
1.2.2: Units of Angular Measurement

A degree is the unit of measurement for angles. The symbol of degree is °. A circle shows a
complete rotation of the end point of a radius.

Figure 2: A circle has 360 degrees

The degree is divided into 60 equal parts called minutes. Each minute is divided into 60 equal
parts called seconds. These further divisions allow greater accuracy when measuring angles.
The auto mechanic is generally concerned with degrees and minutes or degrees and fractional
degrees.

1 circle = 360 degrees (°)
60 minutes (‘)
1 degree (°) = 60 seconds (“)

1 minute (‘) =

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Information Sheet 1.2: Geometrical Shapes
1.2.3: Kinds of Angles

A right angle is a 90° angle, it is, therefore, one fourth of a complete rotation of 360° (1/4 x
360° = 90°). An acute angle measures between 0° to 90°. An obtuse angle measures between
90° to 180°. A straight angle measures 180°, it is one half a complete rotation and form a
straight line (1/2 x 360° = 180°).

1.2.4: The protractor

The protractor is an instrument used to measure angles and to draw angles. The protractor
has two scales. Each scale is graduated from 0° to 180° and can be read from either right or
left side. The vertex of the angle to be measured is place at the middle of the base of the
protractor. This point is marked on the protractor.

Figure 3: A Protractor

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Information Sheet 1.2: Geometrical Shapes
1.2.5: Addition of Angular Measurements

In order to find the tolerance limits on angles, the mechanic adds and subtracts angular
measurements. When adding angular measurements, add like units and simplify. Remember
that 1 degree (°) = 60 minutes (‘); 1 minute (‘) = 60 seconds (“).

Examples: 4. 7° 30’ 50”
+ 2° 30’ 40”
1. 4° ------------------
+ 20° 9° 60’ 90” = 10° 1’ 30”
-----------
24° 5. 5° 40”
+ 4° 20’ 40”
2. 15° 10’ --------------------
+ 16° 40’ 9° 20’ 80” = 9° 21’ 20”
--------------
31° 50’

3. 10° 30’
+ 5° 40’
---------------
15° 70’ = 16° 10’

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Information Sheet 1.2: Geometrical Shapes
1.2.6: Subtraction of Angular Measurements

When subtracting angular measurements, subtract like units. Exchange when necessary, but
keep in mind if you exchange 1 degree, it equals 60 minutes. And if you exchange 1 minute, it
equals 60 seconds.

Examples: 4. 23° 10’ = 22° 70’
- 16° 20’ = 16° 20’
1. 75° --------------- -----------
- 20° 6° 50’

----------
55°

2. 34° 40’ 5. 75° 20’ 10” = 74° 79’ 70”
- 14° 20’ - 24° 30’ 30” = 24° 30’ 30”
--------------------- -----------------
--------------- 50° 49’ 40”
20° 20’
In example 5, 1 degree is exchanged and
3. 45° 44° 60’ also 1 minute is exchanged so that seconds
can be subtracted. In this case the steps are
- 30° 40’ = 30° 40’ 75° 20’ 10” = 74° 80’ 10” = 74° 79’ 70”.

--------------- ------------
14° 20’

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Information Sheet 1.2: Geometrical Shapes

1.2.7: The Triangle

A triangle is a plane figure bounded by three straight sides. Every triangle contains three
angles whose sum equals 180°; consequently, if we know the value of any two angles of a
triangle, we can find the value of the third angle. Figure 4 shows a triangle with the value of
each angle. The sum of these three angles equals 180°.

Figure 4

The altitude of a triangle is the perpendicular drawn from the vertex to the base. The base is
the side upon which the triangle rests. Since the triangle can be moved to make any side the
base, a triangle can have three altitudes, each upon appropriate base.

1.2.7.1: The Isosceles Triangle
An isosceles triangle has two equal sides. In the isosceles triangle, figure 5, side AB = side
AC. Furthermore, in an isosceles triangle the angle B is equal to the angle C. The altitude
drawn from the vertex to the third side of the isosceles triangle bisects the third side and forms
two equal right triangles.

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Information Sheet 1.2: Geometrical Shapes

Figure 5

1.2.7.2: The Right Triangle
A right triangle contains a right angle. Figure 6 shows a right triangle. The side AB opposite the
right angle is called the hypotenuse of the right triangle. The side CB is the altitude, and the
side AC is the base of the right triangle. By the definition of altitude, either BC or AC could be
the altitude.

Figure 6

The right triangle is very important in the shop practice and design. In figure 6, the angle C is
the right angle; consequently, it is equal to 90°. Since the sum of the three angles of any
triangle equals 180°, angle A and angle B would be acute angles whose sum equals 90°.

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Information Sheet 1.2: Geometrical Shapes

If we are given the value of one acute angle of a right triangle, we can find the value of the
angles of the right triangle. If the value of one acute angle of a right triangle equals 40°, the
value of the other angles would be 90° and 50°.
To find the relation between the three sides of a right triangle, we employ the Pythagorean
Theorem, which one of the most widely quoted theorem of geometry. This theorem states that
the square of the hypotenuse equals the sum of the squares of the other two sides.
In figure 7, squares are constructed upon the hypotenuse, altitude, and the base of the right
triangle ABC. The hypotenuse contains 5 units, the base 3, and the altitude 4. Hence,
according to the Pythagorean Theorem,

5² = 4² + 3²
25 = 16 + 9

Figure 7

Figure 8

In figure 8, let us call the hypotenuse c, the altitude a, and the base b. Then, the relation can
be expressed thus, (hypotenuse)² = (altitude)² + (base)², or c² = a² + b².
Extracting the square root of both sides of the equation, we obtain c = √ a² + b². Like roots (in
this case the square root) of equals are equal.

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Information Sheet 1.2: Geometrical Shapes

This formula will enable us to find the hypotenuse of a right triangle when the altitude and base
are given.
By transposing the equation c² = a² + b² so that we have a² = c² – b², the formula

a = √ c² – b² is obtained. This formula will enable us to find the altitude of a right triangle when
the hypotenuse and base are given. In similar fashion we can obtain the formula
b = √ c² – a².

The three formulas are as follows:

hypotenuse (c) = √ a² + b²

altitude (a) = √ c² – b²

base (b) = √ c² – a²

Example 1:
Find the hypotenuse of a right triangle whose
altitude is 9 ft. and whose base is 12 ft. (fig 9)
hypotenuse (c) = √ a² + b²

c = √ 9² + 12²
c = √ 81 + 144
c = √ 225
c = 15 ft.

Figure 9

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Information Sheet 1.2: Geometrical Shapes

Example 2:

Find the altitude of a right triangle whose

hypotenuse is 13 ft. and whose base is 5 ft.
(fig 10)

altitude (a) = √ c² – b²

a = √ 13² – 5²

a = √ 169 – 25

a = √ 144

a = 12 ft.

Figure 10

1.2.7.3: The Equilateral Triangle

The equilateral triangle has three sides equal and three angles equal. Since the sum of the
three angles in any triangle equals 180°, each angle of an equilateral triangle equals 60°. The
altitudes drawn from any vertex of an equilateral triangle divides the triangle into two equal
right triangles, and at the same time the altitude bisects the side to which it is drawn. In figure
11, the altitude AD divides the equilateral triangle into two equal right triangles. CD is equal to
DB, and the angle CAB is bisected, thus making angle CAD and angle BAD each equal to 30°.
The altitude of an equilateral triangle can be found when one side is known, figure 12. If a
stands for the altitude and s stands for a side, then:

s²
a = s² – -------

2

s²
a = s² – -------

4

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Information Sheet 1.2: Geometrical Shapes

3 s²
a = -------

4

a=½s√3

Figure 11 Figure 12

When the altitude of an equilateral triangle is given, we can find a side by using the formula
a = ½ s √ 3 thus:
a = ½s√3

2a √ 3
s = ---------- × ---------- = 2/3 a √ 3

√3 √3

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Information Sheet 1.2: Geometrical Shapes
Example:
Find s in an equilateral triangle if a equals 12 cm.

2/3 a √ 3 = 2/3 × 12 × 1.732 = 13.856 cm

1.2.7.4: The Scalene Triangle
In a scalene triangle no two sides are equal, and none of the angles of the triangle equals 90°.
See figure 13.

Figure 13

1.2.7.5: Special Triangles

There are three special right triangles used in the shop practice and engineering design.
1. The 30° – 60° – 90° triangle. See figure 14 (a). The side opposite the 30° angle is one half

the hypotenuse. The draftsman’s triangle is a 30° – 60° – 90° triangle.

2. The 45° – 45° – 90° triangle. See figure 14 (b). The two sides which included the 90° angle
are equal. It is often called the isosceles right triangle.

3. The 3-4-5 right triangle. The three sides must be in ratio of 3, 4, and 5. For example, 6, 8,
and 10 are in the proper ratio. Check, using the Pythagorean Theorem, 3² + 4² = 5². This
can be applied when it is required to square the corners of a plate or board. For example, in
figure 14 (c), it is required to square the plate at C. First, with C as a center mark off 4 units
on CD. With C as a center and a radius equal to 3 units, draw arc E. With A as a center and

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Information Sheet 1.2: Geometrical Shapes

a radius equal to 5 units, draw arc F intersecting arc E at point B. Draw CB extended to the
top, forming the 90 angle CD. Triangle ABC is a 3-4-5 right triangle.

Figure 14

1.2.8: Area of Triangle
The area of a triangle can be found by taking one half the product of the base and altitude.
Stated in a formula it would be written thus:

Area = ½ × attitude × base or A = ½ ab
By the use of this formula we can find the area of any triangle if we know the value of the
altitude and base.

Example 1:
Find the area of a triangle whose base equals 8 cm, and altitude equals 6 cm.

A = ½ ab
A = ½ × 6 × 8 = 24 cm²

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Information Sheet 1.2: Geometrical Shapes Figure 15

Example 2:
Find the area of the right triangle, figure 15.
By Pythagorean theorem:

a = √ 10² – 5²
a = √ 100 – 25

a = √ 75
a = 8.66 in.
Then:
A = ½ ab
A = ½ × 8.66 × 5
A = 21.65 sq. in.

1.2.9: The Quadrilateral
A quadrilateral is a plane figure bounded by four sides. The quadrilaterals are the
parallelogram, rectangle, square, and trapezoid.

1.2.9.1: Parallel lines
Parallel lines are in the same plane and do not meet if they are extended.
1.2.9.2: Parallelogram
A parallelogram is a quadrilateral whose opposite sides are parallel. See figure 16.
The opposite sides and angles of a parallelogram are equal, and the sum of the angles equals
360°.
The diagonal of a parallelogram joins any two opposite vertices and divides the parallelogram
into two equal triangles.
The altitude of a parallelogram is the perpendicular distance between the base and opposite
side.

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Information Sheet 1.2: Geometrical Shapes

The formula which gives the area of a parallelogram when the base and altitude are given is
stated:

Area = altitude × base Figure 16
A = ab

Then:
b=A÷a
a=A÷b

1.2.9.3: The Rectangle
A rectangle is a parallelogram which has four angles, each equal to 90° (a right angle). Figure
17.
The diagonal of a rectangle divides the rectangle into two equal right triangles.
The following formulas apply to the rectangle:

Area = altitude × base
or A = ab

Then:

b=A÷a Figure 17
a=A÷b
d = √ b² + a²

In this formula d is the diagonal.
See figure 17.

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Information Sheet 1.2: Geometrical Shapes

1.2.9.4: The Square
A square is a parallelogram whose angles are all right angles and whose sides are equal. See
figure 18.
The diagonal of a square divides the square into two equal 45° – 45° – 90° right triangles.
The following formulas apply to the square in
which s denotes the side and d denotes the
diagonal:

Area = s²
d = s√2
s = ½d√2

Figure 18

1.2.9.5: The Trapezoid
A trapezoid is a quadrilateral which has two sides parallel. The two parallel sides are called the
bases. The two nonparallel sides are called the legs.
The altitude of a trapezoid is the perpendicular distance between the bases. See figure 19.
The sum of the four angles of a trapezoid equals 360°. We can see that the sum of the four
angles of the quadrilaterals thus far discussed equals 360°. This is true in any quadrilateral.
The formula for the area of a trapezoid is given below, where A stands for the area of the
trapezoid, a the altitude, b the bottom base, and b' the top base.

A = ½ a ( b + b' )

Figure 19

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Information Sheet 1.2: Geometrical Shapes

Example:
Find the area of the trapezoid whose altitude is 1.5 dm and bases 4 dm and 2 dm.

A = ½ a ( b + b' )
A = ½ × 1.5 ( 4 + 2)
A = 0.75 × 6
A = 4.50 dm²

1.2.10: The Circle
A circle is a plane figure bounded by a curved line every point of which is equally distant from a
point called the center.
The curve line is called the circumference of the circle
A line drawn through the center of a circle and terminating in the circumference is called the
diameter. The diameter of a circle divides the circle into two equal parts is called semicircles.
A line drawn from the center of a circle to the circumference is called the radius. The radius of
a circle is equal to one half the diameter.
In figure 20, o is the center of the circle, AB is the diameter, oc is the radius, and the closed
curved line is the circumference.

Figure 20

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Information Sheet 1.2: Geometrical Shapes

1.2.10.1: The Measurements of the Circle
The ratio between the diameter and circumference of a circle is equal to a constant called pi
( π ). This constant π is equal to approximately 22/7. For more accurate computations the
value of π is taken as 3.1416. This means that the circumference of a circle is 3.1416 times the
diameter. Since the diameter of a circle is 2 times the radius, the circumference of a circle
would also be 2 times the radius times 3.1416.
If c stands for the circumference of a circle, r for the radius, and d for the diameter, the
following formulas show the relation between the circumference, diameter, and radius:

d = 2r
c = πd
c =2πr
From these three formulas, the following relations can be obtained:
r=d÷2
d=c÷π
r = c ÷ (2π)

Example 1:
Find the circumference of a circle whose diameter is 5 cm.

c = πd
c = 3.1416 × 5
c = 15.708 cm

Example 2:
Find the radius of a circle whose circumference is 43.98 cm.

r = c ÷ 2π
r = 43.98 ÷ ( 2 × 3.1416 )
r = 7 cm

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Information Sheet 1.2: Geometrical Shapes

1.2.10.2: The Area of a Circle
The area of a circle is found by multiplying π by the square of the radius. Another method for
finding the area of a circle is used when the diameter of the circle is given. In the second
method one fourth of π (0.7854) is multiplied by the square of the diameter.
When A stands for the area of a circle, r for the radius, and d for the diameter, the following
formulas show the relation between the area, diameter, and radius:

A = π r²
A = ( π d² ) ÷ 4 = 0.7854 d²
r = √A÷π
d = 2√A ÷ π

Example 1:
Find the area of a circle whose radius is 100 mm.

A = π r²
A = 3.1416 × 100² = 31416 mm²

Example 2:
Find the area of a circle whose diameter is 60 cm.

A = ( π d² ) ÷ 4
A = ( 3.1416 × 3600 ) ÷ 4
A = 2827.44 cm²

Example 3:

Find the diameter of a circle whose area is 78.54 dm².
d = 2√A ÷ π

d = 2 √ 78.54 ÷ 3.1416

d = 2 √ 25

d = 10 dm.

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Information Sheet 1.2: Geometrical Shapes

1.2.10.3: A Sector of a Circle

A sector of a circle is the part of a circle bounded by two radii and an arc. Figure 21 shows the
sector AOB.

The angle AOB is called a central angle, since Figure 21
by definition a central angle of a circle is the
angle included between the radii. The central
angle contains the same number of degrees
as the arc which it intercepts. If, in the figure,
arc AB contains 60°, the central angle AOB
also contains 60°. The circumference of a
circle contains 360°; consequently, the area of
the sector is the same part of the area of the
circle as the arc is of the circumference. If the
angle AOB is 60°, the area of the sector AOB
would be 60/360 of the area of the circle.
If α denotes the number of degrees in the
central angle, r the radius of the circle and d
the diameter of the circle,

the area of a sector is expressed by the following formulas:
A = ( π r² ) × ( α ÷ 360 )
A = [( π d² ) ÷ 4 )] × ( α ÷ 360 )

Example:

If a circle whose diameter is 60 mm contains a sector with a central angle of 45°, what is the
area of the sector?

A = [( π d² ) ÷ 4 )] × ( α ÷ 360 )

A = [( 3.1416 × 60² ) ÷ 4 )] × ( 45 ÷ 360 )

A = [( 3600 ) ÷ 4 )] × ( 0.3927 )

A = 900 × 0.3927

A = 353.43 mm²

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Worksheet 1.1.1: Geometrical Shapes
Learning outcomes:
1 Perform calculations involving different geometrical shapes.
Learning Activity:
1.2 Performing calculations involving different geometrical shapes.

Add the following.

1. 11° 40' 20" + 17° 30' 30"

2. 23° 36' 7" + 4° 48'
3. 100° 39' + 68° 42' 25"
4. 17° 43' + 11° 21'
5. 51° 40' 40" + 3° 30' 45"

Subtract the following.

1. 10° – 4° 30'

2. 17° 31' – 10° 45'
3. 72° 9' 16" – 60° 8' 11"
4. 26° 40' 16" – 11° 50' 30"
5. 17° 40' 10" – 5° 40' 20"

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Worksheet 1.1.1: Geometrical Shapes
1. Find the value of the third angle in the following triangles.
(a) (b)

(c) (d)

(e) (f)

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Worksheet 1.1.1: Geometrical Shapes
2. If one acute angle of a right triangle is 18° 42', what is the value of the other angle?
3. In the isosceles triangle in figure 1, find
the value of each angle marked (?).

Figure 1
4. In figure 2, the plate is to be burned off along line AB, which can be marked off by

measuring BC. How long is BC?

Figure 2

5. Find the area of the parallelogram whose base is 2 ft. 9 in. and altitude is 5 ft. 4 in.

6. The area of rectangle is 108 sq. ft. and the base is 6 ft. 6 in. What is the value of the
altitude?

7. Find the area of a trapezoid whose bottom base is 6 dm, top base is 4 dm and the
altitude is 3 dm.

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Worksheet 1.1.1: Geometrical Shapes

8. Find the radius of a circle whose circumference is 12.75 cm.
9. The central angle of a sector is 30° and the radius is 203.2 mm. Find the area of the

sector.
10. In a circle of 365 cm diameter, what is the area of the sector whose angle is 60°?

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Information Sheet 2.1: Calculation of Triangle

Learning outcomes:
1 Perform calculations involving right triangles
Learning Activity:
1.1 Performing calculations involving right triangles

Objectives:
 Identify the parts of a triangle.
 Learn the history of trigonometry.
 Identify the six trigonometric functions.
 Apply the Pythagorean Theorem in solving the unknown side.

2.1.1: A triangle (right or non- right triangle) has six parts; three sides and three angles. To
solve a triangle is to find the unknown parts from the parts that are known.

C

ab

Right Triangle AB
c

Non-Right Triangle

Right Triangle – is a triangle with right angle.

In solving a concrete problem on right triangles we may proceed as follows:

1. Draw a sketch of the required figure and triangle as accurately as possible according to
the data given. Label the parts. This drawing will show the practical situation. Show the
given and unknown quantities in relation to one another. And provide a rough scale
drawing which may serve as a rough check for the answer obtained.

2. State the given and required parts.

3. From the six trigonometric function, use the one which contains the unknown and the

known parts.

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Information Sheet 2.1: Calculation of Triangle

2.1.2: Trigonometry

Trigonometry is a branch of mathematics that deals with the relationships between the sides
and angles of triangles and with properties and applications of the trigonometric functions of
angles. Trigonometry comes from three Greek words; Tri (three), gonia (corner or angle), and
metron (measures). The earliest applications of trigonometry were in the fields of navigation,
surveying, and astronomy, in which the main problem generally was determine an inaccessible
distance, such as the distance between the earth and the moon, or of a distance that could not
be measured directly, such as the distance across a large lake. Other applications of
trigonometry are found in Physics, Chemistry, and almost all branches of Engineering.

2.1.3: Trigonometric Functions

In angle A: In the figure, Acute angle A (less than 90°)
determine by initial ray AC and terminal ray
θ = angle “theta” AB. A line dropped from point B to C.
c = hypotenuse
a = opposite side
b = adjacent side

 Tangent (tan) Function CB

The tangent function is the ratio of lines Tan A = ---------

CB and AC. AC

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Information Sheet 2.1: Calculation of Triangle

 Sine (sin) Function CB

It is the ratio of lines opposite side CB Sin A = ---------

and hypotenuse AB of a right triangle. AB

 Cosine (cos) Function AC
It is the ratio of adjacent side AC and Cos A = ---------
hypotenuse AB of aright triangle.
AB

 Cotangent (cot) Function AC
It is the ratio of adjacent side AC and Cot A = ---------
opposite side CB of a right triangle.
CB

 Secant (sec) Function AB
It is the ratio of hypotenuse AB and Sec A = ---------
opposite side AC.
AC

 Cosecant (csc) Function AB
It is the ratio of hypotenuse AB and Csc A = ---------
opposite side CB.
CB

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Information Sheet 2.1: Calculation of Triangle
Example:

1. From the top of the tower at C the angle of depression of point A is 56°. If the distance
AB is 180 meters, how high is the tower?

Given:
AB = 180 m
Angle of Depression
α = 56°

Required:
Height of the tower (H)

Solution:
tan 56° = H / 180

H = 180 × tan 56°

H = 267 m

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