Electronics
AC Waveforms
1
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Specific Objectives





Understand how a sine wave of alternating
voltage is generated.
Explain the three ways to express the amplitude
of a sinusoidal waveform and the relationship
between them.
Calculate the RMS, average, and peak-to-peak
values of a sine wave when the peak value is
known.
Calculate the instantaneous value of a sine wave.
Convert peak, peak-to-peak, average, and RMS
voltage and current values from one value to
another.
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2
Specific Objectives (Continued)
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
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Explain the importance of the .707 constant and
how it is derived.
Define frequency and period and list the units of
each.
Calculate the period when the frequency is
known and frequency when period is known.
Explain the sine, cosine, and tangent
trigonometric functions.
Calculate the value of the sine of any angle
between 0° and 360°.
3
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Introduction



AC and the characteristics of a sinusoidal
waveform
Time and frequency measurement of a
waveform
Trigonometric functions
4
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What is AC?

Alternating Current (AC) is a useful form of
voltage.


Direct Current (DC) is another useful form of
voltage.


AC is a flow of electric charge that periodically
changes direction.
DC is the unidirectional flow of electric charge.
AC and DC are both useful power sources for
heating, lights, and motors.
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AC vs DC

DC can power the same things AC can power.


The devices may have to be constructed
differently, but they would work the same.
AC cannot power some of the things that DC
can power.


Example: electronic devices
AC motors were not invented at the time Thomas
Edison built his first DC power station in 1882.
6
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Why Use AC Instead of DC?

AC is relatively easy to produce.



It is created through rotational motion.
AC generation can produce large amounts of
power economically.
AC voltage can be changed from one value to
another relatively easily.

DC voltage cannot be changed from one value to
another easily.
7
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Why Change Voltage?




AC voltage can be stepped up to a higher
value using a transformer (up to 760 kV).
The basic concept is that “power in” equals
“power out” of a transformer.
Power is voltage times current, so if voltage is
made to increase, then current will decrease.
After the voltage is increased, AC is much
more efficient to transmit over long distances.
8
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Why is AC More Efficient?


Electrical power (voltage and current) is sent over
transmission wires from a generator to users.
Lower current sent through those transmission
wires will produce less heat (Joule’s Law).

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Fewer moving electrons reduce the amount of friction.
Voltage does not create heat.
Less heat means less power is lost during
delivery.
We can use thinner wires for power transmission.

Thin wires are more economical.
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9
How Do You Make AC?



Most electricity is produced by induction.
In a generator, induction occurs when a
conductor moves through a magnetic field.
AC is the type of electricity generated by a
conductor moving in a circle through a static
magnetic field.


Circular motion is easy to produce.
Like a wheel going around, it is a simple and
efficient process.
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Rotary Motion


The process of using a water wheel as a
mechanical power source for milling and
sawing has been used for thousands of years.
A water wheel produces circular motion.



This is also called a water-driven turbine.
This process was applied to creating electricity
in the late 1800s.
Two more things were necessary to make AC
power viable: the alternator and the
transformer.
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Mechanical Power Source

An electrical power plant has a capacity, but
the actual amount of power produced is a
function of user demand.


Higher user demand creates a larger load on the
generator.
The generator then needs to draw more
mechanical power from the prime mover.

If the prime mover cannot provide the additional
mechanical power, the plant will shut down.
12
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Hydroelectric Power


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The Ames Hydroelectric Generating Plant was
the world's first commercial system to produce
and transmit AC electricity for industrial use.
In 1890, Westinghouse Electric supplied the
station's generator and motor.
The AC was proven to be effective as it was
transmitted two miles (3 km) at a loss of less
than 5%.

The maximum distance for DC was about a mile.
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AC Generation


In an AC generator, one rotation of the rotor
shaft creates one cycle of voltage.
This voltage is not steady over the cycle, it
changes and reverses polarity depending on
the direction of motion of the conductor
through the magnetic field.


The magnetic field goes in a line from the north
pole to the south pole.
The rotor that contains the conductors move in a
circle through the linear magnetic field.
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AC Generation

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

Electricity is produced by induction.
Induction occurs when a conductor cuts
through a magnetic field line.
A conductor must move perpendicular to the
magnetic field line to cut through it.
Conductor motion parallel to the magnetic
field does not cut through any magnetic field
lines.
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Sine Wave Generation

During a rotation, the motion of a conductor
changes from perpendicular to parallel.

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Perpendicular is a 90° angle.
Parallel is a 0° (or 360°) angle.
Between these examples, the conductor
motion is at some other angle relative to the
direction of the magnetic field.

Rotation goes through 0 to 360 degrees and then
repeats.
16
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Voltage Amplitude
The amount of voltage produced at any point is
proportional to the sine of the angle of motion
relative to the direction of the magnetic field
lines.
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Voltage is a Function of Angle


There is a magnet in a generator.
Field lines go from north to south.
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Voltage is a Function of Angle

A conductor is placed in the magnetic field.

This is actually a single conductor that loops
back through the magnetic field.
19
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Voltage is a Function of Angle

Arrows show the direction of motion for the
conductor in this position.

This motion is perpendicular to the direction
of the field lines (at 90°).
20
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Voltage is a Function of Angle

The sine of a 90° angle equals 1.

This represents the maximum or peak voltage
out of the generator.
21
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Voltage is a Function of Angle

We are here on the sine wave.

This represents the maximum or peak voltage
out of the generator
22
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Voltage is a Function of Angle

Here is the position of a conductor after a
rotation of 45° (actual angle = 135°).

The motion is at an angle of 45° to the
direction of the magnetic field lines.
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Voltage is a Function of Angle

The sine of 45° (or 135°) is 0.707.


We are here on the sine wave.
The amount of voltage produced at this angle
is 0.707 of the peak voltage.
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Voltage is a Function of Angle

This shows a conductor after another rotation
of 45° from the previous example.

This motion is parallel to the magnetic field
lines and represents an angle of 180°.
25
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Voltage is a Function of Angle

The sine of 180° (or 0°) is zero.


We are here on the sine wave.
At this instant the voltage produced by the
generator is zero volts.
26
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Voltage is a Function of Angle

The conductor rotates another 45°.


Polarity starts to reverse.
This motion is now down through the
magnetic field lines (the opposite direction).
27
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Voltage is a Function of Angle

The Sine of 225° is


0.707.
We are here on the sine wave.
Polarity is opposite because the direction of
motion is going the opposite way.
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Voltage is a Function of Angle

Here is the position of the conductor after
another 45° rotation.

This more clearly shows that the direction of
motion (red circle) is down.
29
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Voltage is a Function of Angle

The sine of 270° is negative one.


We are here on the sine wave.
This is the negative peak with equal but
opposite amplitude of the positive peak.
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AC Generation Example
To see animation, copy the link into your computer browser.
http://commons.wikimedia.org/wiki/File:Dynamo.wechsel.wiki.v.1.00.gif
Copyright © Texas Education Agency, 2014. All rights reserved.
Waveform Plot



At each point in the rotating cycle, the amount of
voltage produced is equal to the sine of the angle
created by the direction of motion of the conductor
relative to the direction of the magnetic field.
This produces the
Peak
changing voltage used for
RMS
electricity known as the
sine wave.
As the angle between the
conductor and the
magnetic field changes,
the voltage changes.
Copyright © Texas Education Agency, 2014. All rights reserved.
To see animation, copy link into your computer browser.
https://phet.colorado.edu/en/simulation/generator
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Common Terms


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A plot of the voltage vs time produces a wave
form in the shape of a sine wave.
This wave form has many terms we need to
learn.
There are terms for the amplitude or size of the
wave:


Peak, peak to peak, RMS, instantaneous, and average
voltage
There are terms for how fast the wave changes:

Frequency, period
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34
Waveform Values Graphically
0.637a
VAVG
The angle represents time. An angle is used
because a time measurement would change
with a change in frequency.
35
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Angular Measurement



We use an angle to measure instantaneous
voltage values at an instant of time.
At a given angle, the relative amplitude at that
point is the same for any sine wave regardless
of the frequency.
A sine wave is created due to a conductor
moving in a circle through a magnetic field.


A circle always has 360 degrees.
A sine wave always has 360 degrees for one cycle.
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Radians
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There is another way to measure the angle of
a circle or sine wave.
It is called the radian measurement because
of the relationship between the radius and
the circumference of a circle.
C = 2πr
There are 2π radians in 360 degrees.
37
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Degrees to Radians
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Time-based Waveform Terms

Wave- a disturbance traveling through a
medium.

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For AC electricity, the movement of the electrons
back and forth in the wire
Waveform- a graphic representation of a
wave.
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Waveform depends on both movement and time.
Example: ripple on the surface of a pond
A change in the vertical dimension of a signal is a
result of a change in the amount of voltage.
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Time-based Waveform Terms

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Frequency (f)- the number of cycles of the waveform
that occur in one second of time.
 Measured in hertz (Hz)
Period (P)- the time required to complete one cycle of
a waveform.
 Measured in seconds, tenths
of seconds, milliseconds, or
microseconds
40
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Frequency and Period

There is an inverse relationship between
frequency and period.
1
f=
P

and
1
P=
f
Example: A frequency of 100 Hz gives a period
of 0.01 sec (10 ms)
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Waveform Terminology

Amplitude- height of a wave

Expressed in one of the following methods

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Peak (PK, pk, or Pk)
Peak-to-peak (P-P, PP, or pp)
Root-mean-square (RMS)
Average (AVG)
42
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Waveform Amplitude Specification

Peak- the maximum positive or negative
deviation of a waveform from its zero
reference level.



Sinusoidal waveforms
are symmetrical.
The positive peak value
of sinusoidal will be
equal to the value of
the negative peak.
Measured at an instant
of time
43
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Waveform Amplitude Specification
(continued

Peak-to-peak is the measurement from the
highest amplitude peak to the lowest peak.

Sinusoidal waveform


If the positive peak value
is 10 volts in magnitude,
then the negative peak is
also 10; therefore, peak-topeak is 20 volts.
Non-sinusoidal waveform

It is determined by adding
magnitude of positive and
negative peaks.
44
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Waveform Amplitude Specification
(Cont’d.)

Root-mean-square (RMS)
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Measured over one (or more) cycles of the wave.
Allows the comparison of AC and DC circuit values.
RMS values of AC create the same heat as that
same numerical voltage value of DC.
VRMS = VDC = heat
RMS is most common method of specifying the
value of sinusoidal waveforms.
Almost all voltmeter and ammeters are calibrated
so that they measure AC values in terms of RMS
amplitude.
45
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RMS Heating Effect

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VRMS = 0.707 Vpeak
IRMS = 0.707 Ipeak
A sinusoidal voltage with peak amplitude of
1 volt has the same heating effect as a DC
voltage of 0.707 volts.


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AC creates slightly more heat than DC.
Comparing RMS value to average voltage.
Due to this, the RMS value of voltage is also
referred to as the effective value.
46
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Determining the 0.707 Constant

To determine the 0.707 constant, you must
use the mathematical procedure suggested by
the name, root-mean-square.
VRMS =

Vpk
2
This has nothing to do with the sine function
even though this is the same value as sin 45°.
47
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Average Voltage


Average voltage is the DC equivalent voltage.
The average voltage of AC over a full sine
wave equals zero.


The positive half cycle is equal to the negative half
cycle.
Because of this we only look at half the wave
(the positive half cycle).

This means we are looking at an angle of π
radians.
48
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Average Voltage

Average voltage over a half cycle:
2 Vpk
Vpp
VAVG =
=
π
π


Or, VAVG = .637 VPK
VAVG is produced by rectifying then filtering
the AC in a voltage regulator.

This is the process used in a DC power supply.
49
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Instantaneous Voltage

This is an important value when sampling a
wave at several points.


Example: analog to digital conversion
Different types of waves might have the same
peak or average voltage but different
instantaneous voltages

Examples: square wave, triangle wave
50
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Instantaneous Voltage

Instantaneous voltage is the voltage at a
single point or instant of time.
VINST = Vpk sin θ

To find the angle, take the inverse sine.

Usually a 2nd function on a calculator
θ = sin
−1
(
VINST
Vpk
)
51
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Relationship Circles
P–P
2
PK
INST.
Sine° PK
RMS
0.707 PK
AVG.
0.637 PK
52
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Relationship Calculations
EXAMPLES
120 VAC = 170 Vpk
Formula: PK = RMS  0.707
120  0.707 = 169.7
(round off to 170 Vpk)
18 V @ 72  = 19 Vpk
Formula: PK = Instantaneous  Sine
18  Sine(72°) = 18.9
(round off to 19)
30 Vpk = 21.2 VAC
Formula: RMS = 0.707 X PK
0.707 X 30 = 21.2
350 V @ 23.5 = 30 Vpk
Formula PK = Instantaneous  Sine
350  Sine(23.5°) = 877.7
(round off to 878)
50 Vpp = 17.7 Vrms
Step 1: Need to find PK
Formula: PK = P-P  2
50  2 = 25m
Step 2: Find RMS
Formula: RMS = 0.707 X PK
0.707 X 25 = 17.675
(round off to 17.7 Vrms)
Find the angle with 454 V instantaneous and a
PK of 908 V
Formula: Sine (θ) = Instantaneous  PK
454  908 = 0.5
2nd Sine (.5) = 30
20 V Average
= 22.2 Vrms
= 31.4 Vpk
= 62.8 Vp-p
Step 1: Find PK, Formula: PK = Average  0.637
= 20 0.637 = 31.39 (round off to 31.4)
Step 2: Find RMS, Formula: RMS = 0.707 X PK
= 0.707 X 31.4 = 22.19 (round off to 22.2)
Step 3: Find P-P, Formula: PK = 2 X PK
= 2 X 31.4 = 62.8
53
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Calculate peak voltage given RMS
voltage

VRMS = 120 V



Formula: PK = RMS  0.707
120  0.707 = 169.7
(round up to 170 Vpk)
120 VRMS = 170 Vpk
P–P
2
PK
INST.
Sine° PK
RMS
0.707 PK
AVG.
0.637 PK
54
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Calculate RMS voltage given peak
voltage

Vpk = 30 V



Formula: RMS = 0.707 X PK
0.707 X 30 = 21.2 V
30 Vpk = 21.2 VRMS
P–P
2
PK
INST.
Sine° PK
RMS
0.707 PK
AVG.
0.637 PK
55
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Calculate RMS given p-p

Vpp = 50 V


Step 1: Find Vpk
 Formula: PK = p-p  2
 50  2 = 25 V
Step 2: Find RMS
 Formula: RMS = 0.707 X PK
 0.707 X 25 = 17.675
(round off to 17.7 Vrms)
 50 Vpp = 17.7 Vrms
P–P
2
PK
INST.
Sine° PK
RMS
0.707 PK
AVG.
0.637 PK
56
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Calculate VRMS, Vpk, and Vpp given
VAVG

VAVG = 20 V



P–P
Step 1: Find Vpk,
2
PK
 Formula: PK = Average  0.637
 = 20  0.637 = 31.39
(round up to 31.4)
INST.
Step 2: Find RMS,
Sine° PK
 Formula: RMS = 0.707 X PK
= 0.707 X 31.4 = 22.19
(round up to 22.2)
Step 3: Find P-P,
 Formula: P-P = 2 X PK = 2 X 31.4 = 62.8
RMS
0.707 PK
AVG.
0.637 PK
57
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Calculate peak voltage from
instantaneous voltage

VINST = 18 V @ 72°




Formula:
PK = Instantaneous  Sine
of the angle
= 18  Sine(72°)
= 18  .951 = 18.9 V
(round up to 19)
18 V @ 72  = 19 Vpk
P–P
2
PK
INST.
Sine° PK
RMS
0.707 PK
AVG.
0.637 PK
58
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Calculate peak voltage from
instantaneous voltage

VINST = 350 V @ 23.5°





Formula: PK = Instantaneous
 Sine(angle)
= 350  Sine(23.5°)
= 350  0.4
= 877.7
(round up to 878)
350 V @ 23.5 = 878 Vpk
P–P
2
PK
INST.
Sine° PK
RMS
0.707 PK
AVG.
0.637 PK
59
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Calculate phase angle given
instantaneous voltage and peak
voltage

454 VINST at unknown
angle with a Vpk of 908 V



Formula: Sine θ =
Instantaneous  PK
454  908 = 0.5
2nd Sine (.5) = 30
P–P
2
PK
INST.
Sine° PK
RMS
0.707 PK
AVG.
0.637 PK
60
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Relationship Exercise
#
rms
peak
pk-to-pk
average
200mV
instantaneous
________ ____ V @ 72º
113V
________ V @ 90º
96.4
________ V @ 235º
1.5V @ 122º
689V
________ V @ 35º
Go to
Answers
61
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Sine Wave and Sine Trigonometric
Function




The term sinusoidal has been used to describe a
waveform produced by an AC generator.
The term sinusoidal comes from a trigonometric
function called the sine function.
Sines, cosines, and tangents are numbers equal
to the ratio of the lengths of the sides.
The sine function is used in AC because the
opposite side is the direction of motion of the
conductor through a magnetic field relative to
the angle.
62
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Right-Triangle:
Side and Angle Relationships


A right triangle has a 90° angle.
Each side is named with respect to the angle
you are using (called the angle theta, or θ).




The side of the triangle across from the angle
theta is called the opposite side.
The longest side of a right triangle is called the
hypotenuse.
The remaining side is called the adjacent.
Each of these sides are commonly abbreviated
to their initials, O, H, and A.
63
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Right Triangle
Hypotenuse
Opposite
θ
Adjacent
64
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Basic Trigonometric Functions
In trigonometry, there are three common ratios
used to study right triangles.
1.
Sine
A.
B.
The sine of the angle theta is equal to the ratio formed by
the length of the opposite side divided by the length of
the hypotenuse.
opposite
Sine θ =
hypotenuse
65
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Basic Trigonometric Functions
(Continued)
2.
Cosine
A.
The cosine of the angle theta is equal to the ration formed by
length of the adjacent side divided by the length of the
hypotenuse.
B.
3.
adjacent
Cosine θ =
hypotenuse
Tangent
A.
The tangent of the angle theta is equal to the ratio formed by
length of the opposite side divided by the length of the
adjacent side.
B.
opposite
Tangent θ =
adjacent
66
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Simple Memory Aid

Remember the acronym SOH-CAH-TOA

Pronounced “sock ah toa”
Sine equals opposite over hypotenuse
 Cosine equals adjacent over hypotenuse
 Tangent equals opposite over adjacent

67
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Basic Trigonometric Functions
(Continued)
Opposite
Sine
Adjacent
Hypotenuse
Cosine
Hypotenuse
Opposite
Tangent
Adjacent
68
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Trigonometric Exercise
Triangle #1
Triangle #2
Hypotenuse?
Hypotenuse?
8’ Opposite
10’ Adjacent
5.3 Rods
Opposite
6.8 Rods Adjacent
What is the Hypotenuse? ___________
What is the Hypotenuse? ___________
Triangle #3
Triangle #4
125 Miles
Hypotenuse
Go to
Answers
56’
Hypotenuse
Opposite?
85 miles Adjacent
23.2’
Opposite
Adjacent?
What is the Opposite? ____________
What is the Adjacent? _____________
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Presentation
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Answer Keys
Follow This Slide
End of
Go to Answers
Presentation
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Relationship Key
#
rms
peak
pk-to-pk average
instantaneous
1.
200mV
283mV
566mV
180mV
269mV @ 72º
2.
80v
113V
226V
72V
113V @ 90º
3.
96.4
136V
272V
87V
111V @ 235º
4.
1.25V
1.77V
3.54V
1.13V
1.5V @ 122º
5.
764µV
1.08mV
2.16mV
689V
619µV @ 35º
Go to the
Trigonometric
Functions
71
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Trigonometric Exercise Key
Triangle #1
Triangle #2
Hypotenuse?
Hypotenuse?
8’ Opposite
10’ Adjacent
5.3 Rods
Opposite
6.8 Rods Adjacent
Hypotenuse = 12.8’
Hypotenuse = 8.62 Rods
Triangle #3
Triangle #4
125 Miles
Hypotenuse
Go to the
Calculations
56’
Hypotenuse
Opposite?
85 miles Adjacent
Opposite = 91.7 miles
23.2’
Opposite
Adjacent?
Adjacent = 51’
End of
Presentation
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Trigonometric Exercise Calculations
Triangle #1
Given Adjacent side = 10’ and Opposite side = 8’.
What is the Hypotenuse?
Step 1 - Find the degree angle
Tangent = Opposite
Adjacent
= 8_ = 0.8
10
= enter 2nd tangent (0.8) on your calculator
= 38.65980825 round up to 38.66º
= 38.66°
Step 2 - Change the degree angle to cosine
Hypotenuse = Adjacent =
10
(take 38.66°, enter cosine on your calculator, your answer is 0.780866719)
cosine 38.66
Cosine
= 10’____________
.780866719
73
= 12.8’
Copyright © Texas Education Agency, 2014. All rights reserved.
Trigonometric Exercise Calculations
(Continued)
Triangle #2
Given Adjacent side = 6.8 rods and Opposite side = 5.3 rods.
What is the Hypotenuse?
Step 1 - Find the degree angle
Tangent = Opposite
Adjacent
= 5.3 rods= 0.779411765
6.8 rod
enter 2nd tangent 0.779411765 on your calculator
= 37.93°
Step 2 - Change the degree angle to sine
Hypotenuse = Opposite
Sine
= 5.3 rods
sine 37.93 (enter 37.93, enter sine on the calculator )
= 5.3 rods
0.6146982793
= 8.63
74
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Trigonometric Exercise Calculations
(continued)
Triangle #3
Given Hypotenuse side = 125 miles and Adjacent side = 8.5 miles What is the Opposite
side?
Step 1 - Find the degree angle
Cosine = Adjacent = 85 = 0.68
Hypotenuse
125
enter 2nd function button on calculator enter cosine 0.68
= 47.15635696 (round off to 47.16º)
Step 2 - Change the degree angle to sine
Opposite = Sine x Hypotenuse
enter sine 47.16 on calculator = 0.7332553462
= 0.7332553462 x 125 miles
= 91.65691828 (round off to 91.7)
= 91.7 miles
75
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Trigonometric Exercise Calculations
(continued)
Triangle #4
Given Hypotenuse side = 56’ and Opposite side = 23.2’
What is Adjacent side?
Step 1 - Find the degree angle
opposite
23.2 = 0.4142857143
Sine =
=
hypotenuse
56
= enter 2nd sin 0.4142857143
= 24.47º
Step 2 - Change the degree angle to cosine
Adjacent = Cosine x Hypotenuse
enter cosine 24.47 on the calculator
= 0.910178279 x 56’
= 50.96998362 (round off to 51’)
= 51’
76
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