Electronics and Your Slide Rule, Part III — Combined Operations with Electronic Applications (CIE Electronics Slide Rule No. 515 / Pickett N-515-T)
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electronics
CLEVELAND INSTITUTE OF ELECTRONICS / CLEVELAND, OHIO
Electronics and
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An AUTO-PROGRAMMED. Lesson
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ABOUT THE AUTHOR
Through over 15 years experience in helping students learn through home
study, Mr. Geiger has obtained an intimate understanding of the prob-
lems facing home-study students. He has used this knowledge to make
many improvements in our teaching methods. Mr. Geiger knows that
students learn fastest when they actively participate in the lesson, rather
than just reading it. Accordingly, you will find many “What Have You
Learned?" sections in this lesson, to assist you in getting a firm grasp of
each topic.
Mr. Geiger edits much of our new lesson material, polishing up the manu-
scripts we receive from subject-matter experts so that they are easily read-
able, contain only training useful to the student in practical work, and are
written so as to teach, rather than merely presenting information.
Mr. Geiger's book, Successful Preparation for FCC License Examinations
(published by Prentice-Hall), was chosen by the American Institute of
Graphic Arts as one of the outstanding text books of the year.
) Lesson exclusive with the
Cleveland Institute of Electronics, It is your assur-
ance of the finest in technical education and training.
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z +, of the National Home Study Council ge
3 3 Tix Goeretting Somntechn ben ten erent ee E>
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“Ones ot 3:
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Electronics and Your
Slide Rule
Combined Operations with
Electronic Applications
By DARRELL L. GEIGER
Senior Project Director
Cleveland Institute of Electronics
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25. Reactance and Resonance Decimai Point Locator... Page 53
26. Decimal Point Locator, Resonance Problems... Page 54
27. Reactance Problems... Page 58
28. The 27 Scale... Page 60
29. Using the H Scale... Page 64
Using the Trigonomettic Scales... Page 68
30. The Trigonometric Functions... Page 69
31. Finding Sines and Cosines With the S Scale... Page 71
32. Finding Tangents by Means of the T Scale... Page 75
33. Finding Cotangents With the T Scale... Page 77
34. Combined Operations - Multiplication... Page 78
35. Combined Operations - Division... Page 82
36, Combined Operations for the Tangent and Cotangent. Page 84
Examination... Page 85a
x i
¥
© Copyright 1966, 1965, 1962, 1958 Cleveland Institute of Electronics. : :
AU Rights Reserved/Printed in the United States of America. ¥ :
FOURTH EDITION /Pifth Revised Printing/October, 1966. ge
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A chat with your instructor
In this lesson you learn how to use the special electronic scales on
your slide rule, and also the trigonometric scales. Practical ap-
plications for the latter scales will be given in the next lesson.
In studying this lesson be sure to work all the practice problems.
Practice is very important in learning slide rule operation. Don't
limit your practice to the problems in your slide rule lessons. Al-
ways keep your rule handy, and use it for any needed calculations
(whether business, personal or electronics) that comes up.
Electronics and
Your Slide Rule
Part ITI
REACTANCE AND RESONANCE DECIMAL POINT LOCATOR...
Resonance frequéncy problems and problems concerning inductive
and capacitive reactance are among the most commonly occuring
problems that electronic technicians are required to work. Be-
cause of the rather involved formulas used for these problems and
also because of the very large and very small values normally
employed, locating the decimal point with certainty is a difficult
task. For that reason special scales are provided on the back of
your Electronics rule to enable the decimal point to be quickly and
surely located in all problems involving reactance and resonance.
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Atthe same time the scales also provide approximate solutions
to these problems. This approximation is close enough for
many requirements, Whena more accurate answer is required
it can be quickly obtained by the use of the H scale or the 27 scale
on the front of the rule. The use of these two scales will be ex-
plained later.
The Decimal Point Locator scales have been designed so that
it is unnecessary to convert from one unit to another in using these
scales. This not only saves time but also eliminates a common
source of errors in decimal point location. The next two sec-
tions will provide sufficient examples for you to easily under-
stand the use of the decimal point locator scales.
26 DECIMAL POINT LOCATOR, RESONANCE PROBLEMS...To
locate the decimal point in resonant frequency problems where
C and L are given, set the value of either C or L (it makes no
difference which one) on the Resonance Problems Scale of the
slide opposite the other given value on the upper body scale.
Then read the approximate frequency on the bottom body scale
opposite the appropriate arrow on the slide.
Asan example of how the decimal! point is located in a resonant
frequency problem, suppose an inductance of 40 mh is connec-
tedin parallel witha capacity of 0.03 uf. At approximately what
frequency will this parallel circuitbe resonant? Fig. 20 shows
the setup for working this problem. Set the hairline over 0.03
uf on the upper body scale of the rule. This would be found on
the scale marked Cyf and would be somewhere between the
main division marks 0.01 and 0,1 on that scale. To assist in
locating positions between main division marks, the space be-
tween each of the main divisions is marked by two subdivision
marks, 2 and 5. Forthe space between the 0.01 and 0,1 marks
on the Cuf scale, the 2 and the 5 marks would represent 0. 02
uf and 0.05 yf respectively. Therefore, to represent 0.03 pf
the hairline would be set between the 2 and 5 marks, and closer
to the 2 mark than to the 5, as can be seen in Fig. 20.
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To find resonant frequency of 40mh with 0.03 pf: 55
set O.O3 on sca/fe Cuf opposite YO on scale Lmh
Fig, 20, Showing that 40 mh will resonate with 0.03 jf at approximately
5 ke or 5000 cps.
Now move the slide so that 40 mh on the section of the slide
marked "Resonance Problems" is under the hairline. The
scale used for this purpose would be the one on the slide marked
Lmh- 40 on this scale would be located between the main di-
vision marks 10 and 100. The two subdivision marks, 2 and
5, between these two main divisions represents 20 mh and 50
mh on the Lmh scale. To represent 40 mh the hairline should
therefore be between the 2 and 5 marks but closer to the 5 than
to the 2. See Fig. 20. A high degree of accuracy is never
necessary in setting values on the Decima! Point Locator scales.
The approximate resonant frequency is now read on the f scale
of the lower body of the rule under the arrow marked fj, if the
answer is wanted in kilocycles, or under the arrow marked
fops if the answer is wanted in cycles per second. To read the
answer in kilocycles refer to dashed line A of Fig. 20. On
scale f the dashed line is between the main divisions 1 and 10,
which shows that the frequency lies somewhere between 0 and
10 ke.
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To pin the frequency down closer notice that the dashed line is
between the subdivision marks 3 and 6 on scale f. Hence, the
resonant frequency of this circuit is between 3 and 6 ke. Since
the dashed line is nearer the 6 subdivision than the 3, the fre-
quency must be in the neighborhood of 5 ke. This is close
enough for accurately locating the decimal point and also for
many practical purposes. A more accurate answer can be ob-
tained quickly by the use of the H scale on the front of the rule,
as will be explained later.
Dashed line B shows where to read if the answer to the above
problem is wanted in cycles per second. This dashed line
shows that the arrow fcps lies between 1, 000 and 10, 000 cps.
Narrowing down the range, the arrow is between the subdivision
marks 3 and 6, and the frequency is therefore between 3, 000
and 6, 000 cps. Since it is nearer to 6 than to 3 a good estimate
of the frequency would be 5,000 cps,
In working the above problem the value of L could equally well
have been set onthe upper body scale andC onthe slide. It makes
no difference which of the scales is used for L and which for C.
Arrows are printed before and after each of the Decimal! Point
Locator scales. The purpose of these arrows is to indicate the
direction of progression of thescales. For example, the values
on the Xe scaleon the lower part of the body become progress-
ively larger from left to right. Hence, the associated arrows
point ffi that dtrection. ©n the other hand the vaiues on the Xj
scale directly below become progressively larger from right to
left, so the associated arrows point in the opposite direction
from the arrows for the Xc scale. Knowing the direction of
progression of the scales makes it easier to set and read values.
In the example above a parallel resonant circuit is involved. If
the inductance and capacity were in series the problem would
have been worked the same way, and the same answer obtained.
In working any resonant frequency problem using either the
Decimal Point Locator or the LC scale on the front of the rule
the method is exactly the same whether series or parallel
resonance is involved.
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Example...31
Approximately what value of capacity should be used with an
inductance of 350 wh in order to resonate at 200 kc?
Solution,.. Refer to Fig, 21
(1) Move the slide so that the slide arrow desig-
nated f-ke is opposite 200 on scale f on the
lower body of the rule.
(2) Place the hairline over 350 on scale Luh on
the upper body of the rule.
(3) Under the hairline read 0.0030n scale Cuf on
the slide. Hence, the answer is 0, 003 yf.
To find C required with 350 yh to resonate at 200k:
Sz qe or ly {
Lma bana 2------- == *
t ‘\ 1,000; 100
Seri pot
| qT l Ly ei q | |
press ‘ 0.g61 001
fot Lyn ‘ \
r 1 fac ai E lead \
S T T a a CG | eS T
*¢ ie ' : ‘
*L ; ft
H
f
then opposite ‘350 on scale “iph teod OOOF pi ov on seele Cpt
Fig. 21. Showing thot for 350 yh to resonote ot 200 kc, a copacitance
of approximately 0.003 yf is required.
WHAT HAVE YOU LEARNED?
Using the Decimal Point Locator scales find the approximate
frequency at which the following values of L and C will resonate:
1...L=3mbh; C = 175 puf
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58-2... L = 340 wh; C = 0.0034 pf
3...L=0,025h; C=2yf
4,..L= 700 ph; C = 500 pyf
Using the Decimal Point Locator scales find the approximate
value of L or C needed to resonate at the given frequency:
5...f= 240 ke; C= 0,08 pf
6...f=7,000 cps; L=0.5h
7...f= 40 me; C = 550 pf
8...f= 6500 kc; L= 425 ph
ANSWERS
1. 250ke 2. 200ke 3. 700 cycles 4, 900 ke
5. B8uh 6. 0.001 uf 7. 0.08 uh 8. 2 wef
Pag i REACTANCE PROBLEMS...In solving reactance problems
the frequency on the part of the slide marked 'Reactance Prob-
lems" is placed opposite the given value of inductance or ca-
pacity on the upper body of the rule. The reactance is then
readon the proper scale on the lower body opposite the appro-
priate arrow of the slide.
|
Example. ..32
Find the reactance of a 0,005 pf capacitor when used at 3 mc.
=
ze Solution...See Fig. 22.
22 (1) Set hairline over 0.005 on scale Cys on the upper
33 body.
3s (2) Adjust slide so that 3 on scale fyyc is under
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To find recetance of 0.005 pt of Sme:
aktasadlneeunineiasenacirs joan.
$ Cyt oF hp oN 0.001
s Lint i H
od : H
Sut ast Ce ee ee 2 2 es ee ee
oe TAT 1
fhe eae
fae Von
Xe TT CT Lf 5 T r<
sS x 2 cs
f ‘
J '
porte anne wees
wal.
‘
then opposite “orrew *csL read TOA on scole Xe
Fig. 22. Finding thet the approximate reactance of a 0.005 yf capacitor
ot 3 mc is 10 ohms.
the hairline.
(3) Opposite arrow on slide marked X,,% read 10
ohms on scale X¢ on lower body of rule.
Hence, the reactance of this capacitor at this frequency is ap-
proximately 10 ohms.
Example. ..33
What approximate value of inductance will have a reactance of
4 megohms at a frequency of 3, 000 kc?
Solution. ..
(1) Move slide so that arrow on slide marked
XLM 2 is opposite 4 on scale Xj, on lower
body of rule.
(2) Set hairline over 3,000 on scale fe on slide
of rule.
(3) Under hairline on scale marked Ly, on upper
rule body read 0.3 henry, the approximate
inductance value required.
Example. ..34
At what frequency will a 200 pyf capacitor have a reactance of
25 ohms?
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ep
Solution, ..
(1) Move slide so that slide arrow marked X, — 2
is opposite 25 on scale X,_ on lower bodyof rule.
(2) Set hairline over 200 on scale Cuyf on upper
body of rule.
(3) Under hairline read 30 on scale Fy on slide.
Hence, the required reactance will be obtained at approximate-
ly 30 megacycles.
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WHAT HAVE YOU LEARNED?
Use the Decimal Point Locator scales to find approximate
answer to the following problems:
1...f = 2,300 ke; L= 250 mh; XL = ?
2...f=25 me; C= 150 pf; Xe =?
3...f=0.8me; L=0.45h; XL=?
4...f=7,500 cps; C=3.5 yf; Xe=?
5...X_ = 3,000 9; L = 850 ph; f=?
6...X_ = 6, 500 9; C = 500 pul: f= ?
7...£= 480 kc; XL, = 4002; L=?
8...f£=2.4 me; X¢ = 30,0002; C=?
ANSWERS
1, 3megohms 2. 30ohms 3. 2megohms 4. 6 ohms
5. 700kc 6. 60kce 7. 0.2 mh 8. 2 mue
THE 2m SCALE...If the hairline is over a certain value on the
2m scale, 27 times that value will appear under the hairline on
the D scale. Or conversely, any number under the hairline on
the D scale will be divided by 27 by merely reading under the
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hairline on the 2r scale. This scale is very useful because
the factor 27 widely occurs in electronics.
ee) ee ey .
Example... 35
How many degreesareina radian? 27 radians are equal to 360°.
Solution...
Set the hairline over 360 on scale Dand read 57. 3°, the answer,
under the hairline on scale 2r.
Example... 36
The armature of a generator is rotating at 1,800 revolutions
per minute, which is 30 revolutions per second. What is its
angular velocity, w, in radians per second? The formula is
w = 2rr, where r is revolutions per second.
Solution...
Set hairline over 30 on 27 scale, Read 188.8 radians per second,
the answer, under the hairline on scale D,
ee Oe "eee ew nwe
The most importantuse of the 27 scale is in working capacitive
and inductive reactance problems where better accuracy is re-
quired than is possible with the Decimal Point Locator scales.
When using the 2mr scale for this purpose the problem should
also be worked on the Decimal Point Locator scales. so as to
locate the decima! point.
In working reactance problems in conjunction with the 27 scale,
frequency is always set or read on scale 27. To help you to re-
member this, the 27 scale is also marked (f,). Inductance or ca-
pacity is always set or read on scale Cl, which is shown by this
scale also being identified as L, or C,, Capacitive reactance is
always set or read opposite the appropriate index of scale D,
which you can remember by the arrow at the index labeled Xc.
Inductive reactance is always set or read opposite the appropriate
index of scale C, which you can remember by the arrow at the in-
dex labeled X, .
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62 To find reactance of GOmh choke at S000 kc
opposite 5000 on scale (f) #1 60 on scale Ly
——"< --
(tan
“4
then opposite fades ry x recd 1885 on ecole D
Fig. 23. Finding thet a 60 mh choke ot a frequency of 5000 ke has o
reactance of 1.885 megohms,
Example... 37
Find accurately the reactance of a 60 mh choke coil at 5,000 kc.
Solution...See Fig. 23
(1) Set the hairline over 5,000 hin is the frequency)
on scale (f,)27.
(2) Move the slide so that 60 (which is the inductance)
on scale (Ly or Cy) Clis under the hairline.
(3) Opposite the index of scale X; or f, - C read
1, 886 (which is the inductive reactance) on scale D.
(4) Use the Decimal Point Locator scales on the
back of the rule to obtain 1.5 megohm as the
approximate reactance value.
Hence, the accurate reactance of the choke is 1.885 megohms.
Example...38
Find a more accurate value for the reactance of the capacitor in
Example 32.
Solution...
enon (1) Set hairline over 3 (which is the frequency) on
scale (f,).
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(2) Move slide so that 0. 005 (which is the capacity) on
scale C, is under the hairline.
(3) Opposite the left index on scale X_ read 1061
(which is the capacitive reactance) on scale C,
The approximate reactance has been foundto be
10 ohms.
Hence, the exact reactance is 10.61 ohms.
Example...39
Find a more accurate value for the inductance in Example 33.
Solution...
(1) Set hairline over 3, 000 on scale fy
(2) Move slide so that right index of scale X\_is oppo-
site 4 on scale D.
(3) Read 212 under the hairline on scale Ly.
The approximate inductance was previously found as 0,3 hen-
ry. Hence, the accurate value of the inductance is 0.212 hen-
ry.
Any reactance problem can be worked without confusion if you re-
member these three points:
(1) Frequency is always set or read under the hair-
line on scale 27, also marked (f,).
(2) Inductance or capacity is always set or read under
the hairline on scale CI which is therefore marked
(Lx or Cx).
(3) Inductive reactance ts always set or read opposite
the C index on scale D, and this is indicated on the
rule by the symbol x, and an arrow on the C index
pointing to the Dscale. Capacitive reactance is
always set or read opposite the D index on scale
C and this is indicated by the symbol x and an ar-
row pointing to the C scale.
In working any reactance problem set up the known values in
the positions on the rule indicated by the two applicable rules
above, and then read the unknown value in accordance with the
remaining rule.
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64 Example,..40
Find a more accurate value for the frequency in Example 34.
Solution...
Set up capacity and the reactance in accordance with Rules 2
and 3 above:
(1) By Rule 3 set the left index of scale D opposite
25 on scale C,
(2) ByRule 2 set the hairline over 200 on scale Cl.
(3) By Rule 1 read the frequency, 318, under the hair-
line on scale 2%, The approximate frequency has
previously been found to be 30 mc,
Hence, the accurate frequency is 31.8 mc.
WHAT HAVE YOU LEARNED?
1-8... Use the 27 scale to obtain accurate answers to the WHAT HAVE
YOU LEARNED? section of Topic 27.
9...The velocity of propagation on a transmission line is given by
the formula v = io , where vy is the velocity of propagation, w =
2rf, and 8 is the wavelength constant (phaseshift permile). Find
v when the frequency is 60 cps and f is equal to 0.00213,
ANSWERS
1, 8.61 megohms 4, 6.06o0hms 7. 0,133 mh
2. 42.5 ohms 5. 562 ke 8. 2.2 ppf
3, 2.26 megohms 6. 49ke - 9, 177,000 miles/se
USING THE H SCALE... The H scale is used ir, solving reso-
nant frequency problems when greater accuracy is required
than is obtainable from the Decimal! Point Locator scales,
However, the problem must also be worked with the latter
scales in order to locate the decimal point, This is much
faster than the use of a rough calculation for this purpose.
It is well known that the frequency at which a circuit will
resonate is determined entirely by the product of the circuit
to
Ne)
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inductance and the circuit capacitance (called the LC product),
If the hairline is set over the frequency on scale D, the required
LC product for the circuit to resonate at this frequency is read
under the hairline on scale H, and vice versa.
When using the H scale, frequency is always set or read opposite
the index of scale C, on the D scale. For this reason the symbol
f, and an arrow on the C index pointing to the D scale appears on
your rule,
Example, ,,41
What must be the LC product in order that a circuit will
resonate at 20 mc?
Solution...
First use the Decimal Point Locator scales on the back of the
rule to obtain an approximate answer, as follows:
(1) Set arrow marked f-Mc on slide opposite 20
on scale f on lower body of rule.
(2) The LC product is now found by multiplying
together any two opposite values on the upper
body of the rule and the Resonance Problems
portion of the slide. If 1 is one of the two
opposite valttes no multiplication is required.
If we take 1 on the scale Cuyf on the upper
body of the rule then the value opposite on
scale Luh on the slide will be 70. Hence, the
approximate value of the LC product is 70
whenC is expressedin micromicrofarads and
Linmicrohenries. Now the front of the rule
is used to obtain a more accurate value for
the LC product.
2 (3) Set the hairline over 200n scale D. Then
#2 read 633 under the hairline on the H scale.
= Since the approximate value is 70, the accu-
3% rate value will be 63. 3, where C is in micro-
ge microfarads and L is in microhenries.
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Explanation...
When the hairline is over any frequency value on the D scale, the
required LC product is under the hairline on the H scale.
a a ae | ee)
The reverse of the above problem (Example 41) is to find the
frequency of resonance when the LC product is known, This, of
course, would be done by setting the hairline over the LC product
on scale H, and then reading the frequency under the hairline on
scale D, However, any value can be set on scale H in two dif-
ferent positions, giving two different frequency readings on scale
D.
Obviously only one of these readings can be correct. You should
first determine the approximate frequency for the given LC prod-
uct by the use of the Decimal Point Locator scales. Then the cor-
rect setting on the H scale is the one that gives a frequency value
on the D scale that is near the approximate value previously de-
termined.
The required value of C to go with a given value of L in order to
resonate at a certain frequency can be found by dividing the LC
product by the given L value. Or if C is known it can be divided
into the LC product to give the required value of L to resonate at
a certain frequency. However, the following examples show sim-
pler ways of solving resonant frequency problems where the
CC value is neither known or wanted, and this is generally the
case.
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Example., 42
Find a more accurate value for the capacity in the problem of
Example 31.
Solution...
(1) Set the left index of scale C (which is marked f,)
opposite 200 (which is the frequency) on scale D,
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(2) Set the hairline over 350 (which is the inductance) 67
on scale H(L,y).
(3) Read 181 (which is the required capacity) under
the hairline on scale B,
The approximate value of C has been previously found to be
0.003 uf. Hence, the accurate value is 0.00181 uf.
When C is given and L is to be found the method is the same
as in Example 42 above. Whether finding L or C an index of
the C scale is placed opposite the frequency on the D scale. Then -
with the hairline over the given value of L on scale H, the unknown
value of C is read on scale B under the hairline. Or alternately,
the known value of L or C can be set on scale B and the unknown
value then read on scale H.
The following example illustrates the method when L and C are
known and the frequency is to be found:
Example... 43
An inductance of 40 mh is connected in paralle! with a capacitance
of 0.03 uf. At what frequency will this circuit resonate?
Solution. . .
(1) Find the approximate resonant frequency using the
Decimal! Point Locator scales. This was done in
Topic 26 and was found to be 6000 cps.
(2) Set hairline over 40 on scale H,
(3) Adjust slide so that 3 on scale B is under the hair-
line.
(4) Read opposite the index of scale C on scale D. De-
pending upon which halves of scales H and B were
used for setting the values, the reading obtained
on scale D may be either 1452 or 459. Since the
approximate frequency is known to be 5000 cps,
= the accurate frequency is 4590 cps. The value
i: 1452 is spurious because it is far different from
32 the known approximate frequency. If the reading
iy 1452 is obtained, move the slide so that 3 on the
Be other half of the B scale is under the hairline.
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Then 459 will read opposite the C index on scale
D.
Discussion...
The B scale and the H scale are both of the repeating type, so
any value can be set on either of these two scales in two dif-
ferent positions. In problems in which the frequency is known
and L or C is to be found, as in Example 42, the correct an-
swer is obtained no matter which of the possible positions are
used on the H and B scales. However, when L and C are
known and the frequency is to be found, there are two different
results possible, depending upon the sections of the H and B scales
used,
Only one of the results can be correct. To determine if the
result obtained is correct, see if it is in agreement with the
approximate value, which should have been previously deter-
mined. If not, move the slide so that the other half of the B
scale is used for setting the B scale value.
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WHAT HAVE YOU LEARNED?
1-8...Usethe Hscale tofindaccurate answers to the WHAT
HAVE YOU LEARNED? section of Topic 26.
9...What is the required LC product to resonate at 5 mega-
cycles, L being in microhenrifes and C in microfarads?
ANSWERS
1. 220ke 2. 148ke 3. 710cycles 4, 269ke 5. 5.5 uh
6. 0.00103 Hf 7. 0.0287 rh 8. 1l.4l mee 9. 0.001014
Using the Trigonometric Scales
It is necessary to be familiar with logarithms and trigonometry
before studying the remainder of this manual, If you are not
familiar with these subjects, your slide rule training is now
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c . Ce
4 r5 C90 8- rs c=90°
(oe) {b)
Fig. 24. The right triangle shown in two positions.
completed. However, the author hopes you will eventually be
able toacquire a knowledge of these subjects and then complete
the remainder of this manual.
THE TRIGONOMETRIC FUNCTIONS...A review of trigonometry,
particularly those sections dealing with the right triangle. will
help you understand the following sections. The trigonometric
functions aré defined by the sides of a right triangle. Figure 24(a)
shows a right triangle with sides of length a, b, and c and two
acute angles labeled A and B (angles less than 90 degrees are
called acute angies), and also the right angle C, The sine, cosine,
tangent, and cotangent of angle A are defined as follows:
— c i = b
wows placent 21g c
a
A = Opposite side _
= adjacent side b
cot A = adjacent side =}
opposite side
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The tangent and cotangent are reciprocal quantities:
tan A= —_, cot A= 1,
cot A’ tan A
If you know one of the two acute angles, the other can be found by
subtracting the known angle from 90°. Thus in Fig. 24(a) A = 90°
- Band B=90°- A. For example, if A were 30°, then B would
be 90° - 30° = 60°. When A is 45° then B = 90° - 45° = 45° and the
acute angles are then equal.
Since the angles A and B of the right triangle depend upon each
other (their sum must be 90°), then the trigonometric functions of
these two acute angles also depend upon each other, Figure 24(b)
Shows the same triangle as in Fig. 24(a), but turned over so that
the functions of angle B can be found more easily, From Fig.
24(b) the sine of angle B is found as:
sin B= nppos ite side -b
vpotenuse Cc
Since cos A is aiso equal to 8 , Sin B equals cos A, In the same
way, you can find that cos B = 2 = sin A. This type of relation-
ship is also true for the tangent and cotangent, and so the trigo-
nometric functions of the acute angles of a right triangle are re-
iated as follows: The sine of one angle equals the cosine of the
other and the tangent of one angle equals the cotangent of the other,
Thus in Fig. 24, if A = 30° and B = 60°, then sin 30° = cos 60°, cos
30° = sin 60°, tan 30° = cot 60°, and cot 30°= tan 60°. This prin-
ciple can be extended a bit further. Since angle B is always 90°
minus angle A, the previous rule can be written as:
sin A= cos B= cos (90° - A)
cos A= sin B= sin (90° - A)
tan A=cot B= cot (90° - A)
cot A= tan B= tan (90° - A)
These formulas relate the functions of angle A to those of 90° - A
and can be used, for example, to find the sine of an angle when
only cosines are available, If the sin 10° was needed, it could be
obtained by finding cos (90° - 10°) = cos 80°.
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WHAT HAVE YOU LEARNED?
1...In Fig. 24(a), let B= 36.8°, a= 4, b=3, andc= 5.
(a) What is angle A?
(>) Find sin B, cos B, tan B, and cot B from the sides
of the triangle,
(c) Find sin A, cos A, tan A, and cot A from the sides
of the triangle andcompare these results with part
(b). Which functions of angle A equal which func-
tions of angle B?
2...Some slide rules only give the value of the tangent of angles
between 0 and 45°. How could you use such a slide rule to
find cot 65°?
ANSWERS
1...(a) A = 53. 2°
hypotenuse 65
cos B = 0.80; tan B = 0.75; cot B = 1.33
(c) sin A = 0,80; cos A = 0,60; tan A = 1.33; cot A= 0.75
Sin A = cos B; cos A = sin B; tan A = cot B; cot A = tan B.
2...The cot 65° can be found by looking up tan 25°, since cot 65°
= tan (90° - 654 = tan 25°.
FINDING SLNES AND COSINES WITH THE S SCALE. . .The sine and
cosine of any angle between 5, 75° and 90° can be found by means of
the S scale on your slide rule. Notice that each major division of the
S scale has a pair of numbers side by side. To find the sine of an
angle, ignore the red number of each pair and find the value of the
angle on the S scale reading only the black numbers. Place the
hairline over the angle on the S scale and read the sine of the an-
gle below on the C scale. When finding the sine by means of the
S scale (angles between 5.75 and 90 degrees), always place the
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72
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Read O S58 on C scale
Fig. 25. Finding sin 35.5° by means of the C scale and black § scale.
decimal directly in front of the first digit of the value read on the
C scale, Figure 25 shows the § scale with the left numbers of
each pair omitted and the hairline set on 35.5°, The sin 35. 5° is
then read on the C scale as 0.581.
To find the cosine of an angle, ignore the black numbers of each
pair and find the angle on the S scale reading only the red num-
bers. The cosine of the angle can then be read directly below on
the C scale. The decimal again should be placed in front of the
first digit. Figure 26 shows the S scale with the black numbers
of each pair omitted and the hairline set on 49.5°, (Note that when
you are reading the red numbers of each pair, the scale begins at
84, 25° and decreases to 0° at the right index.) From the C
scale cos 49, 5° is found to be 0.649.
Opposte ¢92.5* on red S scole
— = 5 f EE:
{
|
Reod on € scole
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Fig. 26. Finding cos 49.5° by means of the C scale and red § scale.
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WHAT HAVF YOU LEARNED?
Find the following values by means of the S scale.
1...sin 6,.3°= ? 6...cos 31. 6° = ?
2...8in 75.8° = ? 7...sin 25.3° = ?
3...cos 18.9° = ? 8...cos 54,7" = ?
4...sin 43, 6°= ? 9,..sin 48.6° = ?
5...cos 45° = ? 10...cos 28,7°= ?
ANSWERS
1,..0.109 2,.,,0.970 3...0.946 4...0,690 5...0.707
6,..0.852 7,,.0.427 8...0.578 9.,..0.750 10...0.877
You may have noticed that the right number of each black and red
pair on the § scale is exactly 90° minus the left number of that
pair. In the preceding section it was shown that cos A = sin (90° -
A). When finding a cosine of an angle by the method given above,
you are in reality finding the sine of 90° minus that angle by means
of the black numbers which you ignore. This is why only one scale
is needed to find both sines and cosines.
Often it is desired to work backwards and find the angle when the
sine or cosine of that angle is known. If the value of the sine or
cosine has its decimal in front of the first digit, you know that the
angle is between 5, 75° and 90° and so can be found on the S scale,
Knowing the sine of the angle, set the hairline over this value on
the C scale and read the angle on the S scale, ignoring the red
numbers of each pair. When the cosine of the angle is given, the
process is identical except that the black numbers, rather than the
red, must now be ignored in reading the angle.
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74
y
u,
Abelo—O
Fig. 27. Figure for use with Example 44.
Example. ..44
Given the right triangle shown in Fig. 27, what is angle A?
Solution...
(1) Find cos A from the sides of the triangle.
= adjacent side _ 1
cos. A ypotenuse 2
(2) Knowing cos A, find A by means of the S scale.
Set the hairline over 500 on the C scale. Ignore
the black numbers and read 60° (not 30°) on the S
scale.
Hence, A= 60°,
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WHAT HAVE YOU LEARNED?
Find the angle A in each of the following problems.
1...sin A= 0.242 6...sin A = 0.895
2...cos A= 0,888 7...cos A = 0,818
3...cos A= 0,954 8...sin A= 0.488
4...8sin A= 0,648 9...cos A = 0.233
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5...cos A= 0,578 10...sin A= 0,985
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ANSWERS
1...14,0° 2...27.4° 8...17.4° 4..,.40.4° 5...54.7°
6...63.5° 7...35.1° 8...29.2° 9...76.5° 10...80.0°
FINDING TANGENTS BY MEANS OF THE T SCALE...The tangents
of angles between 5.71° and 84. 29° can be found by means of the T
scale. Like the S scale, the T scale also is made up of pairs of
red and black numbers. The black numbers on the right of each
pair form a scale which starts with 5. 71° at the left index and in-
creases to 45° at the rightindex. To find the tangent of an angle
between 5. 71° and 45°, ignore the red numbers and place the hair-
line over the value of the angle on the scale formed by the black
numbers, The tangent of the angle can then be read below on the
C scale. The decimal again goes in front of the first digit of the
value read on the C scale. For example, to find tan 35°, place
the hairline over 35 on the black numbered portion of the T scale
and read tan 35° = .700 on the C scale.
The red numbers of the T scale form a scale which starts with
84, 29° at the left index and decreases to 45° at the right index.
To find the tangent of an angle between 45° and 84. 29°, place the
hairline over the value of the angle as found on this red scale. In
this case, the value of the tangent must be read below on the CI
scale, For angles between 45° and 84, 29°, the value of the tan-
gent lies between 1 and 10, hence, the decimal must be placed be-
tween the first and second digit of the value read on the CI scale.
Refer to Fig. 28 which shows the hairline set at 71° on the red T
scale, The value 290 is read on the CI scale and since the deci-
mal point goes after the first digit, tan 71° = 2.90.
Here again your slide rule is using one scale for two purposes.
The black numbers are 90° minus the red numbers as was the case
with the § scale. When you find the tangent af say, 70°, you place
the hairline on the red value 70. Since the black number next to
70 is 90° - 70° = 20°, the hairline automatically sets up the tan 20°
on the C scale. From the formulas previously given you can see
that
1 1
tan A= ot A” tan (90° - A)
- 1
so that tan 70° = fan 20°
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76 Opposite 7/* on red T seek
eset cues ot Yare vate ece hes icons
Read cot Tit = 344 on C scale Reod ton 7)*= 250 on Cl aco
Fig. 28. Finding ton 71° ond cot 71° by means of the red T scole.
The final answer which appears on the CI scale is the reciprocal
of tan 20° and hence the tan 70°.
WHAT HAVE YOU LEARNED?
Find the tangents of the following angles by means of the T scale.
1...7.38° 2...45.0° 3...69.1° 4...19.6° 6&...76.3°
C5224 (T...38S. 8% SATE 95552676") 1055.68;4°
ANSWERS
1...0.1287 2...1.00 3...1.671 4...0.356 5...2.79
6,..1.160 7..-0.654 8...3.655 9...0.499 10,..2.53
To find the angle when its tangent is known, you must reverse the
above process. First inspect the value of the tangent, noting the
position of the decimal point. Ifthe decimal is in front of the first
digit, the angle must lie between 5.71° and 45". Place the hair-
line over the value of the tangent on the C scale and read the an-
% leon the black T scale. For example, if the value of the tangent
& is 0.415, set the hairline over 415 on the C scale and read the an-
: gle as 22.5° on the black T scale.
$
£ If the decimal lies between the first and second digit_of the value
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of the tangent, the angle lies between 45° and 84, 29°. In this case, 77
place the hairl