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COS A

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282. To solve a triangle when the three sides are given. [See Arts. 165, 166.]

62 + co – a? We have

2bc The factor b2 + c2 – a2 has here to be calculated by multiplications: hence this formula is not well adapted for logarithmic calculations. On the other hand, the formula

А (8 - b) (8 - c)) tan

gives 2 V8 (8-a) log tan {A = 1 log (8 b) + 1 log (8 – c) – 4 log (8 – a) – } log 8. This is well adapted. Or we may use A (8-5) (8-0)

A

(s a)
sin
2
bc

2

bc If all the angles are to be found the tan formula is the most convenient, because only the four logarithms of s, 8 Q, 8 b, 8-c will be required for the three angles.

Example. Given a=35742 yards, b=29813 yards, c=47265 yards, find A. 35742 56410 56410

56410
29813
35742
29813

47265
47265
20668
26597

9145
2)112820

=56410 .:. From the tables we find log s=4.7513561

log (s b)=4:4248327 log (s a)=4:3152985

log (8 – c)=3.9611837 (adding)=9.0666546

(adding)=8-3860164
8:3860164
2).6806382

-3403191 = 1 log 8 + 1 log (8-a) - } log (8-b) - log (8 -c). .: L tan ? A=10 – 3403191=9.6596809. Now

9.6597076=L tan 24° 33'
9.6593733=L tan 24° 32

*0003343= diff. for l'

.. •0003076=diff. for 11 of 60”, i.e. 54" •5, .:: A=24° 32' 54" •5,

.. A=49° 5' 49".

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=8-6

=S-C

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=(6—c) {1

+

2 /(bc)
b-C

b-0

283. To solve a triangle when two sides and the included angle are given. (See Arts. 167, 168.]

I. To find a, we have a= 12 + C2 – 2bc cos A.

This is not well adapted to logarithmic calculation. We may, however, transform it as follows:

ao = 62 +62 2bc (1 – 2 sin 1 A),
= (b c)2 + 4bc sino 1 A,

4bc

sin?

(6 - c)
Now there must be some angle, o say, whose tangent

А
sin

2. By the help of the tables this angle may be found. Thus since

bc

A tan 0 =

2 .. log tan 6 = log 2 + 1 log b + 1 log c – log (6 c) + log sin , A. Having found 0, we have

a= (b -c) (1 + tan) = (b c) seca 0,
... a = (6c) sec 0,
:. log a = log (6 - c) + log sec 0.
II. To find B and C, we have

B-C

А
tan

cot
2 btc 2
B-C

A
.. log tan = log (6 - c) – log (6 + c) + log cot
2

2. Having found B and C we may find a from either of the equations

a = b sin A cosec B,

(b + c) sin 1 A sec 1 (B-C),

or a = (b c) cos } A cosec } (B - C). The two latter of these (which the student may easily prove) have the advantage over the first that they require only two new logarithms. Thus

log a=log (b+c) + log sin } A + log sec 1 (B-C).

or a =

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b-C

Now a=

Example. Given b=723, c= = 259, A = 35° 18', solve the triangle.

2/(bc) A Let tan A=

sin

2 Then L tan 0=log 2+ } log b + } log c – log (6 –c)+L sin 17° 39' =:3010300+1.4295691+1.2066499

- 2:6665180+9:4817315 =9.7524625, .: 0=29° 29' 23" by reference to the tables.

(b-c) sec 0, .. log a=log (b-c)+log sec 0

=2:6665180+.0602592

= 2:7267772, .: =53360614=533:06 approximately. Next L tan $ (B-C)=log (6 - c) – log (b+c) + L cot A

=2:6665180 - 2.9921115 +104973279

= 10:1717344, .. } (B-C)=56° 2' 39", and 1 (B+C) = 72° 21', .. B=128° 23' 39'' and C=16° 18' 21". Hence log a=log (b+c)+log sin 17° 39' +log sec 56° 2' 39",

= 2.9921115 +1:4817315+ •2529351,

= 2:7267781,
.: a=533-0625=533:06 approximately (as above).

284. To solve a triangle when two sides and a non-included angle are given. [See Arts. 169, 170.]

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.. L sin C = log c-log b + L sin B. This formula is immediately adapted to logarithmic calculation and need not be further discussed.

EXAMPLES XI.

1. Prove from definition (the indices being positive integers):

mnyam = m/al; "/apn = al; "/(ap)=(a)p;
m/aP m/a9 = mm/apm+nq; ap = m/al = mm/apm-ng;

m/{(ma)}" = mn/apr. 2. Interpret a', a-5, aš, a *; and write down the logarithms of each expression to the base a.

3. Find the values of 3-4, 278, 128-1, 219o.

4. Express the identities in Ex. 1, in terms of powers instead of roots of a. 5. Express as powers of w:

1 1

2c2 ,

Naco x 3/xt,

1,

2039

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and write down the logarithms of each expression to the base x. 6. Find the values of

log: 243, log, 343, log, 64, logi 10000, log: 81,
logi 3/11"?, log: 127, log, 174, logs /125,
log13 Thy, log, 2, log, 3, log, 781, log: 32,

1
loge V , log10 log

71000 7125'
logio .00001, log, cos 60°, log, cos 45°,

log, 5, logs .04, logs, 128.
7. Find the characteristics of

log10 3245, log10 123, log10 3.45, log10 ·1234,
log10 :001234, log10 57000, log: 90, log, 100,

logs 200, loge 50, log10 (123 x 109).
Prove the following statements : (8—23).
8. alogb -= bloga. 9. alogo

alog b. alog c.
10. qlogbo = (alogb) .
11. logab x logoc x log. a = 1.
J. T.

14

loga e.

mn

12. loga b log, d=loga d x log.b.
13. loga b log c loge d x loga e =
14. logama * loging x logeri = logopee x log.my x logan.
15. If logan = logom, then each = logam

16. If loga x = logo y = loge 2, then each = log XPy4xt to the base aPb4c". .

17. log to base a" = log x/x to base a.
18. 2 log cos A = log (1 + sin A) + log (1 – sin A).
19. 2 log tan A = log (sec A - 1) + log (sec A + 1).
20. log sin 2A = log 2 + log cos A + log sin A.
21. log cos 2A = log (cos A + sin A) + log (cos A - sin A).
22. log (1 + sin 2A) = 2 log (cos A + sin A).
23. log tan (A + 45°) = log (1 + tan A) - log (1 – tan A).

Logarithms to base 10.

24. Given log 2 = -3010300, find the logarithms of

1
4, 8, 128, 32, 1024, 1, •25, •03125, 200, .002.

'
25. Given log 2 = -3010300, find the logarithms of

1
5, 25, 50, 2, .008, 5000, •0005.

4125'
26. Given log 3 = .4771213, find the logarithms of

9, 3, 30, 243, 03, .i. 27. Given log 2 and log 3 (as above) find the logs of

6, 18, 15, 135, 144, 750, 2, .004. 28. Given log 7 = •8450980, find the logs of

343, 343000, -343, .00343, 3.43. 29. Given log 2, log 3, and log 7 (as above) find the logs of

35, 210, 245, 28, 42, 63, 441, 504. State in each case the possible error in the result calculated.

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