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

62 + ca – a We have

2bc The factor b2 + c2 – aż 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

s (8 a) log tan ; A = { log (8 b) + } log (8 – c) – 1 log (s a) – 1 log s. This is well adapted. Or we may use A ( b) (

A

(s

- a)
sin
2
bc

bc si If all the angles are to be found the tan formula is the most convenient, because only the four logarithms of 8, 8 a, 8-6, 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

=s-6 s=56410 .:. From the tables we find log s=4.7513561

log (8 b)=4:4248327 log (s a)=4:3152985

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

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

•3403191 = { log 8 +3 log (s a) – } log (8 6) – } log (8 — c). .. I 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".

=8-- a

=8-C

=

=

= (6—c) {1

{1+16–054

sino A}

2 [(bc)

.

-C

=

b-C

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C

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= + co – 2bc cos A.

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

a2 = 62 +62 2bc (1 – 2 sin1 A),
= (b - c)2 + 4bc sin } A,

4bc
+

() 2
Now there must be some angle, o say, whose tangent

A

sin

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

2 J(be)

A tan A=

sin

2 .. log tan 6 = log 2 + } log b + { log c - log (6 -c) + log sin A. Having found 0, we have

ao = (b -c)'(1 + tana) = (b -c) seca 0,

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

B-C

А
tan

cot
2 b + c 2'
B-C

A
.. log tan = log (6 -c) - log (b + 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,
or a = (b + c) sin 1 A sec } (B-C),

or a = (b c) cosA cosec 3 (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 (6 + c) + log sin 1, A + log sec } (B-C).

... a

=

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sin

-C

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

2./ (bc) А Let tan A=

b

2 Then L tan 0=log 2+ } log b + log c- log (b-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. Now a=

= (b -c) sec , :: log a=log (6 c)+log sec 0

=2.6665180+.0602592

= 2:7267772, ..d=533:0614=533:06 approximately. Next L tan (B-C)=log (6 -c) - log (b+c)+L cott A

=2:6665180 – 2:9921115+104973279

= 10:1717344, .: (B-C)=56° 2' 39", and } (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=5330625=533:06 approximately (as above).

284. To solve a triangle when two sides and a non-included angle are given. (See Arts. 169, 170.1

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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.

n

=

mn

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x2

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2039

و به

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

mnapn="/ap"; "apn = a; "/(a)=(a)";

/ap mal = mm / apm+ng; apmal = min/apm-ng;

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

3. Find the values of 3-4, 274, 128-4, 219o.
4. Express the identities in Ex. 1, in terms of

powers instead of roots of a. 5. Express as powers of x :

1 1 ತಿ,

1, x?'

Natx /at,

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

log: 243, log, 343, log. 64, logi, 10000, log: 81,
logo1 5/11", log: 127, log, V64, logs 1125,
logis Tē, log, 2, log,7 3, log, 81, log: 32,

1

35
loga V}, logio logo

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

log2 ·õ, log: :04, log32 128.
7. Find the characteristics of

logo 3245, log10 123, logo 3.45, logio 1234,
logio .001234, log10 57000, log, 90, log, 100,

log: 200, logo 50, log10 (123 x 109).
Prove the following statements : (8—23).

8. alog b = bloga. 9. alog be alog b. alogc.
10. alog da = (alogb)

.
11. logab x logoc x logo a = 1.
J. T.

14

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bc

b

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b

=

mn

12. loga b x loge d=loga d x log. b.
13. logab x logo c x loge d x loga e = loga e.
14. logama x logony x logo' = logope* x log.my x logana.
15. If logan= log,m, then each = loga."

16. If loga x = logo y = logom, then each = log XPy4mm to the base aPb4c".

17. log a to base a” = log n/ac 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).

=

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Logarithms to base 10.

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

1
3/2, , •

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

1
5, 25, 50, -2, .008, 5000, .0005.

4, 8, 128, 12, 1024, }, 764, 25, 03125, 200, -002.

1125)

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

9, }, 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

7 35, 210, 245, 28, 42, 63, 441, 504. State in each case the possible error in the result calculated.

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