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the angle increases from 0° to 90°, yet the cosine, cosecant, and cotangent diminish as the angle increases.

Hence, when any angle is not exactly contained in the tables, we must add the difference in the case of a sine, secant, or tangent; but subtract it in the case of a cosine, cosecant, or cotangent.

And, conversely, when the given logarithm is not contained exactly in the tables, we must in the case of the sine, secant, or tangent take out the next lower tabular logarithm as corresponding to the angle next lower; but in the case of a cosine, cosecant, or cotangent, we must take out the next higher tabular logarithm as corresponding to the angle next lower in the tables.

We shall assume that small differences in the angles are proportional to the corresponding differences of the logarithmic trigonometrical ratios

Ex. 1. Find L sin 56° 28' 24".
Referring to tables, we have
L sin 56° 28'

= 9.9209393 Tab. diff. for 60" or D 836

24 diff. for 24" or d

x 836

334
60
:: L sin 56° 28' 24"

9.9209727
Ex. 2. Find L cos 29° 31' 28".
Now L cos 29° 31'

= 9.9396253 Tab. diff. for 60" or D = - 716

28
... diff. for 28"

334
60
.:. L cos 29° 31' 28"

= 9.9395919 Ex. 3. Find the angle A, when L tan A = 9.8658585

We have 9.8658585 L-tan A.
Next lower, 9.8657702 L tan 36° 17',

883 difference or d,
Also, 2648 tab. diff. for 60" = D,
d

883
And

x 60" = 20".
D

2648
Hence, 9.8658585 = L tan 36° 17' 20".

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x 716

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x 60"

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L cot A,

Ex. 4. Find the angle A, when L cot A 10.0397936.

We have, 10.0397936
Next higher, 10.0399770 L cot 42° 22',

1834 = difference or d,
Also, 2537 tab. diff. for 60" or D,
d

1834 And

x 60" = 43". D

2537
Hence, 10.0397936 L cot 42° 22! 43".

ndo * 60"

Ex. V.

1. Given log 47582 = 4.6774427, and log 47583 = 4:6774518, find log 47.58275.

2. Given log 5-2404 = 7193644, and log 524.05 = 2.7193727, find log ·5240463.

3. Given log •56145 = 1:7493111, and log 56·146 = 1.7493188, find log 3.05614581.

4. Given log 61683 = 4.7901655, and log 616-84 = 2.7901725, find the number whose logarithm is 2.7901693.

5. Find the value of (1.05)15, having given log 1.05 = ·0211893, log 20789 = 4:3178336, and log 20790 = 4.3178545.

6. Find the compound interest of £120 for 10 years at 4 per cent. per annum, having given log 1.04 = •0170333, log 14802 4.1703204, and log 14803 = 4.1703497.

7. A corporation borrows £8,630 at 45 per cent. compound interest, what annual payment will clear off the debt in 20 years ? Log 1.045 = .0191163, log 4.1464 = .6176712, and log 4.1465

.6176817. 8. Find the value of (1•032)10 x $37.62

having given

(-347215)
Log 1•032 ·0136797, log 34722 4.5406047.
Log 3762 3.5754188, log 26202 •4183344.
Log 34721

4.5405922, log 26203 = .4183510.

9. Find L sin 32° 28' 31", having given L sin 32° 28' 9-7298197, and L sin 32° 29' = 9.7300182.

10. Find L cosec 43° 48' 16'', having given L sin 43° 48' = 9.8401959, and L sin 43° 49' = 9.8403276. 11. Required the angle whose logarithmic cotangent is 10.1322449, having given L cot 36° 25' = 10.1321127, L cot 36° 26' = 10.1318483.

12. Construct a table of proportional parts, having given 163 as the tabular difference.

13. In what time will a sum of money double itself at 5 per cent. per annum, compound interest?

14. Find a when 1.03% = 1.2143, having given that log 1.03 = .0128372, and log 12143 = 4.0843260.

15. Solve the equation 222-1 – 40 = 9.27, having given log 2 = 3010300.

16. Given L cos 32° 45' = 9.9341986, D = 752, find L cos 32° 45' 12", and L sec 32° 45' 20".

17. Given L tan 28° 38' = 10.2628291, D = 3003, find L tan 28° 37' 15", and Lot 28° 38' 42".

18. Find the angle whose logarithmic cosine is 9.9590635, having given

L cos 24° 29' 9.9590805,
L cos 24° 31' = 9.9589653.

CHAPTER VII.

PROPERTIES OF TRIANGLES.

30. The sines of the angles of a triangle are proportional to the opposite sides.

We shall designate the sides opposite to the angles A, B, C, by the small letters a, b, c, respectively.

Draw AD perpendicular to BC, or to BC produced.

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It follows, therefore, from the symmetrical nature of this

sin A sin B sin C equation, that

Q.E.D. 6

a + 72 ca 31. In any triangle, cos C =

2 ab
Taking the figures of the last article, we have-
(1.) When C is an acute angle-
By Euc. II., 13, ABP = BC + AC2 – 2 BC.CD.

CD
Now
AC
= cos C, or CD

AC cos C.

2 BC. AC cos C,

Hence we have, AB2 BO2 + AC or c = a + 62 - 2 ab cos C.

... cos C

al + 72

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or, c?

(2.) When C is an obtuse angle, as in the second figure

By Euc. II., 12, AB? BCS + AC? + 2 BC.CD; and CD AC cos ACD AC cos (180° – C) = AC cos C. Hence, ABP = BCP + AC + 2 BC (- AC cos C);

a + 72 2 ab cos C.
a +

ca
.: cos C

as before.

2 ab From the form of this result we have also—

63 + c Cos A =

2 bc

a2 + c2 62 Cos B

2 aC

32. To express the sine of any angle of a triangle in terms of the sides. We have Sin’ A = 1 - cos* A = 1

172 + ca

2 bc
(2 bc)2 – (b2 + c – a)

(2 bc)
{(b 2
+ c)– a} {a? (6 – c)*}

(2 bc)
(a + b + c)(b + c a) (a + C-6) (a + bc).

(2 bc) Hence, taking the square root, and taking the positive sign, because (Art. 11) the sin A is always positive when A is the angle of a triangle, we have

1 Sin A Na + b + c)(b + C 2 bc

a) (a + c b)(a + b -c). Let a + b + c = 2 s, then b + c a = 2 (s – a), 6 2 (8 - b), a + b c = 2 (s

c). 1 Hence, sin A

2 bc

28.2 (8 – a). 2 (8 6). 2 (s 6)

a + c

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