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

L cot A,
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".

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

x 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 V-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), 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 41 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 coseo 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 « when 1.034 = 1.2143, having given that log 1.03 ·0128372, and log 12143 = 4.0843260. 15. Solve the equation 224-1 - 40 = 9.2*, having given

•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° 3715", and L cot 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.

log 2

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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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a

с

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, AB = BC + AC - 2 BC.CD.

CD
Now

= cos C, or CD AC cos C.

AC Hence we have, ABP = BC + AC 2 BC. AC cos C, or c = a? + 72 2 ab cos C.

.. cos C

a2 + 62 - c

or, c2

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

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

a2 + 72 2 ab cos C.

a + 62 ca :: cos C

as before.

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

72 + ca aa Cos A

2 bc

a2 + c2 62 Cos B

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2 ac

2

32. To express the sine of any angle of a triangle in terms of the sides. We have

172 + ca Sin? A = 1 - cos” A = 1

2 bc
(2 bc)2 – (62 + 02 – a?)?

(2 bc)
{(b + c)* – ay}{a? (6 – c)"}

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

(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) (6 + c - a) (a + c – b) (a + b - c).

2 bc Let a + b + c = 2 s, then b + c 2 (s a),

2 (8 - b), a + b c = 2 (s – c).

1 Hence, sin A

128.2 (s – a).2 (8 6). 2 (8 – c) 2 bc

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a =

a + c

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ас

From the form of this result we have also

2 Şin B = 18 (8 – a) (8 - ) (8 – c), sin C =

2 ab

18 (8 - a) (8 6) (8 – c).
33. To find the area of a triangle.
Using the figures of Art. 30, we have-
A ABC 1 BC. AD, or, since AD AC sin C,

BC. AC sin C
ab sin C......

.(1). From the form of this result we have alsoA ABC 1 ac sin B ....

:(2). and A ABC bc sin A......

(3). The results in (1), (2), (3) express the area of a triangle in terms of two sides, and the included angle.

We will now express the area in terms of the three sides.
We have
Δ ABC 1 ab sin C, or, by last Art,

2
18 (8

a) (8 - 0) (8 -. c)
V8 (8 - a) (8 – b) (8

c). 34. To express the sine, cosine, and tangent of half an angle of a triangle in terms of the sides. We have, Art. 18, A

62 + c
1 2 sino
= COS A

2 bc
A
72 + c

a (6 - c)
... 2 sin? 1
2
2 bc

2 bc
(a + c b) (a + b c) 2 (8 6). 2 (8 c);
2 bc

2 bc
A (8 - ) (8 -

c)
or,
sin
2

bc
A
b) (8 c)

... (1). 2

bc 5

= 1 ab.ab

8

.... (4).

a

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sin

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