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For co-sines and co-tangents, the degrees are given at the bottom of the page, and the minutes in the righthand column.

The columns marked D 1" contain the difference for 1".

56. Problem.

Find the logarithmic sine of 48° 25′ 30′′.

log sin 48° 25' D 1".19. ... Correc. for 30".19×30 · .. log sin 48° 25' 30" 9.87396

57. Examples.

1. Find the logarithmic sine of 75° 35'.

=

In case of co-sine or co-tangent, the correction must be subtracted, since between 0° and 90°, the greater the angle, the less the co-sine and co-tangent.

9.87390. 6

2. Find the logarithmic sine of 25° 40' 24".

Ans. 9.98610.

58. Problem.

Ans. 9.63673.

3. Find the logarithmic co-sine of 29° 55′ 55′′.

Ans. 9.93782.

4. Find the logarithmic tangent of 50° 50′ 50′′. Ans. 10.08927.

5. Find the logarithmic co-tangent of 65° 45′ 30′′. Ans. 9.65349.

To find the angle corresponding to a given logarithmic sine, co-sine, tangent, or co-tangent.

Find the angle whose logarithmic sine
For next less we have sin 44° 30'

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5", .21)1.00(5.

In case of co-sine and co-tangent, the correction for seconds must be subtracted, since the greater the cosine or co-tangent, and consequently the greater the logarithm, the less the angle for values between 0° and 90°.

59. Examples.

1. Find the angle whose logarithmic sine is 9.98437. Ans. 74° 43′ 17′′. 2. Find the angle whose logarithmic co-sine is 9.78456. Ans. 52° 29′ 19′′. 3. Find the angle whose logarith. tangent is 10.12346. Ans. 53° 02′ 11′′. 4. Find the angle whose logarith. co-tangent is 9.99999. Ans. 45° 00' 03".

60. Problem.

Given any natural function, to find the corresponding logarithmic function.

1st SOLUTION.

Find from the natural function the corresponding angle; then, from the angle, the corresponding logarithmic function.

2d SOLUTION.

Let a denote any arc or angle, ƒ(a), any function of a to the radius 1, and f(a) the corresponding

R

function of a to the radius R. Then, by article 49 we have,

ƒ (a)n = ƒ (a), × R.

Substituting the value of R in the second mer ber, f(a)n = f(a), X 10,000,000,000.

R

.. log f(a) = log f (a), + 10.

R

Hence, Add 10 to the logarithm of the natural function.

61. Examples.

1. Given nat. sin a = .98457, required a and log sin a. Ans. a 79° 55' 25", log sin a 9.99325. 2. Given nat. cos a www.xcomm = .63878, required a and log cos a. Ans. a= 50° 17′ 52", log cos a = 9.80536. 1.68685, required a and log 59° 20 23", log tan a 10.22708. and log 10.15225.

3. Given nat. tan a tan a. Ans. a 4. Given nat. cot a cot a. Ans. a

1.41987, required a 35° 09′ 24′′", log cot a

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62. Problem.

Given any logarithmic function, to find the corresponding natural function.

1st SOLUTION.

Find from the logarithmic function the corresponding angle; then, from the angle, the corresponding natural function.

2d SOLUTION.

From article 49 we have,

ƒ (α) R
f(a)1 =
1
R
... log f (a), = log f (a)-10.

Hence, Subtract 10 from the logarithmic function, and find the number corresponding to the resulting logarithm.

1. Given log sin a sin a.

cos a.

2. Given log cos

tan a.

or h

.. (1)

3. Given log tan a

10.22708, required a and nat. Ans. a= 59° 20′ 23′′, nat. tan a= = 1.68685. 4. Given log cot a = 10.15225, required a and nat. Ans. a= 35° 09′ 24′′, nat. cot a = 1.41987.

cot a.

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63. Examples.

9.87654, required 9.87654, required a and nat. Ans. a= - 48° 48′ 44′′, nat. sin a== .75255.

b

PB PK :: HB : MK,
: 1 psin P.

or h

BP: BR :: HP: SR,
: 1 :: b : sin B.

=

a = a = 9.84877, required a and nat. Ans. a=45° 05′ 41′′, nat. cos a=.70595.

=

RIGHT TRIANGLES.

64. Principles.

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1. Either side adjacent to the right angle is equal to the sine of the opposite angle multiplied by the hypotenuse.

2. The sine of either acute angle is equal to the opposite side divided by the hypotenuse.

Since the angles P and B are complements of each other, sin P= cos B, and sin B = cos P; ... (1) and (2) become,

(3)

p=h cos B.

b =

... (5)

(7)

PH: PN: HB
BH: BT :: HP:

p

h cos P.

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3. Either side adjacent to the right angle is equal to the co-sine of the adjacent acute angle multiplied by the hypot

enuse.

4. The co-sine of either acute angle is equal to the adjacent side divided by the hypotenuse.

b tan P.

and (4)

b cot B.

p cot P.

NL, or b 1

TQ, or p:

1

cos B

... (6)

cos P

and (8)

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b=p tan B.

5. Either side adjacent to the right angle is equal to the tangent of the opposite angle multiplied by the other side.

6. The tangent of either acute angle is equal to the opposite side divided by the adjacent side.

Since the angles P and B are complements of each other, tan P=cot B, and tan B=cot P; .'. (5) and (6) become,

tan P

cot B

p

h

p

tan P.

: : b : tan B.

b

h

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tan B:

cot P =

[ocr errors]

p

p

b

b

p

7. Either side adjacent to the right angle is equal to the co-tangent of the adjacent acute angle multiplied by the other side.

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