Find the angle whose logarithmic sine = 9.84567
For next less we have sin 44° 30′

9.84566

1" X =5", .21)1.00(5.

... Angle 44° 30′ 05′′.

.21

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′ 25′′.

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′ 01′′.

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, f(a), any function of a to the radius 1, and f(a) the corresponding

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 member,

f(a) = f(a) X 10,000,000,000.

... log ƒ(a)n = log ƒ (a), + 10.

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

61. Examples.

=

1. Given nat. sine a .98457, required a and log Ans. a= = 79° 55′ 25′′, log sin a

sin a.

9.99325. 2. Given nat. co-sine a .63878, required a and log Ans. a 50° 17′ 56", log co-sin a="

co-sin a.

=

=

9.80535.

3. Given nat. tan a 1.68685, required a and log Ans. a= 59° 20′ 23′′, log tan a= 10.22708.

tan a.

4. Given nat. co-tan a = 1.41987, required a and log co-tan a. Ans. a=35° 09' 33", log cot a= 10.15224.

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.

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

63. Examples.

1. Given log sin a

sin a.

Ans. a 48° 48′ 44", nat. sin a = .75255.

2. Given log cos a = 9.84877, required a and nat. Ans. a=45° 05′ 40′′, nat. cos a= = .70594.

cos a.

=

3. Given log tan a 10.22708, required a and nat. Ans. a = 59° 20′ 23′′, nat. tan a=1.68685.

tan a.

4. Given log cot a = 10.15225, required a and nat. Ans. a = 35° 09′ 24′′, nat. cot a

cot a.

= 1.41987.

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

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.

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

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

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

9. Either side adjacent to the right angle is equal to the hypotenuse divided by the secant of the adjacent acute angle.

10 The secant of either acute angle is equal to the hypotenuse divided by the adjacent side.

Since the angles B and P are complements of each other sec B =cosec P, sec P =cosec B; ... (9) and (10) become,

11. Either side adjacent to the right angle is equal to the hypotenuse divided by the co-secant of the angle opposite that side.

12. The co-secant of either acute angle is equal to the hypotenuse divided by the side opposite that angle.

Scholium. By some authors, principles 2, 4, 6, 8, 10, and 12, have been given in the form of definitions.

Introducing radius into these formulas, by substituting for any function to the radius 1, the corresponding function to the radius R divided by R, and reducing, we have:

S. N. 5.