To find the arc corresponding to any logarithmic function.

36. This is done by reversing the preceding rule: Look in the proper column of the table for the given logarithm; if it is found there, the degrees are to be taken from the top or bottom, and the minutes from the left or right hand column, as the case may be. If the given logarithm is not found in the table, then find the next less logarithm, and take from the table the corresponding degrees and minutes, and set them aside. Subtract the logarithm found in the table, from the given logarithm, and divide the remainder by the corresponding tabular difference. The quotient will be seconds, which must be added to the degrees and minutes set aside, in the case of a sine or tangent, and subtracted, in the case of a cosine or a cotangent.

Tabular difference 7.68) 391.00 (51", to be added.

Hence, the required arc is 15° 19′ 51′′.

Tabular difference 7.58) 131.00 (17, to be subt.

Hence, the required arc is 74° 28′ 43′′.

37. In what follows, we shall designate the three angles f every triangle, by the capital letters A, B, and C, denoting the right angle; and the sides lying opposite the angles, by the corresponding small letters a, b, and c. Since the order in which these letters are placed may be changed, it follows that whatever is proved with the letters placed in any given order, will be equally true when the letters are correspondingly placed in any other order.

Let CAB represent any triangle, right-angled at A. With C as a centre, and a radius CD, equal to 1,

E

describe the arc DG, and draw GF

and DE perpendicular to CA: then

FD

will FG be the sine of the angle C, CF will be its cosine, and DE its tangent.

Since the three triangles CFG, CDE, and

similar (B. IV., P. XVIII.), we

CAB are

may write the propor

Translating these formulas into ordinary language, we have the following

PRINCIPLES.

1. The perpendicular of any right-angled triangle is equa to the hypothenuse into the sine of the angle at the base.

2. The base is equal to the hypothenuse into the cosine of the angle at the base.

3. The perpendicular is equal to the base into the tan gent of the angle at the base.

4. The sine of the angle at the base is equal to the perpendicular divided by the hypothenuse.

5. The cosine of the angle at the base is equal to the base divided by the hypothenuse.

6. The tangent of the angle at the base is equal to the perpendicular divided by the base.

base

Either side about the right angle may be regarded as the ; in which case, the other is to be regarded as the perpendicular. We see, then, that the above principles are sufficient for the solution of every case of right-angled triangles. When the table of logarithmic sines is used, in the solution, Formulas (1) to (6) must be made homogeneous, by substituting for sin C, cos C, and tan C, respectively,

Making these changes, and reducing, we have,

Given the hypothenuse and one of the acute angles, to find

the remaining parts.

38. The other acute angle may be found by subtracting

1. Given a = 749, and C = 47° 31' 10"; required B, b, and C.

OPERATION.

B = 90° 47° 03' 10" 42° 56' 50".

Applying logarithms to Formula (7), remembering that the logarithm of R is equal to 10, we have,

Applying logarithms to Formula (8), we have,

28

4050

505.882

Ans. B = 42° 56′ 50′′, b = 510.91, and c

= 548.955

505.884

554.39+

2. Given a = 439, and B = 27° 38′ 50′′, to find

C, b, and c.

OPERATION.

C = 90° 27° 38′ 50′′ = 62° 21′ 10′′;

Ans. C 62° 21′ 10′′, b = 203.708, and c = 388.875.

3. Given a = 125.7 yds., and B = 75° 12', to find the other parts.

Ans. C 14° 48', b = 121.53 yds., and c 32.11 yds.

4. Given a = 325 ft., and C 27° 34', to find the other parts.

Ans. B = 62° 26', c = 150.4 ft., and b 288.1 ft.