come to the column designated at the top by sine, cosine,

tang, or cotang, as the case may be; the number there found is the logarithm required.

log sin 19° 55'

log tan 19° 55'

Thus,

45°, look for the degrees at for the minutes in the right

If the angle is greater than the bottom of the page, and hand column; then follow the corresponding horizontal line backwards till you come to the column designated at the bottom by sine, cosine, tang, or cotang, as the case may be; the number there found is the logarithm required.

Thus,

To find the logarithmic functions of an arc which is expressed in degrees, minutes, and seconds.

35. Find the logarithm corresponding to the degrees and minutes as before; then multiply the corresponding number taken from the column headed "D," by the number of seconds, and add the product to the preceding result, for the sine or tangent, and subtract it therefrom for the cosine or cotangent.

EXAMPLES.

1. Find the logarithmic sine of 40° 26' 28".

The same rule is followed for decimal parts, as in Art. 12.

2. Find the logarithmio cosine of 53° 40' 40".

If the arc is greater than 90°, we find the required function of its supplement (Arts. 26 and 28).

3. Find the logarithmic tangent of 118° 18' 25".

4. Find the logarithmic sine of 32° 18′ 35′′.

Ans. 9.727945.

5. Find the logarithmic cosine of 95° 18' 24".

Ans. 8.966080.

6. Find the logarithmic cotangent of 126° 23′ 50′′.

Ans. 10.132333.

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′′.

3. Find the arc corresponding to the logarithmic Ans. 49° 20′ 50′′.

sine 9.880054.

4. Find the arc corresponding to the logarithmic cotangent 10.008688.

Ans. 44° 25′ 37′′.

5. Find the arc

cosine 9.944599.

corresponding to the logarithmic Ans. 28° 19' 45".

t

SOLUTION OF RIGHT-ANGLED

TRIANGLES.

37. In what follows, we shall designate the three angles of every triangle, by the capital letters A, B, and C, A 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,
describe the arc DG, and draw GF

E

G

C

FD

and DE perpendicular to CA then
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 CAB are similar (B. IV., P. XVIII.), we 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.

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.

Either side about the right angle may be regarded as the base; 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,