Ex. 2. What is the arc whose cotangent is 10.219684?

The arc is, therefore, 31° 5' 24".

Ex. 3. Required the arc the cosine of which is 9.764227. Ans. 54° 28' 27".

Ex. 4. Required the arc the tangent of which is 10.876429. Ans. 82° 25' 44".

Ex. 5. What is the arc the cotangent of which is 11.562147? As this corresponds to an arc less than 2°, take it from 20.000000: the remainder, 8.437853, is the tangent. The arc is found as follows:

8.437853

8.437732

1° 34' 10" tang.

Diff. to 1"

76.8) 121.0 (1.6"
768

44.20

The angle is, therefore, 1° 34' 11.6".

Ex. 6. What are corresponds to the cotangent 8.164375?

Ans. 89° 9' 48.6".

135. Table of Chords. This table contains the chords of arcs to 90° for every 5 minutes. It is principally used in laying off angles, as explained in Art. 120, and in protracting surveys by the method of Art. 343.

SECTION IV.

ON THE NUMERICAL SOLUTION OF TRIANGLES.

136. Definition. THE solution of a triangle is the determination of the numerical value of certain parts when others are given. To determine a triangle, three independent parts must be known,-viz.: either the three sides, or two sides and an angle, or the angles and one side. The three angles are not of themselves sufficient, since they are not independent,—any one of them being equal to the difference between the sum of the others and 180°.

In the solution of triangles several cases may be distinguished; these will be treated of separately. These cases are applicable to all triangles. But as there are special rules for right-angled triangles, which are simpler than the more general ones, they will first be given.

A.-THE NUMERICAL SOLUTION OF RIGHT-ANGLED TRIANGLES.

137. The following rules contain all that is necessary for solving the different cases of right-angled triangles.

1. The hypothenuse is to either leg as radius is to the sine of the opposite angle.

2. The hypothenuse is to one leg as radius is to the cosine of the adjacent angle.

3. One leg is to the other as radius is to the tangent of the angle adjacent to the former.

DEMONSTRATION.-Let ABC (Fig. 43) be a triangle right-angled at B. Take AD any radius, and describe the arc DE; draw EF and DG perpendicular to AB. Then EF will be the sine, AF the cosine, and DG the tangent, of the angle A. Now, from similar triangles we

have

Fig. 43.

EG

A

FD

B

EXAMPLES.

Ex. 1. In the triangle ABC, right-angled at B, there are given the base AB = 57.23 chains, and the angle A 35° 27′ 25", to find the other sides.

For facility of calculation, the proportions are generally written vertically, as below.

Ex. 3. Given the two legs AB = 59.47 yards, and BC = 48.52 yards, to find the hypothenuse and the angles. Ans. A 39° 12′ 36′′, C 50° 47′ 24′′, and AC 76.75 yds.

=

Ex. 4. Given the hypothenuse AC 97.23 chains, the perpendicular BC 75.87 chains, to find the rest.

=

Ans. A 51° 17′ 22′′, C 38° 42′ 38", AB 60.81 ch.

Ex. 5. Given the angle A= 42° 19' 24", and the perpendicular BC= 25.54 chains, to find the other sides.

Ans. AC 37.932 ch., AB 28.045 ch.

Ex. 6. Given the angle C72° 42′ 9", and the hypothenuse AC 495 chains, to find the other sides.

=

Ans. AB 472.612 ch., BC 147.18 ch.

Ex. 7. In the right-angled triangle ABC we have the base AB = 63.2 perches, and the angle A 42° 8′ 45′′, to find the hypothenuse and the perpendicular.

Ans. BC 57.20 p., AC 85.24 p.

138. When two sides are given, the third may be found by (47.1); thus,

1. Given the hypothenuse and one leg, to find the other.

Rule. From the square of the hypothenuse subtract the square of the given leg: the square root of the remainder will be the other leg; or,

Multiply the sum of the hypothenuse and given leg by their difference: the square root of this product will be the other leg. This is evident from (47.1) and (cor. 5.2.)

2. Given the two legs, to find the hypothenuse.

Rule. Add the squares of the two legs, and extract the square root of the sum: the result will be the hypothenuse.

EXAMPLES.

Ex. 1. Given the hypothenuse AC = 45 perches, and the leg BC= 29 perches, to find the other leg.

Rule 1. AB = √AC2 — BC2 = √2025 — 841

=

✓1184 =

✔(AC + BC).(AC — BC)

=

√74 × 16 =

Ex. 2. The two legs AB and AC are 6 and 8 respectively: what is the hypothenuse?

Ans. 10.

Ex. 3. The hypothenuse AC is 47.92 perches, and the leg AB is 29.45 perches: required the length of BC. Ans. 37.8 perches. Ex. 4. The hypothenuse of a right-angled triangle is 49.27 yards, and the base 37.42 yards: required the perpen

dicular.

Ans. 32.05.

B. THE NUMERICAL SOLUTION OF OBLIQUE-ANGLED

TRIANGLES.

CASE 1.

139. The angles and one side, or two sides and an angle opposite to one of them, being given, to find the rest.

RULE.

1. As the sine of the angle opposite the given side is to the sine of the angle opposite the required side, so is the given side to the required side.

2. As the side opposite the given angle is to the other given side, so is the sine of the angle opposite to the former to the sine of the angle opposite the latter.

DEMONSTRATION.-Both the above rules are combined in the general proposition. The sides are to one another as the sines of their opposite angles.

Let ABC (Fig. 45) be any triangle. From C let fall CD perpendicular to AB. Then (Art. 137) AC: CD :: r sin. A, and CD: CB:: sin. B: r. Whence (23.5) AC:

CB sin. B: sin. A.