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We have before found it 77° 5′, in the solution of the first case, and this difference arises from the quantities neglected in the valua tion of the logarithms.

112. This case may also be solved a single proportion. For retaining the notation used in the last paragraph of the preceding case, let a', b', c', A', B, C', represent the sides and angles of the supplemental triangle ; and call a+b+c=S. Then according to

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2

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Similarly sin b=sin (180o—B)=sin B; and sin c'=sin C.

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A and we

Substitute these values in the above expression for cos.

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sin

A+B+C 2

The expression under the radical sign is always positive; for =S has been shewn (70) to be greater than 90°, and less than 270o, and therefore cos S is negative, and consequently -cos S, positive.

Again, S-A= B+C-A, but since in every spherical triangle

2

the sum of any two sides is greater than the third side, we have

b'+c' a'; therefore 180°-B+180°-C 180°-A, and consc

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Given the three angles of an oblique spherical triangle respectively 61° 25′, 82° 36′, and 82° 4′; required one of the sides; for example that opposite the angle 82° 36'.

By the preceding formula we have

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EXAMPLES OF THE CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES.

1. In the oblique-angled spherical

triangle ABC, given BC=114o 30', AC 56° 40′, and the angle BAC=

A

125o 20'; required the other parts of the B

triangle ?

C

Answer, B=48° 30′, ▲ C=62o 54, side AB=83o 11'. 2. In the oblique-angled spherical triangle ABC, given AB=45°30′, BC=72° 40′, and the angle A=135o; required the rest?

Answer, C=31° 53′ ‍35′′, ▲ B=24o 31 25′′, AC=34° 4′ 46′′. 3. In the oblique-angled spherical triangle ABC, given BC 114° 30', CA 56° 40', and the angle C=62° 54'; required the rest?

Answer, B=73° 31', ▲ A=125° 20', side AB=83o 12'.

4. In the oblique-angled spherical triangle ABC, given AB=41° 45, BC 730 20', and the angle B=30o 30'; required the rest? Answer, A=130° 8′ 45′′, Z C=31° 33′ 43′′, AC=40° 12′ 59′′.

5. In the oblique-angled spherical triangle ABC, given the angle B-52° 20′, the angle C-63° 40', and the side AB-83° 25′; required the rest? Answer, AC 61° 19′ 53′′, BC=118° 21′ 25′′, LA=127° 26′ 47′′.

6. In the oblique-angled spherical triangle ABC, given the angle C=48° 20', the angle B-125o 20', and the side AC-114° 30'; required the rest?

Answer, AB-56o 40′, BC=83o 11', ▲ A=62o 54'.

7. In the oblique-angled spherical triangle ABC, given the an gles B-53° 30', C-60° 15', and the side BC-115° 40'; required the rest?

Answer, AB-75° 56′ 31′′, AC-67° 2′ 39′′, A=126° 13′ 35′′. 8. In the oblique-angled spherical triangle ABC, suppose the angle A-125° 20′, the angle C-48° 30′, and the side AC 83° 12′, required the rest?

Answer, BC 114° 30', AB-65° 30′, ▲ B=62o 54' 9. In the oblique-angled spherical triangle ABC, given AB=81° 10′, AC=60° 20′, BC=112o 25'; required the angles? Answer, LA=122° 11′ 6′′, ▲ B=52° 42′ 11′′, LC-64° 46′ 36′′.

10. In the oblique-angled spherical triangle ABC, given AB 56° 40′, AC-83o 13', and BC-114o 30'; required the angle A opposite the side BC?

Answer, LA=125° 18′ 56'.

11. In the oblique-angled spherical triangle ABC, given the angle A=125o 20', the angle B-48° 30′, and the angle C-62° 54′ ; required the sides?

Answer, AB-83° 12' 4", BC-114° 29′ 56", AC-56° 39' 29". 12. In the oblique-angled spherical triangle ABC, given the angles A=129° 30', B-54° 35', and C=63o 5'; required the sides? Answer, AB--82° 19′, BC-120° 57′ 4′′, AC-64° 55' 36".

NAPER'S RULES OF THE CIRCULAR PARTS.

The proportions upon which the solution of the various cases of right-angled spherical triangles depend, are simple, and perfectly adapted to logarithmic computation; but they are not easily remembered. All these cases may be solved by Naper's Rules of the Circular Parts, which supply an artificial memory to the computist; and in the whole compass of the mathematical science it will not be easy to find rules equally ingenious and conducive to facility and brevity of computation. The nature and application of these rules will readily be understood from the following explanation:

In a right-angled spherical triangle, the right angle is not considered; the complements of the other two angles, the complement of the hypothenuse, and the two sides, making in all five quantities, are called by Naper circular parts. Any one of the circular parts may be called a middle part (M), then the two circular parts imme. diately on the right and left of M are called adjacent parts; and the other two remaining circular parts, each separated from M, the middle part, by an adjacent part, are called opposite parts: The circular parts are reckoned in order round the triangle, so that to every one considered as a middle part, there are two adjacent and two opposite parts.

In every question proposed for solution, three of the circular parts are concerned, of which two are given, and one required; and of these three the middle part must be such, that the other two may be equi-distant from it, that is, either both adjacent, or both opposite parts.

This arrangement being made, the value of the part required may be found by the following

RULE. The product of the radius and the sine of the middle part, is equal to the product of the tangents of the adjacent, or to the product of the cosines of the opposite parts.

Note. It is obvious that the cosine of a complement is a sine, and the tangent of a complement a cotangent, and vice versa.

Example.

The hypothenuse AB of a right-angled spherical triangle is 63° 56', and the leg BC is 40o, what is the other leg AC, and the angles A and B ?

1. To find the leg AC.

Here the circular parts concerned are the legs and hypothenuse, and it is evident that if AB be made the middle part, BC and AC will be opposite parts; consequently

rad x cos AB=cos BC x cos AC

whence AC 54° 59′ 49′′.

2. To find the angle A.

Here BC is the middle part, and AB and the angle A are opposite parts; therefore

rad x sin BC-sin AB x sin A

whence A is found-45° 41′ 24′′.

3. To find the angle B.

Here the angle B is the middle part, and the leg BC, and the hypothenuse AB are adjacent parts; therefore

rad x cos B-tang BC × cot AB

whence B is found 65° 45′ 57′′.

By the above rule of Naper's we are enabled to solve all the cases of right-angled spherical triangles; and also those cases of oblique-angled spherical triangles in which we have directed a perpendicular to be drawn from an angle to the opposite side, provided that two of the given parts remain in one of the two triangles thus formed.

Trigonometry is a branch of mathematical science which is indispensible on the calculations of remote and inaccessible objects; and hence its use in geography, in ascertaining the various distances and position of places on the earth; in navigation, in directing the course, the latitude and longitude of a ship. Spherical Trigonometry is particularly applied to the sublime science of astronomy, in discovering the positions, magnitudes, and distances of the heavenly bodies.

It may also be applied to the useful arts, as in architecture the theorems of Naper will be found useful in ascertaining the angles which two adjacent planes of a roof at a hip make with each other, the inclination of the planes being given to the horizon.

In short, a catalogue of its applications would be too formidable to be inserted in this place.

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