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1. Given two sides a, c, and an angle C opposite to one of them, to find the other opposite angle A.

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2. Given two angles A, C, and a side c opposite to one of them, to find the other opposite side a.

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3. Given two sides b, c, and the included angle A, to find the other angles B, C.

Tan. (B+C)= cos. (b—c)




(B-C)=sine (b+c)
sine (b-c)

cot. A.

cot. A.

4. Given two angles B, C, and the side between them a, to find the other sides b, c.

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5. Given the three sides a, b, c, to find one of the

angles A.


Tan. A = √


(a+b-c) sine (a+c-b)
(b+c-a) sine § (a + b + c)'

6. Given the three angles A, B, C, to find one of the

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Instead of the last two formulæ the four following are often used.

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Cos. a = √

sine B sine C

sine b sine c

cos. § (A + B — C) cos. } (A + C — B) sine B sine C

It is manifest that the preceding equations are susceptible of many changes, by which they become applicable to the solution of cases when different parts are given. Thus, case 3 becomes, by transposition,

sine (b+c)

cos. (b+c)

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sine (b-c)

cos. (b—c)

tan. (B+C);

by which equation we can find the angle, when the two sides which include it and the other two angles are given.

97. The formulæ in the last four cases deserve particular notice both for their elegance, and for their property of showing whether the arc or angle which they express is less or greater than a quadrant or 90°. For since the sine of an arc is the same as the sine of its supplement, both in value (34 Pl. Tr.) and in sign (97 Pl. Tr.), it is impossible to distinguish whether that arc ought to be less or greater than a quadrant. But when we know the cosine, the tangent, or the cotangent of an arc, and also know that the arc cannot be equal to a semicircle (25), which is the limit of the sides of spherical triangles, and of the arcs which measure their angles, we perceive from the sign of the result whether the arc sought is less or greater than a quadrant. In the first case the cosine, the tangent, and the cotangent have the sign+, in the second the sign(110 Pl. Tr.). If therefore we be careful to give to the known quantities which enter into the preceding formulæ, the signs which ought to affect them according to the values of the arcs to which those signs belong, the sign of the result will always determine the species of the arc or angle required; that is, whether the arc is less or greater than a quadrant, and whether the angle is acute or obtuse.

Napier's Rules of the Circular Parts.

The two rules invented by Lord Napier are very useful in spherical trigonometry, because they reduce all the theorems employed in the solution of right-angled triangles to two, which are simple and easy to remember, and very convenient in practice.

98. Def. 1. In a right-angled spherical triangle, if we omit the right angle, and consider only the five remaining parts, or the three sides and the two oblique angles, then the two sides containing the right angle, and the complements of the two oblique angles and of the hypothenuse, are called the circular parts. See plate, fig. 4.

Thus, in the triangle ABC, having a right angle at B, the circular parts are AB, BC, with the complements of C, AC, and A. These parts are called circular, because, when they are named in the natural order of their succession, they go round the circle.

99. Def. 2. Of the five circular parts if any one be taken for the middle part, the two which are contiguous to it, on the right and left, are called the adjacent parts; and the other two, each of which is separated from the middle part by an adjacent part, are called opposite parts. By some authors they are called adjacent extremes, and opposite extremes.

Thus, in the right-angled triangle ABC, AB, BC, 90°—C, 90°— AC, 90°—A, are the circular parts (Def. 1); and if any one of them, as AB, be reckoned the middle part, then BC and 90°-A, which are contiguous to AB, on different sides, are called adjacent parts; and 90° C and 90° — AC, which are separated from it, are called opposite parts. If BC be taken for the middle part, then AB and 90°. C are the adjacent parts; and 90°- AC and 90°- A are the opposite parts. If 90°-AC be the middle part, then 90°C and 90°- A are adjacent parts, and AB and BC are opposite


This arrangement being made, the two rules of the circular parts are contained in the following proposition.

100. In a right-angled spherical triangle, the rectan gle under the radius and the sine of the middle part is equal to the rectangle under the tangents of the adja

cent parts, or to the rectangle under the cosines of the opposite parts.

Note. The reader must remember, that the cosine of the complement of an arc or an angle is the sine of the arc or angle; and that the tangent of the comp. is the cotangent; and conversely. The demonstration is as follows.

Cases 1, 2. Make AB the middle part.

Sine AB : tan. BC :: R : tan. A (49) :: cot. A : R, therefore R x sine AB= cot. A x tan. BC.

Again, Rs.: C: s. AC: s. AB (48),

therefore Rx s. AB=s. C x s. AC.

If BC be made the middle part, the prop. may be proved in the same manner.

Cases 3, 4. Make the compl. of the angle A the middle part.

In the triangle abC, tan. bC : s. ab :: tan. a: R (49), or cot. AC: cos. A :: cot. AB: R (46) :: R : tan. AB, therefore R x cos. Acot. AC x tan. AB.

Again, R: s. aC :: s. C : s. ab (48),

or R: cos. BC :: s. C: cos. A (46),

therefore R x cos. A=s. C x cos. BC.

If the compl. of C be made the middle part, the prop. may be proved in the same manner.

Case 5. Make the compl. of AC the middle part.

In the triangle abC, s. bC : tan. ab :: R : tan. C (49) :: cot. C: R, or cos. AC: cot. A :: cot. C: R (46), therefore Rx cos. AC=cot. A x cot. C.

Again, R: s. Ca :: s. a: s. bC, (48) or R: cos. BC :: cos. AB: cos. AC, therefore R x cos. AC=cos. AB x cos. BC. Hence the proposition is manifest.

101. The following table contains all the equations arising from these theorems, by means of which all the cases of rightangled spherical triangles may be resolved with facility.

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102. To apply the general prop. to the resolution of any case of a right-angled spherical triangle, consider which of the three quantities named (the two things given and the thing required) must be made the middle part, so that the other two may be equidistant from it, that is, both adjacent or both opposite parts; then one of the theorems contained in the prop. will give the value of the part required. Thus, if AC and CB be given to find A; then, if BC be made the middle part, AC and A will be the opposite parts; therefore R x s. BC= cos. (90°— A) x cos. (90°—AC)=s. A x 5. AC, therefore s. BC s. A = R=1. Again, if AC and A be given to find s. AC' AB, it is obvious that A is in the middle between the adjacent parts AB and 90°-AC; therefore R x cos. A = tan. AB ×

cot. AC, therefore tan. AB =

(42 Pl. Tr.).

cos. A

cot. AC

cos. A x tan. AC

In the same manner may all the other cases be resolved.

But without the consideration of the rule the equation may be immediately taken from the table, by observing which of the five cases contains the two given quantities and the quantity required. For instance, if AB and BC be given to find C; these three quantities are found in case 5, and R x cos. AC

cos. AB X cos. BC, therefore log. cos. AC=log. cos. AB+log. cos. BC-10. The value of the unknown quantity is thus immediately obtained from the equation in the table.

If you want to know the relative values of the quantities, you must convert the equation containing them into a proportion.

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