sine (A+a)

Also, Tang (45°+c)= ±r sine (A- a)*

sine (c+c)

And, Tang (45°+ža)=±r sine (c-c) *

CASE XI. Given a side of a right-angled spherical triangle, and its opposite angle to find the adjacent angle.

CASE XII. Given a side for right-angled spherical triangle, and its opposite angle, to find the hypothenuse.

Also, tang (45°+16)=±√tang ÷ (c+c) . cot 1⁄2 (c−c) ·

And, tang (45°+6)=√ tang (A+a). cot 1⁄2 (A—a)

CASE XIII. Given the two sides, or legs, of a right-angled spherical triangle, to find an angle.

CASE XIV. Given the two sides, or legs, of a right-angled spherical triangle, to find the hypothenuse.

CASE XV. Given two angles of right-angled spherical tri

CASE XVI. Given two angles, of a right-angled spherical triangle, to find the hypothenuse.

A -C

cot

2

2

GENERAL OBSERVATIONS ON THE SPECIES AND AMBIGUITY OF THE CASES.

(P) This subject has been already explained sufficiently to answer every possible case, at pages 171 and 172; but the species of the sides and angles may be determined from the equations produced by Baron Napier's Rules, or from the preceding formulæ, by attending to the signs of the quantities which compose the equations or formulæ.

The sides which contain the right angle are each of the same species as their opposite angles, viz. a is of the same species with A, and b is of the same species with B. (R. 140.)

It may be proper to observe that where a quantity is to be determined by the sines only, and a side or angle opposite to the quantity sought does not enter into the equation, the case will be ambiguous, thus in the XIIth case, where sine b= sine a .r -, the hypothenuse b is ambiguous.

sine A

Again, in the rd case, where sine a=

sine A. sine b

,thesine of a is evidently determinate, because it is of the same species with a which is a given quantity.

(Q) When an unknown quantity is to be determined by its cosine, tangent, or cotangent, the sign of this value will always determine its species; for, if its proper sign be +, the arc will be less than 90°; if the proper sign be the arc will be greater than 90°. (K. 94.)

(R) Again, in Case vith, where rad x cos bcos cx cos a, it is obvious that the three sides are each less than 90°, or that two of them are greater than 90°, and the third less; as no other combination can render the sign of cos cx cos a like that of cos b as the equation requires *.

Legendre's Geometry, 6th edition, page 381.

QUADRANTAL

QUADRANTAL TRIANGLES.

The nature of these triangles and the methods of solving the different cases that may occur have been given at page 187

et seq.

(S) Any spherical triangle of which A, B, C, are the angles, and a, b, c, the opposite sides, may be changed into a spherical triangle of which the angles are supplements of the sides a, b, c, and the sides supplements of the angles A, B, C, (X. 134.) viz. if we call A', B', c' the angles of the supplemental triangle, and a', b', c' the sides opposite to these angles we shall have

A180°-a; B'=180°-b; c'=180°-c

d' = 180° - A ; b'=180°- B ; c' — 180° — c

Hence it is plain, that if a spherical triangle, has a side b equal to a quadrant, the correspondent angle B' of the supplemental triangle will be a right angle, and since there are always three given parts in a triangle, the supplemental triangle will be a rightangled triangle, having two parts given to find the rest; consequently, by finding the required parts in the supplemental rightangled triangle, the different parts of the quadrantal triangle will be known.

(T) Formulæ might have been inserted for solving the different cases of quadrantal triangles, but this would be making an increase of formula to very little purpose, since all quadrantal triangles are easily turned into right-angled triangles.

SOLUTIONS OF THE DIFFERENT CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES.

CASE I. Given two sides of an oblique-angled spherical triangle, and an angle opposite to one of them, to find the angle opposite to the other.

CASE II. Given two sides of an oblique-angled spherical triangle, and an angle opposite to one of them, to find the angle contained between these sides.

SOLUTION. Find the angle opposite to the other given side by Case I.

CASE III. Given two sides of an oblique-angled spherical triangle, and an angle opposite to one of them to find the other side.

SOLUTION. Find the angle opposite to the other given side by Case I.

sine (C+B)

Then, Tang=sine (CB) tang (cut)