CHAPTER IV. ON THE USE OF SUBSIDIARY ANGLES. We have already explained in plane Trigonometry, the meaning of Subsidiary Angles, and the purpose for which they are introduced; we shall now proceed to point out under what circumstances they may be employed with advantage, in Spherical Trigonometry. In the solution of case I., where two sides and the included angle were given, we first determined the two remaining angles, and having found these, we were enabled to find the side also. It frequently happens, however, that the side alone is the object of our investigations, and it is therefore convenient to have a method of determining it, independently of the angle. Thus, for example, let b, c, A be given, and let it be required to determine a, independently of the angles B, c. By (a,) we have Whence cos. a=cos. A sin. b sin. c+cos. b cos. c. From which equation a is determined, but the expression is not in a form adapted to the logarithmic computation; we can, however, effect the necessary transformation by the introduction of a subsidiary angle. Add and subtract sin. b sin. c on the right hand side of the equation. Then cos. a =cos. A sin. b sin. c+cos. b cos. c+sin. b sin. c-sin. b sin. c =cos. b cos. c+sin. b sin. c+sin. b sin. c cos. A-sin. b sin. c =cos. (bc)—sin. b sin. c vers. A 1— cos. a=1— cos. (b—c)+sin. b sin. c vers. A c)+sin. b sin. c vers. A vers. (bc) from which a may be determined by the tables, being known In like manner in case II, where two angles and the included side were given, we first determined the remaining sides, and then we were enabled to find the remaining angle. Now, let us suppose that A, B, c, are given, and that we are required to find C independently of a and b. cos. C+cos. A cos. B From (6) cos. c= sin. A sin. B cos. C=cos. c sin. A sin. B-cos. A cos. B .. 1-cos. C=1—sin. A sin. B (1—vers. c)+cos. A cos. B. =1+cos. (A+B)+sin. A sin. B vers. c. C A+B 2 +sin. A sin. B vers. c In case III, where two sides and the angle opposite to one of them were given, we first determined the angle opposite to the other side, and then the remaining angles and the remaining side in succession. Now, let us suppose the side c, independently of the angle B and of each other, under a form adapted for logarithmic computation. To find C, we have (n.) cot. A=cot. a sin. b cosec. C. - cos. b cot. C or cot. A sin. C=cot. a sin. b-cos. b cos. C or sin. C=cot. a sin. b tan. A cos. b cos. C tan. A ... sin. C+cos. C cos. b tan. A=cot. a sin. b tan. A. .. sin. C cos. +cos. C sin. =cot. a sin. b tan. A cos. sin. (C+4)=cot. a sin. b tan. A =cot. a tan. b sin. sin. cos. b tan. A whence C is known, being previously determined from equa tion tan. =cos. b tan. A. To find c, we have from (a.) cos. A: cos. a-cos. b cos. c sin. b sin. c sin. c sin. b cos. A=cos. a cos. b cos. C sin. c tan. b cos. A= cos. a COS. C cos. b whence c may be found, being previously determined from the equation tan. =tan. b cos. A. In like manner, in case IV, when two angles and the side opposite to one of them were given, we first determined the side opposite to the other angle, then the remaining side and the remaining angle in succession. Now, let A, B, a, be given, and let it be required to determine c and C, independently of b and of each other, and under a form adapted to logarithmic computations. If we take the formula (4.) or or .. sin. c cot. a cot. A sin. B cosec. c+ cos. B cot. c cot. a sin. c=cot. A sin. B+cos. B cos. c sin. c cot. A sin. B tan. a+cos. B cos. c tan. a cos. c cos. B tan. a=cot. A sin. B tan. a. Let cos. B tan. a=tan. 6: sin. cos. A cos. c cot. A sin. B tan. a whence c may be determined, being previously known from equation tan. cos. B tan. a. sin. B sin. C cos. a=cos. A+cos. B cos. C being known from equation In the fifth and sixth cases, any one of the angles or sides required, may be found independently of the rest by the formulæ referred to. EXAMPLES IN SPHERICAL TRIGONOMETRY. Ex. 1. In the right-angled spherical triangle ABC, the hypothenuse AB is 65° 5', and the angle A is 48° 12'; find the sides AC, CB, and the angle B. Ans. AC=55° 7′ 32′′ BC 42 32 19 angle B=64 46 14 Ex. 2. In the oblique-angled spherical triangle ABC, given AB=76° 20′, BC=119° 17', and angle B-52° 5' ; to find AC and the angles A and C. Ans. AC=66° 5' 36" angle A=131 10 42 angle C 56 58 58 Ex. 3. In an oblique spherical triangle the three sides are a=81° 17', b=114° 3′, c=59° 12'; required the angles A, B, C. Ans. A= 62° 39′ 42′′ B=124 50 50 14* APPLICATION OF ALGEBRA TO GEOMETRY. ON THE GEOMETRICAL CONSTRUCTION OF ALGEBRAICAL QUANTITIES. As lines, surfaces, and solids are quantities which admit of increase and decrease, like other quantities, they may, like others, be made the subjects of algebraical operations, either by their numerical representatives, or by symbols expressing such quantities. It is only necessary for this purpose, that their representatives should possess values in relation to each other corresponding to the magnitudes of the quantities which they represent; and that those values should be expressed according to the properties or relations of the lines, surfaces, or solids, to each other; subject to geometrical construction and algebraical notation. We are enabled, also, to express by lines, surfaces, and solids, the solutions furnished by Algebra. This is founded on the known properties of geometrical figures, corresponding to similar properties of the quantities algebraically expressed. In this view of the subject, all quantities under algebraic ex pressions, may be conceived to be susceptible of some kind of geometrical construction. 2. We will proceed to explain the manner of constructing those expressions, and representing, under a geometrical form, the conditions of an equation. This is called constructing the algebraic quantities. Ex. 1. Let it be proposed to construct such a quantity as the value of the letters composing the ab с the following: tity being known. From any point A draw two indefinite lines AM, AN, making any angle with each other; upon one of these lines AM take AB=c, and AD=a; then upon the line AN take AC=b. Having drawn the line BC, draw also DE parallel to BC, this will determine AE as the va- D ab lue of. For the parallels DE, BC, give this C proportion AB: AD:: AC: AE, (Prop. XIV. Cor. 2 B. IV. El. Geom.) or c:a::b: AE. A quan E |