sin. (A+B) tan. (A+B)= But it has just been shewn, that cos. (A+B) sin. (A+B)=sin. Ax cos. B+cos. Ax sin. B, and that cos. (A+B)=cos. Ax cos. B-sin. Axsin. B; therefore, tan. (A+B) = sin. Ax cos. B+cos. Ax sin. B and dividing both the numerator and deno cos. Ax cos. B-sin. AX sin. B' In like manner, tan. (A~~B)=: tan. A tan. B = -tan. A+tan. B 1 tan. Axtan. B' 9. If the Theorem demonstrated in Prop. 3, be expressed in the same manner with those above, it gives 10. In all the preceding Theorems, R, the radius, is supposed =1, because in this way the propositions are most concisely expressed, and are also most readily applied to trigonometrical circulation. But if it be required to enunciate any of them geometrically, the multiplier R, which has disappeared, by being made = 1, must be restored, and it will always be evident from inspection in what terms this multiplier is wanting. Thus, Theor. 1, 2 sin. A x cos. B=sin. (A+B)+sin. (A−B), is a true proposition, taken arithmetically; but taken geometrically, is absurd, unless we supply the radius as a multiplier of the terms on the right hand of the sine of equality. It then becomes 2 sin. A× cos. B=R (sin. (A+B)+sin. (A—B)); or twice the rectangle under the sine of A, and the cosine of B equal to the rectangle under the radius, and the sum of the sines of A+B and A-B. In general, the number of linear multipliers, that is, of lines whose numerical values are multiplied together, must be the same in every term, otherwise we will compare unlike magnitudes with one another. The propositions in this section are useful in many of the higher branches of the Mathematics, and are the foundation of what is called the Arithmetic of Sines. ELEMENTS OF SPHERICAL TRIGONOMETRY. PROP. I. If a sphere be cut by a plane through the centre, the section is a circle, having the same centre with the sphere, and equal to the circle by the revolution of which the sphere was described. FOR all the straight lines drawn from the centre to the superficies of the sphere are equal to the radius of the generating semicircle, (Def. 7. 3. Sup.). Therefore the common section of the spherical superficies, and of a plane passing through its centre, is a line, lying in one plane, and having all its points equally distant from the centre of the sphere; therefore it is the circumference of a circle (Def. 11. 1.), having for its centre the centre of the sphere, and for its radius the radius of the sphere, that is, of the semicircle by which the sphere has been described. It is equal, therefore, to the circle of which that semicircle was a part. DEFINITIONS. 1. ANY circle, which is a section of a sphere by a plane through its centre, is called a great circle of the sphere. COR. All great circles of a sphere are equal; and any two of them bisect one another. They are all equal, having all the same radii, as has just been shewn; and any two of them bisect one another, for as they have the same centre, their common section is a diameter of both, and therefore bisects both. 2. The pole of a great circle of a sphere is a point in the superficies of the sphere, from which all strai ht lines drawn to the circumference of the circle are equal. 3. A spherical angle is an angle on the superficies of a sphere, contained by the arcs of two great circles which intersect one another; and is the same with the inclination of the planes of these great circles 4. A spherical triangle is a figure, upon the superficies of a sphere, comprehended by three arcs of three great circles, each of which is less than a semicircle. PROP. II. The arc of a great circle, between the pole and the circumference of another great circle, is a quadrant. D Let BC be a great circle, and D its pole; if DC, an arc of a great circle, pass through D, and meet ABC in C, the arc DC is a quadrant. Let the circle, of which CD is an arc, meet ABC again in A, and let AC be the common section of the planes of these great circles, which will pass through E, the centre of the sphere: Join DA, DC. Because AD-DC, (Def. 2.), and equal straight lines, in the same circle, cut off equal arcs (28. 3.), the arc AD the arc DC; but ADC is a semicircle, therefore the arcs AD, DC are each of them quadrants. A D B COR. 1. If DE be drawn, the angle AED is a right angle; and DE being therefore at right angles to every line it meets with in the plane of the circle ABC, is at right angles to that plane (4. 2. Sup.). Therefore the straight line drawn from the pole of any great circle to the centre of the sphere is at right angles to the plane of that circle; and, conversely, a straight line drawn from the centre of the sphere perpendicular to the plane of any greater circle, meets the superficies of the sphere in the pole of that circle. COR. 2. The circle ABC has two poles, one on each side of its plane, which are the extremities of a diameter of the sphere perpendicular to the plane ABC; and no other points but these two can be poles of the circle ABC. PROP. III. If the pole of a great circle be the same with the intersection of other two great circles: the arc of the first mentioned circle intercepted between the other two, is the measure of the spherical angle which the same two circles make with one another. Let the great circles BA, CA on the superficies of a sphere, of which the centre is D, intersect one another in A, and let BC be an arc of another great circle, of which the pole is A; BC is the measure of the spherical angle BAC. Join AD, DB, DC; since A is the pole of BC, AB, AC are quadrants (2.), and the angles ADb, ADC are right angles: therefore (4. def. 2. Sup.), the angle CDB is the inclination of the planes of B D the circles AB, AC, and is (def. 3.) equal to the spherical angle BAC; but the arc BC measures the angle BDC, therefore it also measures the spherical angle BAC.* For COR. If two arcs of great circles, AB and AC, which intersect one another in A, be each of them quadrants, A will be the pole of the great circle which passes through E and C the extremities of those arcs. since the arcs AB and AC are quadrants, the angles ADB, ADC are right angles, and AD is therefore perpendicular to the plane BDC, that is, to the plane of the great circle which passes through B and C. The point A is therefore (1. Cor. 2.) the pole of the great circle which passes through B and C. PROP. IV. If the planes of two great circles of a sphere be at right angles to one another, the circumference of each of the circles passes through the poles of the other; and if the circumference of one great circle pass through the poles of another, the planes of these circles are at right angles. Let ACBD, AEBF be two great circles, the planes of which are right angles to one another, the poles of the circle AEBF are in the circumference ACBD, and the poles of the circle ACBD in the circumference AEBF. From G the centre of the sphere, draw GC in the plane ACBD perpendicular to AB. Then because GC in the plane ACBD, at right angles to the plane AEBF, is at right angles to the common section of the two planes, it is (Def. 2. 2. Sup.) also at right angles to the plane AEBF, and therefore (I. Cor. 2.) C is the pole of the circle AEBF; and if CG be produced in D, D is the other pole of the circle AEBF. In the same manner, by drawing GE in the plane AEBF, perpendicular to AB, and producing it to F, it has shewn that E and F are the poles of the circle ACBD. Therefore, the poles of each of these circles are in the circumference of the other. A G Again, If C be one of the poles of the circle AEBF, the great circie ACBD which passes through C, is at right angles to the circle AEBF. For, CG being drawn from the pole to the centre of the circle AEBF, is at right angles (1. Cor. 2.) to the plane of that circle; and therefore, every plane passing through CG (17. 2. Sup.) is at right angles to the plane AEBF; now, the plane ACBD passes through CG. COR. 1. If of two great circles, the first passes through the poles of the When in any reference no mention is made of a Book, or of the Plane Tigonometry, the Spherical Trigonometry is meant. second, the second also passes through the poles of the first. For, if the first passes through the poles of the second, the plane of the first must be at right angles to the plane of the second, by the second part of this proposition; and therefore, by the first part of it, the circumference of each passes through the poles of the other. COR. 2. All greater circles that have a common diameter have their poles in the circumference of a circle, the plane of which is perpendicular to that diameter. PROP. V. In isosceles spherical triangles the angles at the base are equal. Let ABC be a spherical triangle, having the side AB equal to the side AC; the spherical angles ABC and ACB are equal. Let C be the centre of the sphere; join DB, DC, DA, and from A on the straight lines DB, DC, draw the perpendiculars AE, AF; and from the points E and F draw in the plane DBC the straight lines EG, FG perpendicular to DB and DC, meeting one another in G: Join AG. D C G E B Because DE is at right angles to each of the straight lines AE, EG, it is at right angles to the plane AEG, which passes through AE, EG (4. 2. Sup.); and therefore, every plane that passes through DE is at right angles to the plane AEG (17. 2. Sup.); wherefore, the plane DBC is at right angles to the plane AEG. For the same reason, the plane DBC is at right angles to the plane AFG, and therefore AG, the common section of the planes AFG, AEG is at right angles (18. 2. Sup.) to the plane DBC, and the angles AGE, AGF are consequently right angles. But since the arc AB is equal to the arc AC, the angle ADB is equal to the angle ADC. Therefore the triangles ADE, ADF, have the angles EDA, FDA, equal, as also the angles AED, AFD, which are right angles; and they have the side AD common, therefore the other sides are equal, viz. AE to AF (26. 1.), and DE to DF. Again, because the angles AGE, AGF are right angles, the squares on AG and GE are equal to the square of AE; and the squares of AG and GF to the square of AF. But the squares of AE and AF are equal, therefore the squares of AG and GE are equal to the squares of AG and GF, and taking away the common square of AG, the remaining squares of GE and GF are equal, and GE is therefore equal to GF. Wherefore, in the triangles AFG, AEG, the side GF is equal to the side GE, and AF has been proved to be equal to AE, and the base AG is common; therefore, the angle AFG is equal to the angle AEG (8. 1.). But the angle AFG is the angle which the plane ADC makes with the plane DBC (4. def. 2. Sup.), because FA and FG, which are drawn in these planes, are at right angles to DF, the common section of the planes. The angle AFG (3. def.) is therefore equal to the |