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. Q. E. D. DEFINITIONS. 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. II. The pole of a great circle of a sphere is a point in the superficies of the sphere, from which all straight lines drawn to the circumference of the circle are equal. III. A spherical angle is an angle on the superficies of a sphere, contained by the arches of two great circles which intersect one another; and is the same with the inclination of the planes of these great circles. IV. A spherical triangle is a figure, upon the superficies of a sphere, comprehended by three arches of three great circles, each of which is less than a semicircle. PROP. II. The arch of a great circle, between the pole and the circumference of another great circle, is a quadrant. Let ABC be a great circie, and D its pole; if DC, an arch of a great circle, pass through D, and meet ABC in C, the arch DC is a quad rant. D Let the circle, of which CD is an arch, 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 arches (28.3.) the arch AD = the arch A DC; but ADC is a semicircle, therefore the arches AD, DC are each of them quadrants. Q. E. D. E 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 great 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 arch of the first-mentioned circle intercepted between the other two, is the measure of the sphe rical 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 arch 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 the circles AB, AC, and is (def. 3.) equal to the spherical angle BAC; but the arch BC measures the angle BDC, therefore it also measures the spherical angle BAC.* Q. E. D. B A D C COR. If two arches 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 arches. For since the arches 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 (Cor. 1. 2.) the pole of the great circle which passes through B and C. PROP. VI. 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 *When in any reference no mention is made of a Book, or of the Plane Trigonometry, the Spherical Trigonometry is meant. 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 (Cor. 1.2.) C is the pole of the circle AEBF; and if CG be produced to 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. these circles are in the circumference of the other. Again, If C be one of the poles of the circle AEBF, the great circle 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 (Cor. 1. 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. Therefore, &c. Q. E. D. COR. 1. If of two great circles, the first passes through the poles of the 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 great 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 D 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. A Because DE is at right angles to each of the straight lines AE, EG, it is at D right angles to the plane AEG, which I i 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 arch AB is equal to the arch 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 spherical angle ACB; and, for the same reason, the angle AEG is equal to the spherical angle ABC. But the angles AFG, AEG are equal. Therefore the spherical angles ACB, ABC are also equal. Q. E. D. PROP. VI. If the angles at the base of a spherical triangle be equal, the triangle is isosceles. Let ABC be a spherical triangle having the angles ABC, ACB equal to one another; the sides AC and AB are also equal. Let D 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 A plane DBC the straight lines EG, FG perpendicular to DB and DC, meeting one another in G; join AG. Then, it may be proved, as was done in the last proposition, that AG is at right angles to the plane BCD, and that therefore the angles AGF, AGE are right angles, and also that the angles AFG, AEG are equal to the angles which the planes DAC, DAB make D G E B |