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In a spherical triangle the greater angle is opposite to the greater side ;
Let ABC be a spherical triangle, the greater angle A is opposed to the greater side BC.
Let the angle BAD be made equal to the angle B, and then
А. BD, DA will be equal, (6.), and therefore AD, DC are equal to BC ; but AD, DC are greater than AC (7.), therefore BC is greater than AC, that is, the greater angle A is opposite to the greater side BC. The converse is demonstrated as Prop. 19. 1.
B В Elem. Q. E. D.
According as the sum of two of the sides of a spherical triangle is great
er than a semicircle, equal to it, or less, each of the interior angles at the base is greater than the exterior and opposite angle at the base, equal to it, or less ; and also the sum of the two interior angles at the base greater than two right angles, equal to two right angles, or less than two right angles.
Let ABC be a spherical triangle, of which the sides are AB and BC ; produce any of the two sides as AB, and the base AC, till they meet again in D ; then, the arch ABD is a semicircle, and the spherical angles at A and D are equal, because each of them is the inclination of the circle ABD to the circle ACD. 1. If AB, BC be equal to
B a semicircle, that is, to AD, BC will be equal to BD, and therefore (5.) the angle D, or the angle A will be equal A
D to the angle BCD, that is, the interior, angle at the base equal to the exterior and opposite.
2. If AB, BC together be greater than a semicircle, that is greater than ABD, BC will be greater than BD; and therefore (9.), the angle D, that is the angle A, is greater than the angle BCD.
3. In the same manner it is shewn, AB, BC together be less than a semicircle, that the angle A is less than the angle BCD.
Now, since the angles BCD, BCA are equal to two right angles, if the angle A be greater than BCD, A and ACB together will be greater than two right angles. If A be equal to BCD, A and ACB together, will be equal to two right angles ; and if A be less than BCD, A and ACB will be less than two right angles. Q. E. D.
If the angular points of a spherical triangle be made the poles of three
great circles, these three circles by their intersections will form a triangle, which is said to be supplemental to the former ; and the two triangles are such, that the sides of the one are the supplements of the arches which measure the angles of the other.
Let ABC be a spherical triangle ; and from the points A, B, and C as poles, let the great circles FE, ED, DF be described, intersecting one another in F, D and E ; the sides of the triangle FED are the supplements of the measures of the angles A, B, C, viz. FE of the angle BAC, DE of the angle ABC, and DF of the angle ACB : And again, AC is the supplement of the angle DFE, AB of the angle FED, and BC of the angle EDF. Let AB produced meet DE, EF in
I G, M; let AC meet FD, FE in K, L; and let BC meet FD, DE in N, H.
M Since A is the pole of FE, and the
B В circle AC passes through A, EF will pass through the pole of AC (1. Cor. 4.) and since AC passes through C, the pole of FD, FD will pass through the pole of AC ; therefore the pole of AC
K is in the point F, in which the arches
E DF, EF intersect each other.
H samé manner, D is the pole of BC, and E the pole of AB.
And since F, E are the poles of AL, AM, the arches FL and EN (2.) are quadrants, and FL, EM together, that is, FE and ML together, are equal to a semicircle. But since A is the pole of ML,
MŲ is the measure of the angle BAC, (3.), consequently FE is the supplement of the measure of the angle BAC. In the same manner, ED, DF are the supplements of the measures of the angles ABC, BCA.
Since likewise CN, BH are quadrants, CN and BH together, that is, NH and BC together, are equal to a semicircle ; and since D is the pole of NH, NH is the measure of the angle FDE, therefore the measure of the angle FDE is the supplement of the side BC. In the same manner, it is shewn that the measures of the angles DEF,
EFD are the supplements of the sides AB, AC, in the triangle ABC. Q. E. D.
The three angles of a spherical triangle are greater than two, and less
than six, right angles.
The measures of the angles A, B, C, in the triangle ABC, together with the three sides of the supplemental triangle DEF, are (11.) equal to three semicircles; but the three sides of the triangle FDE, are (8.) less than two semicircles ; therefore the measures of the angles A, B, C are greater than a semicircle ; and hence the angles A, B, C are greater than two right angles.
And because the interior angles of any triangle, together with the exterior, are equal to six right angles, the interior alone are less than six right angles. Q. E. D.
PROP. XIII. If to the circumference of a great circle, from a point, in the surface
of the sphere, which is not the pole of that circle, arches of great circles be drawn ; the greatest of these arches is that which passes through the pole of the first-mentioned circle, and the supplement of it is the least; and of the other arches, that which is nearer to the greatest is greater than that which is more remote.
Let ADB be the circumference of a great circle, of which the pole is H, and let C be any other point; through C and H let the semicircle ACB be drawn meeting the circle ADB in A and B ; and let the arches CD, CE, CF also be described. From C draw CG perpendicular to AB, and then, because the circle AHCB which passes through H, the pole of the circle ADB, is at right angles to ADB, CG is perpendicular to the plane ADB. Join GD, GE, GF, CA, CD, CE, CF, CB.
H Because AB is the diameter of the circle ADB, and G a point in it, which is not the centre, (for the centre is in the point where the perpendicular from H meets AB), therefore AG, the part of the diameter in which the centre
A is, is the greatest, (7. 3.), and GB the least of all the straight lines D
F that can be drawn from G to the circumference; and GD, which is nearer to AB, is greater than GE, which is more remote. But the triangles CGA, CGD are right angled at G, and therefore AC2= AG3 +GC2, and DC?=DG2 +6C3 ; but AG: +603-7 DG? +GC2; because AG 7 DG ; therefore AC27 DC?, and AC-7 DC. "And be. cause the chord AC is greater than the chord DC, the arch AC is
greater than the arch DC. In the same manner, since GD is greater than GE, and GE than GF, it is shewn that CD is greater than CE, and CE than CF. Wherefore also the arch CD is greater than the arch CE, and the arch CE greater than the arch CF, and CF than CB; that is, of all the arches of great circles drawn from C to the circumference of the circle ADB, AC which passes through the pole H, is the greatest, and CB its supplement is the least ; and of the others, that which is nearer to AC the greatest, is greater than that which is more remote. Q. E. D.
In a right angled spherical triangle, the sides containing the right angle,
are of the same affection with the angles opposite to them, that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles, and conversely.
Let ABC be a spherical triangle, right angled at A, any side AB will be of the same affection with the opposite angle ACB.
Produce the arches AC, AB, till they meet again in D, and bisect AD in E. Then ACD, ABD are semicircles, and AE an arch of 90°. Also, because CAB is by hypothesis a right angle, the plane of the circle ABD is perpendicular
G to the plane of the circle ACD, so that the pole of ACD is in ABD, (cor. 1. 4.), and is therefore the point E. Let EC be an arch of a great A circle passing through E and C. Then because E is the
B pole of the circle ACD, EC
E is a (2.) quadrant, and the
G plane of the circle EC (4.) is at right angles to the plane of the circle ACD, that is, the spherical angle ACE is a right angle ; and therefore, when AB is less than AE, the angle ACB, being less than ACE, is less than a right
B angle. But when AB is greater than AE, the angle ACB is greater than ACE, or than a right angle. In the same way may the converse be demonstrated. Therefore, &c. Q. E. D.
If the two sides of a right angled spherical triangle about the right angle
be of the same affection, the hypotenuse will be less than a quadrant ; and if they be of different uffection, the hypotenuse will be greater than a quadrant.
Let ABC be a right angled spherical triangle ; according as the two sides AB, AC are of the same or of different affection, the hypotenuse BC will be less, or greater than a quadrant.
The construction of the last proposition remaining, bisect the semicircle ACD in G, then AG will be an arch of 90°, and G will be the pole of the circle ABD.
1. Let AB, AC be each less than 90°. Then, because C is a point on the surface of the sphere, which is not the pole of the circle ABD, the arch CGD, which passes through G the pole of ABD is greater than CE, (13.), and CE greater than CB. But CE is a quadrant, as was before shewn, therefore CB is less than a quadrant. Thus also it is proved of the right angled triangle CDB, (right angled at D), in which each of the sides CD, DB is greater than a quadrant, that the hypotenuse BC is less than a quadrant.
2. Let AC be less, and AB greater than 900. Then because CB falls between CGD and CE, it is greater (13.) than CE, that is than a quadrant. Q. E. D.
Cor. 1. Hence conversely, if the hypotenuse of a right angled triangle be greater or less than a quadrant, the sides will be of different or the same affection.
Cor. 2. Since (14.) the oblique angles of a right angled spherical triangle have the same affection with the opposite sides, therefore, according as the hypotenuse is greater or less than a quadrant, the oblique angles will be of different, or of the same affection.
Cor. 3. Because the sides are of the same affection with the opposite angles, therefore when an angle and the side adjacent are of the same affection, the hypotenuse is less than a quadrant ; and conversely.
PROP XVI. lin any spherical triungle, if the perpendicular upon the base from the op
posite angle fall within the triangle, the angles at the base are of the same affection ; and if the perpendicular fall without the triangle, the angles at the base are of different affection.
Let ABC be a spherical triangle, and let the arch CD be drawn from C perpendicular to the base AB.