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FDE, therefore the measure of the angle FDE is the sup plement of the side BC. In the same manner, it is shown that the measures of the angles DEF, EFD are the supplements of the sides AB, AC, in the triangle ABC. Q.E.D.

PROP. XI. FIG. 7.

THE three angles of a spherical triangle, are greater than two right angles, 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 (10. of this) equal to three semicircles; but the three sides of the triangle FDE, are (7. of this) 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.

All the external and internal angles of any triangle are equal to six right angles; therefore all the internal angles are less than six right angles.

PROP. XII. FIG. 8.

IF from any point C, which is not the pole of the great circle ABD, there be drawn arches of great circles CA, CD, CE, CF, &c. the greatest of these is CA, which passes through H the pole of ABD, and CB, the remainder of ACB is the least, and of any others CD, CE, CF, &c. CD, which is nearer to CA, is greater than CE, which is more remote.

Let the common section of the planes of the great circles ACB, ADB, be AB; and from C, draw CG perpendicular to AB, which will also be perpendicular to the plane ADB (4. def. 11.); join GD, GE, GF, CD, CE, CF, CA, CB.

Of all the straight lines drawn from G to the circumference ADB, GA is the greatest, and GB the least (7. 3.); and GD, which is nearer to GA, is greater than GE, which is more remote. The triangles CGA, CGD, are right angled at G, and they have the common side CG; therefore the squares of CG, GA together, that is, the square of CA, is greater than the squares of CG, GD together, that is, the square of CD: And CA is greater than CD, and therefore the arch CA is greater than CD. In the same manner, since GD is greater than GE, and GE than GF,

Fig. 10.

Fig. 9.

&c. it is shown that CD is greater than CE, and CE than CF, &c. and consequently, the arch CD greater than the arch CE, and the arch CE greater than the arch CF, &c. And since GA is the greatest, and GB the least of all the straight lines drawn from G to the circumference ADB, it is manifest that CA is the greatest, and CB the least of all the straight lines drawn from C to the circumference: And therefore the arch CA is the greatest, and CB the least of all the circles drawn through C, meeting ADB. Q.E.D.

PROP. XIII. FIG. 9.

In a right angled spherical triangle, the sides are of the same affection with the opposite angles; that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles.

Let ABC be a spherical triangle, right angled at A, any side, AB, will be of the same affection with the opposite angle ACB.

Case 1. Let AB be less than a quadrant, let AE be a quadrant, and let EC be a great circle passing through E, C. Since A is a right angle, and AE a quadrant, E is the pole of the great circle AC, and ECA a right angle; but ECA is greater than BCA, therefore BCA is less than a right angle. Q.E.D.

Case 2. Let AB be greater than a quadrant, make AE a quadrant, and let a great circle pass through C, E, ECA is a right angle' as before, and BCA is greater than ECA, that is, greater than a right angle. Q.E. D.

PROP. XIV.

If the two sides of a right angled spherical triangle be of the same affection, the hypothenuse will be less than a quadrant; and if they be of different affection, the hypothenuse will be greater than a quadrant.

Let ABC be a right angled spherical triangle, if the two sides AB, AC, be of the same or of different affection, the hypothenuse BC will be less or greater than a quadrant.

Case 1. Let AB, AC be each less than a quadrant. Let AE, AG, be quadrants; G will be the pole of AB, and E the pole of AC, and EC a quadrant; but by Prop. 12, CE

is greater than CB, since CB is farther off from CGD than CE. In the same manner, it is shown that CB, in the triangle CBD, where the two sides CD, BD, are each greater than a quadrant, is less than CE, that is, less than a quadrant. Q. E.D.

Case 2. Let AC be less, and AB greater than a quadrant; Fig. 10. then the hypothenuse BC will be greater than a quadrant: for let AE be a quadrant, then E is the pole of AC, and EC will be a quadrant. But CB is greater than CE by Prop. 12. since AC passes through the pole of ABD. Q. E. D.

PROP. XV.

Ir the hypothenuse of a right angled triangle be greater or less than a quadrant, the sides will be of different or the same affection.

This is the converse of the preceding, and demonstrated in the same manner.

PROP. XVI.

IN any spherical triangle ABC, if the perpendicular AD from A, on the base BC, fall within the triangle, the angles B and C at the base will be of the same affection; and if the perpendicular fall without the triangle, the angles B and C will be of different affection.

1. Let AD fall within the triangle; then (13. of this), Fig. 11. since ADB, ADC are right angled spherical triangles, the angles B, C, must each be of the same affection as AD.

2. Let AD fall without the triangle, then (13. of this) Fig. 12. the angle B is of the same affection as AD; and by the same the angle ACD is of the same affection as AD; therefore the angle ACB and AD are of different affection, and the angles B and ACB of different affection.

COR. Hence if the angles B and C be of the same affection, the perpendicular will fall within the base; for, if it did not (16. of this), B and C would be of different affection. And if the angles B and C be of opposite affection, the perpendicular will fall without the triangle; for, if it did not (16. of this), the angles B and C would be of the same affection, contrary to the supposition.

PROP. XVII. FIG. 13.

In right angled spherical triangles, the sine of either of the sides about the right angle, is to the radius of the sphere, as the tangent of the remaining side is to the tangent of the angle opposite to that side.

Let ABC be a triangle, having the right angle at A; and let AB be either of the sides, the sine of the side AB will be to the radius, as the tangent of the other side AC to the tangent of the angle ABC, opposite to AC. Let D be the centre of the sphere; join AD, BD, CD, and let AE be drawn perpendicular to BD, which therefore will be the sine of the arch AB, and from, the point E, let there be drawn in the plane BDC the straight line EF at right angles to BD, meeting DC in F, and let AF be joined. Since therefore the straight line DE is at right angles to both EA and EF, it will also be at right angles to the plane AEF (4. 11.), wherefore the plane ABD, which passes through DE is perpendicular to the plane_AEF (18. 11.) and the plane AEF perpendicular to ABD: The plane ACD or AFD is also perpendicular to the same ABD: Therefore the common section, viz. the straight line AF, is at right angles to the plane ABD (19. 11.): And FAE, FAD are right angles (3. def. 11.); therefore AF is the tangent of the arch AC, and in the rectilineal triangle AEF having a right angle at A, AE will be to the radius as AF to the tangent of the angle AEF (1. Pl. Tr.); but AE is the sine of the arch AB, and AF the tangent of the arch AC, and the angle AEF is the inclination of the planes CBD, ABD (6. def. 11.), or the spherical angle ABC: Therefore the sine of the arch AB is to the radius as the tangent of the arch AC to the tangent of the opposite angle ABC.

COR. 1. If therefore of the two sides, and an angle opposite to one of them, any two be given, the third will also be given.

COR. 2. And since by this proposition the sine of the side AB is to the radius, as the tangent of the other side AC to the tangent of the angle ABC opposite to that side; and as the radius is to the cotangent of the angle ABC, so is the tangent of the same angle ABC to the radius (Cor. 2. def. Pl. Tr.) by equality, the sine of the side AB is to the cotangent of the angle ABC adjacent to it, as the tangent of the other side AC to the radius.

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PROP. XVIII. FIG. 13.

IN right angled spherical triangles, the sine of the hypothenuse is to the radius, as the sine of either side is to the sine of the angle opposite to that side.

Let the triangle ABC be right angled at A, and let AC be either of the sides; the sine of the hypothenuse BC will be to the radius as the sine of the arch AC is to the sine of the angle ABC.

Let D be the centre of the sphere, and let CG be drawn perpendicular to DB, which will therefore be the sine of the hypothenuse BC; and from the point G let there be drawn in the plane ABD the straight line GH perpendicular to DB, and let CH be joined; CH will be at right angles to the plane ABD, as was shown in the preceding proposition of the straight line FA: Wherefore CHD, CHG, are right angles, and CH is the sine of the arch AC; and in the triangle CHG, having the right angle CHG, CG is to the radius as CH to the sine of the angle CGH (1. Pl. Tr.): But since CG, HG are at right angles to DGB, which is the common section of the planes CBD, ABD, the angle CGH will be equal to the inclination of these planes (6. def. 11). that is, to the spherical angle ABC. The sine therefore of the hypothenuse CB, is to the radius as the sine of the side AC is to the sine of the opposite angle ABC. Q. E. D.

COR. Of these three, viz. the hypothenuse, a side, and the angle opposite to that side, any two being given, the third is also given by Prop. 2.

PROP. XIX. FIG. 14.

IN right angled spherical triangles, the cosine of the hypothenuse is to the radius as the cotangent of either of the angles is to the tangent of the remaining angle.

Let ABC be a spherical triangle, having a right angle at A, the cosine of the hypothenuse BC will be to the radius as the cotangent of the angle ABC to the tangent of the angle ACB.

Describe the circle DE, of which B is the pole, and let it meet AC in F, and the circle BC in E; and since the circle BD passes through the pole B of the circle DF, DF will also pass through the pole of BD. (13. 15. 1.

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