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the measure of the angle BAC °, because A is the pole of. GH; Sph. Trig.
therefore DE is the supplement of the measure of the angle
BAC. In the same manner, it may be proved, that DF is the C 3. S. P.
supplement of the measure of ABC, and EF of ACB.

Again, because A, B are the poles of DE, DF, each of the
arches AG, BM is the fourth part of the circumference b; b 2. S. P,
therefore AG, BM, that is, AB, GM together, are half of the
circumference: and MG is the measure of the angle EDF ,
because D is the pole of MG; therefore the measure of the
angle EDF is the supplement of the side AB. In the same
manner, it may be proved, that the measures of the angles
DEF, DFE are the supplements of the sides AC, BC, in the
triangle ABC.

Cor. If the circles DF, EF meet again in P, the triangle DEP is femi-supplemental to the triangle ABC; that is, the fide DE, and the measure of the opposite angle DPE, are the supplements of the measure of BAC, and of BC: but the sides DP, PE are the measures of ABC, ACB ; and the measures of PDE, PED are equal to the sides AB, AC.

THE

PRO P. XIV.
THE three angles of a spherical triangle are to.

gether greater than two, and less than fix right
angles.

Because the measure of each of the angles at A, B, C, together with the fide opposite to it of the triangle DEF, is equal to half the circumference a ; the measures of the three angles a 13. S. F. A, B, C, together with the sides of the triangle DEF, are three halves of the circumference: and the three fides of the triangle DEF are less than the circumference b; therefore the measures b 10. s. p. of the angles at A, B, C are together greater than half of the circumference; and the angles are therefore greater than two right angles.

And because all the exterior and interior angles of a triangle are equal to fix right angles, the interior angles are less than fix right angles.

Cor. Because DE, EF are greater than DFC; twice DF, to- C 9. S. P. gether with the measures of the angles at A, B, C, are less than three halves of the circumference a ; and therefore the three angles at A, B, C, together with twice the supplement of the Jeast of them, are less than fix right angles.

PROP.

IF

Sph, Irig,

PROP. XV.
F arches of great circles be drawn to the circum,

ference of any great circle, from a point in the superficies of the sphere, which is not its pole ; the greatest of them is that which passes through the pole, and its supplement is the leaft; and of the others, that which is nearer to the greatest, is greater than that which is more remote.

Let ABD be a great circle, of which E is the pole, and let F be any other point in the superficies of the sphere, and let the great circle AFD pass through the points E, F, and meet the circle ABD in the ameter AD; Of the arches of great circles FB, FC, &c. that can be drawn froin F to the circumference ABD, FA is the greatest, and FD the least ; and of the others, FB which is nearer to FA, is greater than FC which is

Inore remote.

S. P.

b

G

Draw FG perpendicular to AD, and join GB, GC, FA, FB,

FC, FD: And because the circle AFD is perpendicular to the 25. Cor.4 circle ABD a, for it passes through its pole E, and that

E
FG is at right angles to their

common section AD, FG is B4.Def.11. perpendicular to the plane

ABD; therefore FGB, FGC
are right angles. And because
G is a point in the diameter A
AD, which is not the centre,

B
for a straight line from E to

the centre is perpendicular to ci.Cor.3. AD; GA which passes through the centre is the greatest,

and GD the least of all the straight lines that can be drawn & 7. 3. from G to the circumference ABD; and GB which is nearer

to GA, is greater than GC which is more remote : and because GA is greater than GB, the squares of AG, GF are greater

than the squares of BG, GF, that is, the square of AF is € 47. 1. greater

e than that of BF; therefore AF is greater than BF: * 15. 3. and they are in equal circles; therefore the arch AF is greater

than the arch BF. In the same manner, it may be demonstrated, that the arch BF is greater than the arch CF, and the arch CF than the arch DF. Wherefore FA is the greatest, and FD the least, and FB greater than FC.

S. P.

PROP.

IN

PRO P. XVI.

Sph. Trig. N a right angled spherical triangle, the sides about

the right angle are of the fame affection with their opposite angles; that is, if one of the fides be greater than the fourth part of the circumference, the angle opposite to it is greater than a right angle ; and if equal, equal; and if lefs, less.

Let ABC be a spherical triangle, having BAC a right angle, the fide AB is of the fame affection with the oppolite angle ACB.

If AB be the fourth part of the circumference, B is the pole a a 2.Cor.4. of AC, and ACB is a right angle b.

b.Cor.4.

S. P.

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If AB be not the fourth part of the circumference, take AE the fourth part of it, and let a great circle pass through E and C; therefore E is the pole of AC a, and ECA is a right a 2.Cor.4. angle b: But if AB be greater than AE, the angle ACB is greater than ACE ; and if AB be less than AE, the angle ACB is less than ACE ; that is, than a right angle.

S. P.

PRO P. XVII.

IN
N a right angled spherical triangle, if the sides

about the right angle be of the fame affection, the hypotenuse is less than the fourth part of the circumference ; and if they be of different affection, the hypotenuse is greater than the fourth part of the circumference : or if one of the other two be the fourth part of the circumference, the hypotenuse is equal to it,

Let

b

P.

Spb. Trig. Let ABC be a spherical triangle, having the right angle

BAC; if one of the fides AB, AC be the fourth of the circumference, the hypotenuse BC is also the fourth ; but if AB, AC be both less, or both greater, the hypotenuse is less than the fourth part of the circumference; and if one of them be less, and the other greater, the hypotenuse is greater.

Let the circumferences AB, AC meet again in D, and bisect a 2. S. P. the arch ABD in E; therefore E is the pole of ACD a; bisect

also ACD in G, and G is the pole of ABD, and let a great circle pass through E, C: If the point B coincide with E, that is, if AB be the fourth part of the circumference, BC coincides with EC, and is the fourth part of the circumference. If AB, AC be both less than AE or AG; CE is nearer to CGD

which passes through the pole G, than CB is ; therefore CB is b 15. S. less than CE the fourth part of the circumference. In the

same manner, in the triangle CBD, of which the sides CD, BD are each of them greater than CE, the hypotenuse CB is less than CE.

But if AB be greater than AE, and AC less, the arch CB is nearer to CGD than CE is; therefore CB is greater than CE the fourth part of the circumference.

Cor. I. On the contrary, if the hypotenuse be less than the forth part of the circumference, the sides are of the same af. fection ; for if they be of different affection, the hypotenuse is greater than the fourth: and if the hypotenuse be greater than the fourth part of the circumference, the sides are of different affection.

Cor. 2. Hence, in a right angled triangle, if the angles be of the same affection, the hypotenufe is less than the fourth part of the circumference : and if they be of different affection, the hypotenule is greater than the fourth : and conversely. For the angles are of the same affection with their opposite sides.

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N any spherical triangle, if the perpendicular

drawn from the vertex to the base, fall within the triangle, the angles at the base are of the same affection : and if it fall without, they are of different affection.

Let ABC be a spherical triangle, and let the arch CD of a great circle be drawn from C perpendicular to AB.

First,

First, Let CD fall within the triangle ABC; and because Sph. Trig, ADC is a right angled triangle, the angle CAD is of the fame w affection with the opposite fide CD a.

a 16. S. T. For the same reason, the angle CBA is of the same affection with CD; therefore the angles CAB, CBA are of the same affection.

D

3 Next, Let the perpendicular CD fall without the triangle ; and because CDA is a right angle, the angle CAD is of the same affection with CD *; that is, with the angle CBD; but CAD. CAB are of different affection, because they are

C together equal to two right angles; therefore the angles CAB, CBA are of different affection.

Cor. Hence, if the angles at A, B be of the same affection, the perpendi

B

D cular falls within the triangle ; for, if it

A fell without, they would be of different affection. And if the angles at A, B be of different affection, the perpendicular falls without the triangle; for, if it fall within, they would be of the faine affection.

IN

PRO P. XIX. (N a right angled spherical triangle, the fine of

either of the sides about the right angle, is to the radius of the sphere, as the tangent of the other side to the tangent of the angle oppolite to it.

Let ABC be a spherical triangle, having the right angle BAC;

; as the fine of the side AB to the radius, so is the tangent of the side AC to the tangent of the angle ABC.

C Let D be the centre of the

C.
sphere, and join DA, DB,
DC, and from the point A
draw AF perpendicular a to
BD, and froin F draw 6 FE,
in the plane BDC, at right
angles to BD; and let it meet
DC in E, and join AE: and D
because DF is at right angles,
both to AF and FE, it is per-
pendicular to the plane AFE;

FR
Kk therefore

a 12. 1. b 11. I.

64. II.

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