« PreviousContinue »
SOLUTION OF THE CASES OF OBLIQUE ANGLED TRIANGLES.
In an oblique angled triangle, "of the three sides and three angles, any three being given, the other three may be found, except when the three angles are given; in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them.
GIVÉN. A, B, and there. BC, AC. S, C: S, A: : AB: BC, Fig. 16. fore C, and the side
and also, s, C: S, B, AB.
1: AC. (2)
: AB 17,
AC : AB :: S, B : S, C. two sides and angles A and (2.) This case admits of two angle opposite to C. solutions ; for C may be greatone of them.
er or less than a quadrant.
AB, AC, and A, The an- AB+AC: AB-AC : : T, two sides and the glęs B and C+B : T, C_B: 3. the sum included angle.
and difference of the angles C,
BA : AC: : R: T, ABC,
2 ACXCB : ACq+CB9ABq ::R: Co S, C. If ABQ +CBq be greater than ABq. Fig. 16.
2 ACXCB : ABq-AC9
-CBq :: RCo S, C. If AB, BC, CA the A, B, C, ABq be greater than AC9+ three sides. the three CBg. Fig. 17.(4.) angles.
2. If ABq be greater than ACg+CBg. Fig. 17. BC : BA+AC: : BA-AC : BD +DC ; and BC the difference of BD, DC is given, therefore each of them is given. K7.)
And CA : CD ::R: Co s, C. (1.) and C being found, A and B are found by case 12. or 3.
of the sphere, from which all straight lines drawn to the cir.
the centre of the sphere, and whose centre therefore is the
comprehended by three arches of three great circles, each of
contained by two arches of great circles, and is the same with
GREAT circles bisect one another.
As they have a common centre their common section will be a diameter of each which will bisect them.
| PROP. II. FIG. 1. THE arch of a great 'circle betwixt the pole and the circumference of another is a quadrant.
Let ABC be a great circle, and D its pole ; if a great circle DC pass through D, and meet ABC in C, the arch DC will be a quadrant.
Let the great circle CD meet ABC again in A, and let AC be the common section of the great circles, which will pass
through E the centre of the sphere: join DE, DA, DC: hy def. 1. DA, DC are equal, and AE, EC are also equal, and DE is common; therefore (8. 1.) the anglés DEA, DEC are equal; wherefore the arches DA, DC are equal, and consequently each of them is a quadrant. Q. E. D.
PROP. HI. FIG. 2.
IF a great circle be described meeting two great circles AB, AC passing through its pole - A in B, C, the angle at the centre of the sphere upon the circum. ference BC, is the same with the spherical angle BAC, and the 'arch BC is called the measure of the spherical angle BAC.
Let the planes of the great circles AB, AC intersect one an. other in the straight line AD pássing through D their common centre; join DB, DC.
Since A is the pole of BC, AB, AC will be the quadrants, and thë angles ADB, ADC right angles; therefore (6. def. 11.) the angle CDB is the inclination of the planes of the circles AB, AC; that is, (def. 4.) the spherical angle BAC. Q. E. D.
Cor. If through the point A, two quadrants AB, AC be drawn, the point A will be the pole of the great circle BC, passing through their extremities B, C.
Join AC, and draw AE a straight line to any other point E in BC; join DE: since AC, AB are quadrants, the angles ADB, ADC are right angles, and AD will be perpendicular to the plane of BC: therefore the angle ADE is a right angle, and AD, DC are equal to AD, DE, each to each ; therefore AE, AC are equal, and A is the pole of BC, by def. 1. Q. E. D.
PROP. IV. FIG. S.
IN isosceles spherical triangles, the angles at the base are equal.
Let ABC be an isosceles triangle, and AC, CB the equal sides; the angles BAC, ABC, at the base AB, are equal.
Let D be the centre of the sphere, and join DA, DB, DC: in-DA take any point E, from which draw, in the plane AÐC, the straight line EF at right angles to ED meeting CD in F, and draw, in the plane ADB, EG at right angles to the same ED; therefore the rectilineal angle FEG is (6. def. 11.) the inclination of the planes ADC, ADB, and therefore is the same with the spherical angle BAC: from f draw FH perpendicular to DB, and from H draw, in the plane ADB, the straight line HG at right angles to HD meeting EG in G, and join GF. Because DE is at right angles to EF and EG, it is perpendicular to the plane FEG, (4. 11.) and there. fore the plane FEG, is perpendicular to the plane ADB, in which DÉ is: (18. 11.) in the same manner the plane FHG is perpendicular to the plane ADB; and therefore GF the common section of the planes FEG, FHG is perpendicular to the plane ADB ; (19. 11.) and because the angle FHG is the inclination of the planes BDC, BDA, it is the same with the spherical angle ABC ; and the sides AC, CB of the spherical triangle being equal, the angles EDF, HDF, which stand upon them at the centre of the sphere, are equal ; and in the triangles EDF, HDF, the side DF is common, and the angles DEF, DHF are right angles; therefore EF, FH are equal ; and in the triangles FEG, FHG the side GF is common, and the sides EG, GH will be equal by the 47. 1. and therefore the angle FEG is equal to 'FHG; (8. 1.) that is, the spherical angle BAC is equal to the spherical angle ABC.
PROP. V. FIG. 3.
IF, in a spherical triangle ABC, two of the angles BAC, ABC be equal, the sides BC, AC opposite to them, are equal.
- Read the construction and demonstration of the preceding proposition, unto the words and the sides AC, CB,” &c. and the rest of the demonstration will be as follows, viz.
And the spherical angles BAC, ABC, being equal, the recti. lineal angles FEG, FHG, which are the same with them, are equal; and in the triangles FGE, FGH the angles at G are right angles, and the side FG opposite to two of the equal angles is common; therefore (26. 1.) EF is equal to FH; and in the right angled triangles DEF, DHF the side DF is common; wherefore (47. 1.) ED is equal to DH, and the angles EDF, HDF, are therefore equal, (4. 1.) and consequently the sides AC, BC of the spherical triangle are equal.