An Introduction to the Theory and Practice of Plane and Spherical Trigonometry, and the Stereographic Projection of the Sphere

PROPOSITION XIV. Problem. (Plate IV. Fig. 18.)

(Y) Through a given point, in any projected great circle, to draw another great circle perpendicular to the given one.

General Rule. Find the pole of the given circle (N. 155.) and through that pole and the given point draw a great circle (W. 156.). For (I. 130) if one great circle pass through the pole of another, it cuts it at right angles.

Or, I. The pole of the primitive is in the centre, therefore a diameter of the primitive always cuts its circumference at right angles.

II. To draw an oblique circle perpendicular to an oblique circle.

Let BAD be the given oblique circle and a the given point, find p the pole of the oblique circle BAD (N. 155.), and through the points p, and A, draw the great circle ECAP (W. 156.).

III. To draw an oblique circle perpendicular to a right circle.

Let HG be the right circle and a the given point, draw DB at right angles to HG, then в, and D, are the poles of HG; through the three points B, A, D, draw the circle BAD, and it will cut HG at right angles.

PROPOSITION XV. Problem. (Plate IV. Fig. 19, 20, and 21.)

(Z) About any given point, as the pole of a great circle, to describe a small circle at a given distance from that pole. Or, at a proposed distance from a given great circle to describe a parallel circle."

I. If the small circle be parallel to the primitive. With the semi-tangent of its distance from the pole P, (Plate IV. Fig. 19.) or radius PC, describe the circle CDE.

II. If the small circle be parallel to a right circle. Let the given right circle be AB, whose poles are c and D (Plate IV. Fig. 20.). Set off the chord of the small circle's distance from its pole, from c to n and m; or the chord of its distance from the right circle AB, from в to n, and from a to m; and draw AMF and Aon, where these lines intersect the axis PF, in o and F, will give the two extremities of the diameter of the circle mon to be projected, the middle point E will be the centre.

Or, Having found the points m and n, draw Pn, and nɛ at right angles to it, and it will cut the axis PC produced in E, the centre of the parallel circle mon. (A. 150.)

Or, Find the three points m, o, and n, as above, and draw a circle through them.

III. If the small circle be parallel to an oblique circle. Let the oblique circle be DAB (Plate IV. Fig. 21.). Find p the pole of the oblique circle DAB, and through p draw Dpe; make Ev and Ew each equal to the proposed distance from the pole,

and

and draw D and DV, cutting the axis GF produced in o and F, the two extremities of the diameter of the circle mon to be projected, the middle point c will be the centre.

Or, Find p the pole of the oblique circle DAB (N. 155.), and measure its distance pp from the centre of the primitive by a scale of semi-tangents. Then add and subtract this distance to and from the complement of the parallel circle's distance from the oblique circle; set off the sum, by a scale of semi-tangents, from P to F, and the difference from P to o; the middle point between o and F, (viz. c) is the centre. (D. 152.)

PROPOSITION XVI. Problem. (Plate V. Fig. 1.)

(A) About any given point, as the pole of a great circle, to describe a great circle in a given primitive circle.

I. If the given point be (P) in the centre of the primitive, the primitive is the great circle required.

II. If the given point be (c) in the circumference of the primitive, through c, draw a diameter cre, and another brв at right angles to it. Then, 6PB is the great circle required.

III. If the given point be (p) neither in the centre nor in the circumference of the primitive. Through p, and the centre of the primitive, draw a straight line rPB, and cross it at right angles with the diameter ec, through p draw epw, make wm equal to wc, and through m draw emr, then with r as a centre and radius re, describe the required great circle cae.

(B) This problem is easily deduced from X. 149.

PROPOSITION XVII. Problem. (Plate V. Fig. 2.)

General Rule. Find the pole of the given circle (N. 155.), from which draw straight lines through the ends of the arc to be measured, cutting the primitive in two points, the distance between these points applied to a scale of chords, will give the measure of the arc.

Or, I. The pole of the primitive is the centre P, therefore any arc of the primitive is measured by taking the extent of the arc (as be) and applying it to a scale of chords.

II. To measure any part of a right circle as ra, Pp, pv, &c. From c the pole, draw caf, cre, cpg, cvw; then ef applied to a scale of chords is the measure of ra, eg of Pp, and gw

of pv.

Or, Pа, pp, pv, applied to the line of semi-tangents, will give their respective measures. And, it is evident that up is the difference

difference between the measures of pv, and Pp, and that ap is the sum of the measures of pa and pp.

III. To measure any part of an oblique circle as cr, ro, oa, &c. Find p the pole thus: draw caf, make ƒg an arc of 90 degrees, and join gc, then p is the pole. From p through r, o, a, m, draw pR, po, pa, pм. Then RO, applied to a scale of chords, is the measure of an arc on the sphere represented by ro ; OA is the measure of an arc on the sphere represented by and AM is the measure of an arc on the sphere represented by am, &c. (L. 154.).

oa;

PROPOSITION XVIII.

Problem. (Plate V. Fig. 2.)

(D) To cut any number of degrees from the arc of of a great circle; or, from a given point, in a projected great circle, to cut off an arc equal to a given arc.

This proposition is the reverse of Prop. XVII.

I. If the projected great circle be the primitive whose pole is P, and e the given point. Take the number of degrees in the given arc from a scale of chords and apply the extent from e to b, then eb is the arc required.

II. If the projected great circle be a right circle (as xA) whose pole is c, and a the given point. From the pole c, through a, draw caf, make fg equal to the given arc, by a scale of chords, and join cg, then will pa contain the required number of degrees.

Or, Any number of degrees may be cut from A by the scale of semi-tangents, in the same manner as the arcs were measured in the XVIIth Problem.

III. If the projected circle be an oblique circle (as cae) whose pole is p, and o the given point. From p through o draw poo, make or equal to the given number of degrees, and join pR, then or will contain the number of degrees required.

PROPOSITION XIX. Problem. (Plate V. Fig. 3.)`

(E) Any great circle cae in the plane of projection being given: to describe another great circle BAD, which shall cut the given circle CAe, and also the primitive in any assigned angles.

About the pole P of the primitive describe the parallel circle no, at a distance equal to the angle which BAD is required to make with the primitive (Z. 157.). About p, the pole of cae, at a distance equal to the measure of the angle which BAD is required to make with CAC, describe the parallel circle ab(Z. 157.) cutting no in d.

About d as a pole describe the great circle BAD (A. 158.),

cutting

cutting the primitive in B, and Cae in A: then BAD is the great circle required.

(F) This construction is evident from Q. 131.

PROPOSITION XX.

Problem. (Plate V. Fig. 4.)

(G) To measure any spherical angle.

General Rule. Find the poles of the two great circles which form the angle (N. 155.). Straight lines drawn from the angular point through these poles, will cut the primitive in two points; the distance between which, applied to a scale of chords, will give the measure of the required angle.

The truth of this appears from Q. 131.

Or, I. If the angular point be at P, and rpv, the angle to be measured, then Pr, PV are quadrants; and as a spherical angle is always measured on the arc of a great circle at a quadrant's distance from the angular point (R. 132.), rv applied to a scale of chords is the measure of the angle rPv.

II. If the angle be formed by the primitive, and an oblique circle, as OCE.

Find m the pole of cae, and from c draw cnw; and through P, the pole of the primitive, draw cpe, then we applied to a scale of chords gives the measure of OCE. To measure oCP,

the angle formed by the right circle cpe, and the oblique circle CAC; through m, the pole of cae, and r, the pole of cre, draw cw, and cr; then wr applied to a scale of chords is the measure of ocp.

Or, The angle OCE may be measured by the line of semitangents; thus, let oɛ be applied to the scale of semi-tangents from 90 towards the left hand, the number of degrees contained between the points of the compasses, will be the measure of OCE; and Po, applied to the scale of semi-tangents, from the beginning of the scale towards 90, will give the measure of ocp.

III. If the angle be formed by two oblique circles, as cae and

BAD.

Find m the pole of cae, and n the pole of BAD (N. 155.). From the angular point A, through m and n, draw ams and Anv, cutting the primitive in s, and v; then sʊ, applied to a scale of chords, will give the measure of the angle BAC, or of its equal eAD.

CHAP. II.

INVESTIGATION OF GENERAL RULES FOR CALCULATING THE SIDES AND ANGLES OF RIGHT-ANGLED SPHERICAL TRIANGLES, AND THE LOGARITHMICAL SOLUTIONS OF ALL THE CASES.

PROPOSITION XXI.

(H) In any right-angled spherical triangle.

Radius, is to the sine of any side; as the tangent of the adjacent angle, is to the tangent of the opposite side.

DEMONSTRATION. Let AEC be a spherical triangle, right-angled at A. rad: sine AB:: tang 4B: tang AC. For, let D be the centre of the sphere, and draw the radii DC, da, and DB; also from the right-angle A, in the plane DBA, draw AE perpendicular to DB, and it will be the sine of the arc AB.

At the point E, in the plane DBC, draw EF perpendicular to DB, then will the angle FEA be the inclination of the planes DBA and DBC, and consequently (D. 129.) equal to the spherical angle

CBA.

Draw AF from the point A, a tangent to the arc AC, and produce the radius DC to F; then since the arc AC is perpendicular to the plane DBA (for by hypothesis it cuts the arc AB at rightangles), AF will be perpendicular to AE.

Because AC is less than a quadrant, and DA, DB, and AE, are in the same plane DBA, and that EF is at right-angles to DB (by construction); DC cannot likewise be at right-angles to DB. Therefore DC and E being in the same plane DBC, and not parallel to each other, must meet.

The plane triangle FAE is right-angled at A; the perpendicular AF is a tangent to the spherical perpendicular Ac; the base AE is the sine of the spherical base AB; and the angle FEA is equal to the spherical angle ABC.

Hence, rad: AE :: tang FEA: AF.

That is, rad: sine AB::tang ABC: tang AC, Q.E.D.

(I) SCHO

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