THEOREM XI. FPVK In the Stereographic Projection, the Angles made by the Circles on the Surface of the Globe or Sphere, are equal to the Angles made by their Representations in the Plane of the Projection. Demonsiration. Suppose the Eye at A, project the Spherical Angle P BO; draw the Tangents B BK, B1, to the Circles BO, BP; from the Point B, draw the Line B C, parallel to the Diameter DO, A and draw up the Rays AC, AB; let fall the Perpendicular K I, and draw the Line FI; Now (by the Elements ) the Angle KBA= ACB = ABC; CON and the Angle B FK = KBF; Therefore the Line B K=KF; Then in the Triangles I KF, IK B, the Lines or Sides 1 K, KF, are equal to the Sides 1 K, K B, and the Angles I KF, I KB, are both right ones ; therefore I FK = IBK ; Now because the Tangents B K, BI, are projected into the Lines KF, LF; VOL. II. G There Therefore their Angle on the Sphere I B K, is equal to their Representation in the Plane of the Pro jection I FK; But the Angle made by the Tangents, IB K, is equal to that made by the Circles, P BO; Wherefore the Spherical Angle P BO= I FK, are its Representation on the Plane of the Projection. 2. E. D. CH A P. V. Problems of the Stereographic Projection of the Sphere in Plano. T PROBLEM I. Praktice. Pole will be in the Center C. cle, as AB, Р A, or B, each way to the Points E, and D and those Points will be the Poles required. Case 3. Of an Oblique Circle E FD, thro the Point E, draw the Line Ee, and lay off 90 Degrees from e to f; then draw the Line Ef, curtirg the Diameter AB, in the Point P, the Pole required. Or thus. The Half-Tangents of the Angle A EF, set off from the Center C to P, give the Pole required. Note, In all these Cases, the Poles of lesser parallel Circles are the same. PROBLEM II. To describe, or lay down, any Spherical Angle. Pra£tice. In this Problem are four E Case 1. When the An- C B Cafe 2. When the Angle is in the Periphery, as at E; let there be made the Angle A EH= 409 ; fit the Sector to the Radius CE, and set the Secant of 40° from E or D, to C ; then on the Center e, describe the Circle EHD; VOL. II. G 2 then KA e w then shall the Angle A EH= ADH = 40°, as was required. Case 3. When the Angle is in the Primitive, but not in the H4 Center, or Periphery ; as at A, G Comprehended by the Arch of D la a Right Circle D A, and the Arch of an Oblique Circle HA;:. HT and suppose it is required to contain 52° 30'; take the Co-Tangent of C A, and set it from the Center C to G; draw the Line Ge, parallel to the Diameter E F. Then take the Co-Tangent of the given Angle, viz. 37° 30', and set it from the Point G toe; then on the Center e, describe the Oblique Circle H AI, which shall contain, with DA, the required Angle D AH = 529 30'. Cafe 4. When the Angle A is contained under the Arches of two Oblique Circles HA, and K A; let it be required to contain 15°; By the last Case, make the Angle H AD= 52 30' ; also make the Angle KAD = 37%, 30'; so shall che Difference be the Angle HĀK = 15°, co'. as required. PROBLEM III. To draw any Great Circle thro' any Point, so, that it shall contain any given Angle with the Primitive Circle. Pratice, Praflice. Suppose the Angle E AG E and Oblique Circle AGB, Point O; a: passes in C, B Note, The Point O must be so far distant from the Center C, that the Tangent from C, and the Secant from 0 may intersect each other. Otherwise it is impossible. 30'; PROBLEM IV. To draw a Great Circle thro' any two Points given within the Primitive Circle. that nitive Practice. Ericla |