THEOREM XI. 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. Now (by the Elements) the Angle KBA = ACB = and the Angle B F K = K B F ; Therefore the Line B K=KF; Then in the Triangles IKF, IK B, the Lines or Sides 1 K, K F, are equal to the Sides I K, K B, and the Angles IK F, I K B, are both right ones; therefore IFK = IBK ; Now because the Tangents BK, BI, are projected into the Lines KF, LF; VOL. II. G There Therefore their Angle on the Sphere IB K, is equal to their Representation in the Plane of the Projection IFK; 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 F K, are its Representation on the Plane of the Projection. Q, E. D. CHAP. V. Problems of the Stereographic Projection of the Sphere in Plano. T PROBLEM I. O find the Pole of any Great Circle. Practice. In this Problem-are three Cafes ; Cafe 1. Of the Primitive Circle AF BD; By Definition IX, its Pole will be in the Center C. F Cafe 3. Of an Oblique Circle E FD, thro' the Point E, draw the Line E e, and lay off 90 Degrees from e to f; then draw the Line E f, cutting the Diameter AB, in the Point P, the Pole required. Or thus. The Half-Tangents of the Angle AE F, fet off from the Center C to P, give the Pole required. Note, In all thefe Cafes, the Poles of leffer parallel Circles are the fame. PROBLEM II. To describe, or lay down, any Spherical Angle. Practice. In this Problem are four Cafes. Cafe 1. When the Angle is at the Center of the Primitive C. Let there be made the then is the Angle BCF, of the juft Quantity required. Cafe 2. When the Angle is in the Periphery, as at E; let there be made the Angle A EH≈ 40°; fit the Sector to the Radius C E, and fet the Secant of 40° from E or D, to C; then on the Center e, defcribe the Circle E HD; VOL. II. G 2 then = then fhall the Angle AEH ADH = 40°, as was required. Cafe 3. When the Angle is in the Primitive, but not in the Center, or Periphery ; as at A, Comprehended by the Arch of a Right Circle D A, and the Arch of an Oblique Circle H A; and fuppofe it is required to contain 52° 30′;. take the Co-Tangent of CA, K H D and fet it from the Center C to G; draw the Line G e, parallel to the Diameter E F. F G e F f Then take the Co-Tangent of the given Angle, viz. 37° 30', and fet it from the Point G to e; then on the Center e, defcribe the Oblique Circle H A I, which fhall contain, with D A, the required Angle DAH Cafe 4. When the Angle A 52° 30′. is contained under the Arches of two Oblique Circles HA, and KA; let it be required to contain 15o; By the last Cafe, make the Angle HAD = 52 30′; alfo make the Angle KAD = 37°, 30'; fo fhall the Difference be the Angle HAK = 15°, 00′. as required. PROBLEM III. To draw any Great Circle thro' any Point, fo, that it fhall contain any given Angle with the Primitive Circle. Practice, and with the Secant of 58°, Ъ and one Foot in the given Point O, interfect the other Arch in e; E F G on which, as a Center, defcribe AGB, the Oblique Circle required to be drawn. Note, The Point O must be fo far diftant from the Center C, that the Tangent from C, and the Secant from O may interfect each other. Otherwise it is impoffible. PROBLEM IV. To draw a Great Circle thro' any two Points given within the Primitive Circle. Practice. |