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The projection and solution of the 12 Cases in oblique angled spherical triangles in six Cases. See Fig. 7.
THe fundamental circle N H Z M, is alwayes supposed ready drawn, and crossed into Quadrants, and the Diameters produced beyond the Circle.
The three sides, Z P, P Z, and Z S, given, to project the Triangle.
By a line of chords prick off Z P, and
draw the diameter P C T, crossing it at right angles in the center with AE C E, set half the co-tangent P S, from C to G, and he secant P S from C to R, upon the center R, with the extent R G, draw the the ark FGL. Again, set half the co-tangent Z S, from C to D, and the tangent Z S, from D to O, with the extent O D, upon the center O, draw the ark B D P, mark where these two arks intersect each other as at S. Then have you three points T S P, to draw that ark, and the three points N S P, to draw that ark, which make up your triangle.
Given two sides Z S, and Z P, with the com∣prehended Angle P Z S, to project the Tri∣angle.
Prick off Z P, and draw PCT, and AECE, and the ark B D P, by Case 1. Again, set the tangent of half the excess of the angle P Z S above 90, from C to W, and co-se∣cant of that excess from W to K, upon the center K, with the extent K W; draw the ark N W Z, which cuts the ark B D P in S. Then have you the three points T S P, to draw that ark which makes up the triangle.
Two Angles S Z P, and Z P S, with the com∣prehended side Z P, given, to project the Tri∣angle.
Prick off Z P, and draw the lines P C T, and AE C E, by Case 1, and the angle NWZ. by Case 2. Lastly, set half the co-tangent ZPS from C to X, and the secant Z P S, from X to V, upon the center V, with the extent V X, draw the ark T X S P, and the triangle is made.
Two sides, ZP, and PS, with the Angle op∣posite to one of them SZP, given, to project the Triangle.
Prick off ZP, and draw PCT, and AECE, by Case 1. and the angle SZP by Case 2. Lastly, by Case 1. draw the ark FGL, and mark where it intersects NWZ, as at S, then have you the three points TSP, to draw that ark, and make up the triangle.
Two Angles SZP, and ZPS, with the side op∣posite to one of them ZS, given, to project the Triangle.
Draw the ark BDP, by Case 1. and the ark NWZ, by Case 2. at the intersection of these two arks, set S, with the tangent of the angle ZPS, upon the center C. sweep the ark VΔI. Again, with the secant of the ark ZPS upon the center S, cross the ark VΔI, as at the points V and I. Then in case the hypothenuse is less than a quadrant (as here) the point V, is the center, and with the extent VS, draw the ark TSP, which makes up the triangle. But in case the hypo∣thenuse is equal to a quadrant, Δ, is the cen∣ter; if more than a quadrant, I, is the cen∣ter; in which cases the extent from Δ, or I, to S, is the semidiameter of the ark TSP.
Three Angles ZPS, and PZS, and ZSP, given, to project the Triangle. See Fig. 7. and 8.
The angles of any spherical triangle may
be converted into their opposite sides, by taking the complement of the greatest angle to a Semicircle for the hypothenuse, or greatest side. Wherefore by Case 1. make the side ZP, in Fig. 4. equal to the angle ZSP, in Fig. 3, and the side ZS, in Fig. 4. equal to the angle ZPS, in Fig. 3. and the side PS, in Fig. 4. equal to the complement of the angle PZS, to a Semicircle in Fig. 3. Then is your triangle projected where the angle ZPS in Fig. 4. is the side ZS, Fig. 3. Again, the angle ZSP, Fig. 4. is the side ZP, in Fig. 3. Lastly, the complement of the angle PZS to a Semicircle in Fig. 4. is the measure of the hypothenuse, or side P S, in Fig. 3.
The Triangle being in any of the former Cases projected, the quantity of any side or angle may be measured by the following rules.
First, The side Z P, is found by applying it to a line of chords. Secondly, CX, ap∣plyed to a line of tangents, is half the co∣tangent of the angle ZPS. Thirdly, CW applyed to a line of tangents, is half the co-tangent of the excess of the angle SZP, above 90. Fourthly, set half the tangent of the angle ZPS, from C, to Π, a ruler laid to ΠS, cuts the limb at F; then PF, ap∣plyed
to a line of chords, gives the side PS. Fifthly, take the complement of the angle PZS, to a Semicircle, and set half the tan∣gent of that complement from C, to λ, a ruler laid to λS, cuts the limb at B, and ZB, applyed to a line of chords, gives the side ZS. Sixthly, a ruler laid to Sλ, cuts the limb at L. Again, a ruler laid to SΠ, cuts the limb at φ, and L φ, applyed to a line of chords, gives the angle ZSP.