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THE Triangle that I shall make use of in the Solution of the 12 Cases of an Oblique-angled Sphericall Triangle shall be this, A B E: whose Sides and Angles are as followeth.
| degr. | m. | |||
| The Side | A B, the Base, | contains | 30 | 00 |
| B E | 18 | 47 | ||
| A E | 42 | 51 | ||
| The Angle | A | contains | 23 | 30 |
| B | 122 | 36 | ||
| E | 38 | 15 |
The Analogie or Proportion is,
As the Sine of the Angle at E 38 degr. 15 min. is to the Side A B 30 degr.
So is the Sine of the Angle at A 23 degr. 30 min. to the Sine of the Side B E.
FIrst, draw the Quadrant A B C: then out of your Line of Chords take 38 degr. 15 min. the quantity of the Angle at E, and set it upon the Quadrant from B to F. Also take 30 degr. the quantity of the given Side A B, out of your Line of Chords, and set them upon the Quadrant from B to E, and draw the Lines F G and E H both of them parallel to A B.
This done, take in your Compasses the distance A H, and
setting one foot of that extent in G, with the other describe the Arch L, and draw the Line A M so that it may onely touch the Arch L. Then setting one foot of the Compasses in I, with the other take the nearest distance to the Line A M: this di∣stance set from A to K, and draw the Line N K parallel to A B. So shall the distance B N, measured upon the Line of Chords, contain 18 degr. 47 min. the quantity of the enqui∣red Side B E.
The Analogie or Proportion is,
As the Sine of the Base A B 30 degr. is to the Sine of the An∣gle at E 38 degr. 15 min.
So is the Sine of the Side B E 18 degr. 47 min. to the Sine of the Angle at A.
YOur Quadrant being drawn, from your Line of Chords take 30 degr. the quantity of the Side A B, and set them from B to E. Also take 38 degr. 15 min. the quantity of the Angle at E, and set them from B to F. Likewise take 18 degr. 47 min. the quantity of the given Side B E, from your Line of Chords, and set them from B to N; and draw the Lines E H, and F G, and N K, all parallel to A B.
This done, take in your Compasses the distance A H, and setting one foot in G, with the other describe the Arch L, draw∣ing the Line A M onely to touch the Arch. Then take the di∣stance A K in your Compasses, and setting one foot of them upon the Line A C, move it along that Line gently, till the other Point, being turned about, will onely touch the Line A M; and where the Compass-point resteth upon the Line A C, which it will doe at the Point I, through I therefore draw
the Line I D parallel to B A. So shall B D, being measured on your Line of Chords, contain 23 degr. 30 min. the quan∣tity of the enquired Angle at A.
The Analogie or Proportion is,
As the Co-sine of A B 60 degr. is to the Co-sine of B E 71 d. 13 min.
So is the Co-sine of A C 62 degr. 6 min. the place where a Perpendicular would fall from B, to the Co-sine of C E 75 degr. 3 min. Which C E, being added to A C, will give the quantity of the Side A E.
HAving drawn your Quadrant, take 60 degr. the Com∣plement of A B, out of your Line of Chords, and set
of B E, and set them from B to S. In like manner take 62 d. 6 m. and set them from B to Q; and draw the Lines P O, S T, and Q R, all of them parallel to B A.
This done, take in your Compasses the distance A O, and fetting one foot in T, with the other describe the Arch Y; and draw the Line A Z onely to touch the Arch Y. Then take the distance A R, and setting one foot upon the Line A C, move it gently along the same, till the other foot, being tur∣ned about, do onely touch the Line A Z; and where the Com∣pass-point resteth, which it will doe at V, through that Point draw a Line V X parallel to B A. So shall the distance C X, measured upon your Line of Chords, contain 14 degr. 57 min. which added to the former Segment of the Side A C 27. degr. 54 min. the Sum will be 42 degr. 51 min. the quantity of the whole Side A E, which was required.
By the foregoing Cases in Rectangled Sphericall Triangles (if you let fall a Perpendicular from the Angle B upon the Side A E, as B C,) you may finde the Angle A B C 69 degr. 22 min. and the Angle C B E 53 degr. 14 min. which being obtained,
The Analogie or Proportion is,
As the Sine of the Angle A B C 69 degr. 22 min. is to the Sine of the Angle C B E 53 degr. 14 min.
So is the Co-sine of A 66 degr. 30 min. to the Co-sine of the Angle at E.
THE Angles being found and the Quadrant drawn, take 53 degr. 14 min. out of your Chord, and set them from B to E. Likewise take 69 degr. 22 min. and set them from B
to G. Also take 66 degr. 30 min. and set them from B to F; and draw the Lines G H, F I, and E K, all parallel to B A.
This done, take in your Compasses the distance between A and K, and setting one foot in H, with the other describe the Arch M, and draw the Line A N onely to touch the Arch. Again, setting one foot of the Compasses in I, with the other take the least distance to the Line A N, which distance set from A to L. Lastly, draw the Line L D parallel to B A, cutting the Quadrant in D. So shall the Arch B D, measu∣red upon your Chords, contain 51 degr. 45 min. the Com∣plement of the enquired Angle at E; or, D C shall give you 38 degr. 15 min. the Angle it self.
By the preceding Cases of Right-angled Sphericall Tri∣angles, the Perpendicular B C being let fall, finde the Angle A B C, which will be found to be 69 degr. 22 min. This Angle being obtained,
The Analogie or Proportion is,
As the Co-sine of the Angle at A 66 degr. 30 min. is to the Co∣sine of the Angle at E 51 degr. 45 min.
So is the Sine of A B C 69 degr. 22 min. to the Sine of the Angle C B E.
THE Angle A B C being found, and the Quadrant drawn, take 66 degr. 30 min. out of your Line of Chords, and set them from B to F. Also take 51 degr. 45 min. from the Chord, and set them from B to D. Likewise take 69 degr. 22 min. (the quantity of the Angle found A B C) from your
Chord, and set them from B to G, and draw the Lines F I, D L, and G H, all parallel to A B.
This done, take in your Compasses the distance A L, and setting one foot of that extent in I, with the other describe the Arch M, and draw the Line A N onely to touch the Arch. Again, set one foot of the Compasses in H, and with the other take the least distance to the Line A N, which had, it will reach from A to K; through which Point K draw the Line K E parallel to B A, cutting the Quadrant in E. So shall the distance B E, being measured on your Line of Chords, contain 53 degr. 14 min. the quantity of the remaining part of the Angle A B E; which being added to the Angle A B C (before found) 69 degr. 22 min. the Sum of them will be 122 degr. 36 min. the quantity of the Angle A B E, which was required.
These Five last Cases are all that in Oblique-angled Sphericall Triangles are resolvable by Sines onely, except the Three last, which we resolve by another kind of Artifice. The 6. 7. 8. and 9. fol∣lowing have Tangents ingredient in the Proportion, and so do require a different way of resolving, which is as followeth.
The Perpendicular B C being let fall from the Angle at B, you must find the Segment of the Base C E 14 degr. 57 min. by the preceding Cases of Right-angled Sphericall Triangles: Which had,
The Analogie or Proportion is,
As the Sine of the Side C E found 14 degr. 57 min. is to the Sine of A C 27 degr. 54 min.
So is the Tangent of the Angle at A 23 degr. 30 min. to the Tangent of the Angle at E.
YOur Quadrants A B C and C A D being drawn, take 14 degr. 57 min. the quantity of C E found, out of your Line of Chords, and set them from B to E. Also take 27 d. 54 min. the Segment of the Base A E, and set them from B to H; and draw the Lines E F and G H both parallel to A B. Also, take out of your Line of Chords 23 d. 30 min. the quantity of the Angle at A, and set them from C to N; and draw the Line A N, continuing it to I.
This done, take the distance A G, and set it from C to O. Also take the distance A F, and setting one foot in O, with the other de∣scribe the Arch L; and draw the Line C M only that it may touch the Arch L. Then take in your Compasses the distance C I, set one foot upon the Line C D, moving it along the same till the other, being turned about, do onely touch the Line C M; and where the other Point of the Compasses rest∣eth upon the Line C D, which you will find it to doe at K, draw the Line A K, cutting the Quadrant in P. So shall C P,
being measured upon your Line of Chords, contain 38 degr. 15 min. the quantity of the Angle E, which was required to be found.
Having let fall the Perpendicular B C, and found the Angle B C E by the former Cases, then
The Analogie or Proportion is,
As the Co-sine of C B E 38 degr. 46 min. is to the Co-sine of A B C 20 degr. 38 min.
So is the Tangent of A B 30 degr. to the Tangent of B E.
THE Quadrants drawn, out of your Line of Chords take 38 degr. 46 min. the Complement of the Angle C B E, and set them from B to S. Also take from your Chords 20 degr. 38. min. the Complement of the Angle A. B C, and set them from B to Q, and draw the Lines P Q and R S both parallel to C D. Again, take in your Compasses the di∣stance A R, and set it from C to T. Likewise take the distance from A to P, and setting one foot of that distance in T, with the other describe the Arch Y, and by it draw the Line C M onely to touch the Arch Y. Then take 30 degr. from your Chord, and set them from C to X, drawing the Line A X, and prolonging it to Z. Lastly, from the Point T take the nearest distance to the Line C M, which distance set from C to V, and draw the Line A V, cutting the Quadrant in AE. So shall C AE contain 18 degr. 47 min. of your Line of Chords, which is the quantity of the enquired Side B E.
Having let fall the Perpendicular B C, and found the Angle A B C 69 degr. 22 min. by the foregoing Cases of Right-an∣gled Sphericall Triangles,
The Analogie or Proportion is,
As the Tangent of B E 18 degr. 47 min. is to the Tangent of A B 30 degr.
So is the Co-sine of A B C 20 degr. 38 min. to the Co-sine of C B E.
HAVING drawn your Quadrants, out of your Line of Chords take 18 degr. 47 min. the Side B E, and set them from C to M. Also take 30 d. the quantity of the Side A B, from your Chord, and set them from C to N; and draw the Lines A M and A N, continuing them to G and H. Likewise take 20 degr. 38 min. out of your Line of Chords, and set them from B to F, and draw F E parallel to A B.
This done, take in your Compasses the distance C C, and setting one foot in the Point H, with the other describe the Arch K, and draw the Line C L onely to touch it. Then take in your Compasses the distance A E, and setting one foot of the Compasses upon the Line C D, move it along the same gently, till the other, being turned about, do onely touch the Line C L; and where the Compass-point resteth, which you will finde it to doe at O, make a mark, and take the distance O C in your Compasses, and set it from A to P, and draw the Line P Q parallel to A B. So shall B Q, measured upon the Line of Chords, contain 36 degr. 46 min. the Comple∣ment of the Angle C B E; or, the distance C Q, measured upon the Chords, will give 53 degr. 14 min. the Angle C B E it self; which being added to 69 degr. 22 min. the Angle A B C, the Sum will be 122 degr. 36 min. the quantity of the whole Angle A B E required.
The Perpendicular being let fall, and A C 27 degr. 54 min. a part of the Side A E; then
The Analogie or Proportion is,
As the Tangent of the Angle at E 38 degr. 15 min. is to the Tangent of the Angle at A 23 degr. 30 min.
So is the Sine of A C 27 degr. 54 min. to the Sine of E C.
HAving drawn your Quadrants, take out of your Life of Chords 38 degr. 15 min. the quantity of the Angle at E, and set them from C to a. Also take 23 degr. 30 min. the quan∣tity of the Angle at A, and set them from C to b, drawing the Lines A b and A a, and continuing of them to T and V.
This done, take in your Compasses the distance C T, and setting one foot of them in V, with the other describe the Arch X, and draw the Line C Y onely to touch the Arch X. Then take in your Compasses the distance A R, and set it from C to AE, and from the Point AE take the least distance to the Line C Y, which distance set from A to m, and draw the Line m n parallel to A B. So B n, being measured upon your Line of Chords, shall contain 14 degr. 57 min. a por∣tion of the Side A E; which being added to the other portion A C 27 degr. 54 min. the Sum of them is 42 degr. 51 min. the whole Side A E, which was required.
Take the Sum and the Difference of the two Sides A B 30 d. and B E 18 degr. 47 min.
| degr. | m. | |
| Their Sum is | 48 | 47 |
| Their Difference is | 11 | 13 |
FIrst, draw a right Line A O C, and upon the Centre O de∣scribe the Semicircle A B C, drawing the Semidiameter or Perpendicular B O.
Being thus prepared, take out of your Line of Chords 48 d. 47 min. the Sum of the two Sides given, and set them upon the Semicircle from A to E. Also take 11 degr. 30 min. the Difference of the two given Sides, and set them from A to D. Again, take 122 degr. 36 min. the quantity of the given An∣gle, and set them from A to K; [but A B being 90 degr. set the residue, namely, 32 degr. 26 min. from B to K,] and draw the Lines D I, E H, and K L, all of them perpendicu∣lar to the Line A C.
This done, take in your Compasses the distance between I and H, and setting one foot of that extent in C, with the other describe the Arch M, and draw the Line A N so that
t may onely touch the Arch M. Then setting one foot of the Compasses in L, with the other take the nearest distance you can from L to the Line A N, and set that distance from I to G. Lastly, upon the Point G erect the Perpendicular G F, cutting the Semicircle in F. So shall the distance A F, being measured upon your Line of Chords, contain 42 degr. 51 m. the quantity of the Base of the Triangle A E, which was re∣quired.
To work this by the Canon.
The Analogie or Proportion is,
1. As the Radius is to the Co-sine of the Angle given,
So is the Tangent of one of the given Sides to the Tangent of a fourth Arch.
2. As the Co-sine of this fourth Arch is to the Co-sine of the former given Side,
So is the Co-sine of the other given Side, the fourth Arch be∣ing subtracted therefrom, to the Co-sine of the Side sought.
First, find the Sum and the Difference of the Sides B E and A E.
| degr. | m. | |
| Their Sum is | 61 | 38 |
| Their Difference is | 24 | 04 |
THE Sum and Difference of the two Sides being taken, out of your Line of Chords take 24 degr. 4 min. the Difference of them, and set them from A to P. Also from your Line of Chords take 61 degr. 38 min. the Sum of the two Sides, and set them from A to X. Again, take 30 degr.
the quantity of the third Side A B, and set them from A to Q; and draw the Lines P S, Q T, and X Y, all three per∣pendicular to A C.
This done, take the distance between Y and S, and setting one foot of the Compasses in C, with the other describe the Arch Z, and draw the Line A AE so that it may onely touch the Arch Z. Then take in your Compasses the distance be∣tween S and T, and setting one foot thereof upon the Line A C, move it gently along the same, till the other foot, being turned about, do onely touch the Line A AE; and where the Compass-point resteth, which you will find it to doe at the Point Y, upon this Point Y erect the Perpendicular Y R. So shall the distance A R, being measured upon your Line of Chords, give 38 degr. 15 min. the quantity of the Angle at E, which was required to be found.
To work this by the Canon.
The Analogie or Proportion is,
As the Rectangle contained under the Sines of the Sides is to the Square of the Radius,
So is the Rectangle contained under the Sines of the half Sum of the three Sides, and the Difference between this half Sum and the Base, to the Square of the Co-sine of half the An∣gle required.
IF for either of the Angles next the Side required we take its Complement to 180 degr. those Angles will be turned into Sides, and the Sides into Angles: And then will the man∣ner of resolving the Triangle be the same as in the last Case.