The General Use of the Canon and Tables of Logarithms.

CHAP. II.
Concerning the Use of the Lines of Sines and Tangents in the resolving of Spherical Triangles.

Concerning the Use of the Lines of Sines and Tangents I have shewed in general, in the seventh and eightth Chapters of the first book of the Cross-staff, how they might serve for the Resolution of all Spherical Triangles. More particularly in the Use of my Sector, Chap. 5. I reduced that which is commonly required in a Spherical Triagle into 28 Cases. And for these they may be all resolved by my Tables of Artificial Sines and Tangents without the help of Secants or versed Sines.

This manner of work will be always such as in the ordinary Rule of Three, For, here we have three Numbers given, whereby to find a fourth Proportional. And therefore either we may add the Logarithms of the second and third, and subtract the Logarithm of the first:

Or we may take the Difference between Logarithms of the first and second, and apply that Difference to the Logarithm of the third.

The first of these ways is best for the resolution of right angled Triangles where the Radius, viz. 10.000000 is one of the three Numbers given, but the second way by Differences is more convenient for the rest.

the like manner of work may be observed when we are to consider the Sines or Tangent of Degrees, Minutes and Seconds. For the seconds, not expressed in the Canon, will be found by the Part Proportional: as I will shew in the Examples following.

  1. If it were required to find the Sine of 51 gr. 32 min. 15 sec. I should find,
    The Sine of 51 deg. 32 m. is 9.8937452
    The Sine of 51 deg. 33 m.   9.8938455
     
    The Tabular Difference between them 1003
    Then the Difference between 32 m. and 33 m. being 60 Seconds, the Proportion will hold,
    As 60 Seconds unto 1003
    So 15 unto 251
    the part Proportional to be added unto the Sine 51 deg. 32 min.
    So shall we have 9.8937703, for the Sine of 51 deg. 32 min. 15 sec.
  2. If it were required to find Degrees, Minutes and Seconds belonging to this Tangent 10.0999782
    I should find by the Canon that this is somewhat more than the Tangent of 51 deg. 32 min. 10.0999134
    Less than the Tangent of 51 deg. 33 min. 10.1001728
     
    The Tabular Difference between these is 2594
    And the proper Difference is 648
    between the lesser of these Tangents, and the Tangent given: therefore,
    As 2594 unto 60 Seconds.
    So 648 unto 15
    And so, I find this to be the Tangent of 51 deg. 32 min. 15 sec.
  3. If it were required to find the Sine belonging to this Tangent 10.0999782, I should find the Ark to be somewaht more than 51 gr. 31 min. and the Sine correspondent somewaht more than 9.8937452, then taking out the Differences as before, I find that
    As the tabular Difference of Tangent 2594 3.4139700
    is to the proper Difference 648 2.8119750
     
      6023950
     
    So the tabular Difference of Sines 1003 30013009
    To the Part proportional 252 23989059
    This Part proportional added unto the former Sine 9.8937452
    gives 9.8937703 for the Sine required.

These Premisses considered, I come to the 28 Cases before mentiones, wherein I set down a Canon and an Example for each Case, and these for the most part the same which I used before.

Those which have no further use but of Degrees and Minutes, may take that Sine or Tangents which they find to be next in the Canon, and neglect the Seconds.

In a Rectangle Triangle.

1. To find a Side by knowing the Base and the Angle opposite to the required Side.

Rectangle Triangle

As in the Rectangle Triangle ACB, wherein A stands for the Equinoctial point; AB, and Ark of the Ecliptick representing the Longitude of the Sun in the beginning of Aries, BC an Ark of the declination eg the Sun from the Equator, and AC an Ark of the Equator representing the Right Ascension of the Sun in B: Knowing the base AB to be 30 gr. and the Angle BAC 23 gr. 31 min. 30 sec. if it were required to find the Side BC.

  d. m. sec.  
As the Radius, the Sine of 90 00 00 10.0000000
 
Is to the Sine of the Base 30 00 00 9.6989700
So the Sine of the opposite Angle 23 31 30 9.6011352
 
To the Sine of the Side required 11 30 43 19.3001052

And so writing the Sine 9.6001052 in a paper by itself and holding to the Sine of the base in the Canon 1 gr. 2.3.4.5. and so forward, it would be no long work to write the Sum in an Column by itself, and so find the declination for each Degree and Minute of the Ecliptick.

2. To find a Side by knowing the Base and the other Side.

As in the Rectangle ACB having AB 30 gr. and BC 11 gr. 30 m. 43 sec. to find the Side AC.

As the Cosine of the Side given 11 30 43 9.9911740
 
Is to the Radius 90 00 00 10.0000000
So the Cosine of the Base 30 00 00 9.9375306
 
To the Cosine of the Side required 27 53 43 9.9463566

3. To find a Side by knowing the two Oblique Angles.

As in the Rectangle ACB having CAB for the first Angle 23 gr. 31 min. 30 sec. and ABC for the second 69 gr. 20 m. 35 sec. to find the side AC.

As the Sine of the next Angle 23 31 30 9.96011353
 
Is to the Radius 90 00 00 10.0000000
So the Cosine of the opposite Angle 69 20 35 9.5474918
 
To the Cosine of the Side required 27 53 43 9.9463566

4. To find the Base by knowing both the Sides.

As in the Rectangle ACB, having AC 27 gr. 53 m. 43 sec. and BC, 11gr. 30 m. 43 sec. to find the Base AB.

As the Radius 90 00 00 10.0000000
 
To the Cosine of the one Side 27 53 43 9.9463566
To the Cosine of the other Side 11 30 43 9.9911640
 
To the Cosine of the Base 30 00 00 9.9375306

5. To find the Base by knowing one Side and the Angle opposite to that Side.

As if in the former Triangle ACB we draw BD an Ark of the Horizon for the latitude of 51 gr. 30 min. reputing the Amplitude of the Suns rising from the East, we shall have two Triangles more, one Rectangle BCD, the other Obliquadrangles ABD, and so in the Rectangle DCB, having BC 11 gr. 30 m. 43 sec. and BDC 38 gr. 30 min. if it were required to find the base DB.

As the Sine of the Angle 38 30 00 9.7941495
 
To the Sine of the Side 11 30 43 9.3001052
So is the Radius 90 00 00 10.0000000
 
To the Sine of the Base 18 41 56 9.5059556

6. To find an Angle by knowing the other Oblique Angle, and the Side opposite to the Angle required.

As in the Rectangle ACB, having BAC 23 gr. 31 min. 30 sec. and AC 27 gr. 53 min. 43 sec. to find the Angle ABC.

As the Radius 90 00 00 10.0000000
 
To the Sine of the Angle given 23 31 30 9.6011353
To the Cosine of the Side 27 53 43 9.9463566
 
To the Cosine of the Angle required 69 20 35 19.5374918

7. To find an Angle by knowing the other Oblique Angle, and the Side opposite to the Angle given.

As in the Rectangle ACB, having BAC 23 gr. 31 min. 30 sec. and BC 11 deg. 30 min. 43 sec. to find the Angle ABC.

To the Cosine of the Side 11 30 43 9.9911740
 
To the Cosine of the Angle given 23 31 30 9.9613153
So the Radius 90 00 00 10.0000000
 
To the Sine of the Angle required 69 20 35 9.9711413

8. To find an Angle by knowing the base, and the Side opposite to the Angle required.

As in the Rectangle BCD, having BD 18 gr. 41 m. 56 sec. and BC 11 gr. 30 min. 43 sec. to find the Angle BDC.

As the Sine of the Base 18 41 56 9.5050000
 
Is to the Radius 90 00 00 10.0000000
So the Sine of the opposite Side 11 30 43 9.3001052
 
To the Sine of the Angle 38 30 00 9.7941495

These eight Propositions have been wrought by Sines alone; the eight following require joynt help of Tangents.

9. To find a Side, by knowing the other Side, and the Angle opposite to the Side required.

As in the Rectangle ACB, having AC 27 gr. 53 min. 43 sec. and BAC 23 gr. 31 min. 30 sec. to find the Side BC.

As the Radius 90 00 00 10.0000000
 
To the Sine of the Side given 27 53 43 9.6701112
So the Tangent of the opposite Angle 27 53 43 9.6701112
 
To the Tangent of the Side required 23 31 30 19.3089311

10. To find a Side, by knowing the other Side, and the Angle next to the Side required.

As in the Rectangle BCD, having BC 11 gr. 30 min. 43 sec. and BDC 38 gr. 30 min to find DC.

As the Tangent of the Angle 38 30 00 9.9006052
 
To the Tangent of the Side given 11 30 47 9.3089311
So the Radius 90 00 00 10.0000000
 
To the Sine of the Side required 14 50 11 9.4083259

11. To find a Side, by knowing the Base, and the Angle next to the Side required.

As in the Rectangle ACB, having AB 30 gr. 00 min. and BAC 23 gr. 31 m. 30 sec. to find the Side AC.

As the Radius 90 00 00 10.0000000
 
To the Cosine of the Angle 23 31 30 9.9623153
So the Cosine of the Base 30 00 00 9.7614393
 
To the Tangent of the Side required 27 53 43 19.7237546

12. To find the Base by knowing both Oblique Angles.

As in the Rectangle ACB, having BAC 23 gr. 31 min. 30 sec. ABC 69 gr. 20 m. 35 sec. to find the Base AB.

As the Tangent of the one Angle 23 31 30 9.6388199
 
To the Cotangent of the other 69 20 35 9.5763505
So the Radius 90 00 00 10.0000000
 
To the Cosine of the Base 30 00 00 9.9375306

13. To find the Base by knowing one of the Sides and the Angle next to that Side.

As in the Rectangle ACB, having AC 17 gr. 53 min. 43 sec. and BAC 23 gr. 31 min. 30 sec. to find the Base AB.

As the Cosine of the Angle 23 31 30 9.9623153
 
Is to the Radius 90 00 00 10.0000000
To the Tangent of the Side 27 53 43 19.7237547
 
To the Tangent of the Base 30 00 00 9.7614394

14. To find an Angle by knowing both Sides.

As in the Rectangle ACB, having AC 27 gr. 53 min. 43 sec. and BC 11 gr. 30 min. 43 sec. to find the Angle ABC.

A the Sine of the next Side 11 30 43 9.3001052
 
Is to the Radius 90 00 00 10.0000000
To the Tangent of the opposite Side 27 53 43 9.7237547
 
To the Tangent of the Angle 69 20 35 10.4236495

15. To find an Angle by knowing the Base, and the Side next to the Angle required.

As in the rectangle BCD, having BD 18 gr. 41 m. 56 sec. and BC 11 gr. 30 m. 43 sec. to find the Angle BDC.

As the Tangent of the Base 18 41 56 9.5295061
 
To the Tangent of the Side 11 30 43 9.3089311
So the Radius 90 00 00 10.0000000
 
So the Cosine of the Angle 53 00 46 9.7794248

16. To find an Angle by knowing the Base, and the other Oblique Angle.

As in the Rectangle ACB, having the Base AB 30 gr. and BAC 23 gr. 31 m. 30 sec. to find BAC.

As the Cosine of the Base 30 00 00 9.77370000
 
To the Tangent of the Side 11 30 43 9.3089311
Is to the Radius 90 00 00 10.0000000
 
So the Cotangent of the Angle required 69 20 35 10.4236495

These 16 Cases are all that can fall out in a Rectangle Triangle. Those which follow do hold in any Spherical Triangle whatsoever.

In any Spherical Triangle whatsoever.

17. To find a Side opposite to an Angle, by knowing one Side and two Angles, the one opposite to the Side given, the other to the Side required.

As in the Triangle ABD, having AB 30 gr. BDC 38 gr. 30 m. and BAD 23 gr. 31 m. 30 sec. to find the Side BD, which here representeth the Amplitude.

As the Sine of the next Angle 38 30 00 9.7941495
To the Sine of his opposite Side 30 00 00 9.6989700
 
  951795
 
So is the Sine of the oposite Angle 23 31 30 9.6011352
To the Sine of the Side required 18 41 56 9.5059557

Or changing the Side of the two middle Terms.

As the Sine of the next Angle 38 30 00 9.7941495
To the Sine of the oposite Angle 23 31 30 9.6011352
 
To the Sine of the Side given 30 00 00 9.6989700
To the Sine of the Side required 18 41 56 9.5059557

And so writing this Difference 1939143 in a Paper by itself, and holding it to the Sine of the Side in the Canon 1 gr. 2, 3, 4, 5, and so forward, it would be no long work to subtract, and write the Remainder in a Column by itself, and so find the Amplitude for each Degree and Minute of the Ecliptick.

Or instead of subtracting this Difference, we might first take the same out of the Radius, and then add the Complement as I shewed before, in the general explication of the Rule of Three.

18. To find an Angle opposite to a Side, by knowing one Angle and two Sides, the one opposite to the Angle given, the other to the Angle required.

Another Rectangle Triangle

As in the Triangle ZPS representing the Zenith, Pole, and Sun: where ZP is the Complement of the latitude, PS the Complement of the Declination, ZS the Complement of the Suns Altitude, PZS the Azimuth, ZPS the hor of the day from the Meridian, and PSZ the Angle of the Suns Position in regard of the Pole and Zenith; having PZS 130 gr. 3 min. 11 sec. PS 70 gr. and ZS 40 gr. to find the Angle ZPS.

To the Sine of the next Side 70 00 00 9.9729858
Is to the Sine of his opposite Angle 130 03 11 9.8839153
 
  890705
 
So the Sine of the opposite Side 40 00 00 9.8080675
To the Sine of the Angle required 31 34 25 9.7189970

19. To find an Angle by knowing the three Sides.

As in the Triangle ZPS, having ZP 38 gr. 30 min. PS 70 gr. and ZS 40 gr. to find the Angle ZPS, subtending the Base ZS.

  • As the rectangle contained under the Sines of the Sides, is to the Square of the Radius:
  • So 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 Cosine of half the Angle required.
The Base subtended is 40 gr. 00 m.
 
The two Sides including the Angle 38 30
70 00
 
The Sum of the three Sides 148 30
 
The Half-Sum of these three 74 15
The Difference between this and the Base 34 15

Here for the Square of the Radius we take 20.0000000, to this we add 9.9833805 the Sine of 34 gr. 15 min. and 9.7503579 the Sine of 34 gr. 15 min. which make 39.7337384.

Then for the Rectangle of the Sides, we add 9.7941495 the Sine of 38 gr. 30 min. and 9.9729858, the Sine of 70 gr. which make 19.7671353. This we take out of 39.7337384, and there remains for the Logarithm of the Square 19.9666031, the half thereof 9.9833015 we find to be the Cosine of 15 gr. 47 min. 13 sec. And so the whole Angle required is 31 gr. 34 min. 26 sec.

Or for such Numbers as are to be subtracted, we take them out of the Radius, and write down their Complements, and then add them to gether with the rest, the manner of the Work in either way will be such as followeth.

40 gr. 00 m.  
38 30 9.7941495   2058505
70 00 9.9729858   270142
 
148 30 19.7671353  
 
74 15 9.98318905   9.98318905
34 15 9.7503570   9.7503570
  20.0000000  
 
 
  39.7337384  
  19.9666031 gr. m. sec. 19.9666031
  9.9833015 15 47 13 9.9833015
  31 34 26  

In the like manner we may find the Angle PZS to be 130 gr. 3 m. 11 sec. and the Angle ZSP 30 gr. 28 m. 11 sec.

20. To find a Side by knowing the three Angles.

If for either of the Angles next the Side required, we take the Complement to 180 gr. these Angles will be turned into Sides, and the Sides into Angles. Then may the work be the same as in the former Proposition.

As in the Triangle ZPS, knowing the Angle ZPS to be 31 gr. 34 m. 26 sec. PZS 130 gr. 3 m. 11 sec. and ZSP 30 gr. 28 m. 11 sec. if it were required to find the Side ZS opposite to the Angle ZPS, I would take 130 gr. 3 m. 11 sec. out of 180 gr. the Remainder will be 49 gr. 56 m. 49 sec.

Then, as if I had a Triangle of three known Sides, one of 31 gr. 34 m. 26 sec. another of 30 gr. 28 m. 11 sec. and the third of 49 gr. 56 m. 49 sec. I would seek the Angle opposite to the first of these Sides by the last Proposition.

So the Angle which is thus found would be the Side which is here required.

Thus here the Angle opposite is 31 34 26  
 
 
The lesser of the next Angles 30 28 11 9.7050790
The Complement of the other 49 56 49 9.8839153
 
 
The Sum of these three 111 59 26  
The Half-Sum 55 59 43 9.9185490
The Diff. from the opp. Angle 25 25 17 9.6164170
The Sum of double the Radius and   20.0000000
 
The Sines of Half-Sum and Difference is   39.5349660
Take hence the Sines of the next Angles   19.5889943
 
There remains for the Square   19.9459717
The half whereof is   9.9729858

the Cosine of 20 gr. 00 m. and so the Side required, 40 gr. 00 m.

The other Sides may be found in the same sort; but when we know either three Sides and one Angle, or three Angles and one Side, the rest may be found more readily by the 17 or 18 Proposition.

21. To find a Side by the other two Sides and the Angle comprehended.

This and the Proportion following are best resolved by reducing the oblique-angle Triangles given, into two Rectangles.

Two Rectangle Triangles

As in the Triangle ZPS, having ZP 38 gr. 30 m. PS 70 gr, 00 m. and ZPS 31 gr. 34 m. 26 sec. to find the Side ZS.

In that we have ZP and ZPS, we may suppose a Perpendicular ZR to be let down from the Angle at Z upon the greater Side PS: So if ZPS the Angle given be less tahn 90 gr. it will fall within the Triangle; if more thzan 90 gr. it will fall without the Triangle, upon the Side produced, and divide the Triangle given into two Rectangles ZRS and ZRP. Wherein,

  1. We may find the quantity of this perpendicular by the first Proposition of Spherical Triangles.
  2. We may find the Side PR either by the second of tenth, or rather by the eleventh Proposition: which Side PR will give the Side RS.
  3. Having ZR and RS, we may find the Base ZS, by the fourth Proposition, as I shewed in the Use of the Sector.

But here for variety I will shew how the same may be done at two Operations, both in this and the rest of the Cases following, without knowing the quantity of the perpendicular.

  1. As the Radius or Sine of ZRP 90 00 00 10.0000000
     
    To the Cosine of the Angle ZPR 31 34 26 9.9304233
    So the Tangent of the Side ZP 38 30 00 9.9006052
     
    So the Tangent of the Ark PR 34 07 30 19.8310175
  2. As the Cosine of PR 03 07 30 9.9179342
    To the Cosine of ZP 38 30 00 9.9006052
     
      243899
     
    So the Cosine of RS 35 52 30 9.9086438
    To the Cosine of ZS 40 00 00 9.8842539

22. To find a Side by knowing the other two Sides and one Angle next to the Side required.

As in the Triangle ZPS, having ZP, 38 gr. 30 m. and ZS 40 gr. 00 m. and ZPS, 31 gr. 34 m. 26 sec. to find the Side PS.

  1. Find the Ark PR by the eleventh Proposition as before
  2. As the Cosine of PZ 38 30 00 9.8935443
    To the Cosine of PR 34 07 30 9.9179342
     
      243899
     
    So the Cosine of ZS 40 00 00 9.8842539
    To the Cosine of SR 35 52 30 9.9086438

23. To find a Side by knowing one Sides and the two Angle next to the second Side.

As in the Triangle ZPS, having ZP 38 gr. 30 m. ZPS 31 gr. 34 m. 26 sec. and ZSP 30 gr. 28 m. 11 sec. to find the Side PS.

  1. Find the Ark PR as before.
  2. As the Tangent of ZSP 38 28 11 9.7696236
    To the Tangent of ZRS 31 34 26 9.7885746
     
      189510
     
    To the Sine of PR 34 07 30 9.7489617
    To the Sine of SR 35 52 30 9.7679127

24. To find a Side by knowing two Angle and the Side inclosed by them.

As in the Triangle ZPS, having ZP 38 gr. 30 m. ZPS 31 gr. 34 m. 26 sec. and PZS 130 gr. 3 m. 11 sec. to find the Side ZS.

  1. As the Cosine of PZ 38 30 00 9.8935443
     
    Is the Radius   90 00 00 10.0000000
    So the Cotangent of ZPS 31 34 26 10.2114253
     
    So the Tangent of PZR 64 18 50 10.3178810
  2. As the Cosine of SZR 65 44 22 9.6137228
    To the Cosine of PZR 64 18 50 9.6369311
     
      232083
     
    So the Tangent of PZ 38 30 00 9.9006052
    To the Tangent of ZS 40 00 00 9.9238135

25. To find an Angle by knowing the other two Angles and one Side next to the Angle required.

As in the Triangle ZPS, having ZP 38 gr. 30 m. ZPS 31 gr. 34 m. 20 sec. and ZSP 30 gr. 28 m. 11 sec. to find the Angle PZS.

  1. Find the Angle PZR as before.
  2. As the Cosine of ZPS 31 34 26 9.9304223
    To the Cosine of ZSP 30 28 11 9.9354554
     
      50331
     
    So the Sine of PZR 64 18 50 9.9548122
    To the Sine of SZR 64 44 21 9.9598453

27. To find an Angle by knowing tow Sides and the Angles contained by them.

As in the Triangle ZPS, having ZP 38 gr. 30 m. PS 70 gr. and ZPS 31 gr. 34 m. 26 sec. to find the Angle ZSP.

  1. Find the Ark PR as before.
  2. As the Sine of SR 35 52 30 9.7679127
    To the Sine of PR 34 07 30 9.7489617
     
      189510
     
    As the Tangent of ZPS 31 34 26 9.7885746
    To the Tangent of ZSP 30 28 11 9.7696236

28. To find an Angle by knowing two next Sides, and one of the other Angles.

As in the Triangle ZPS having ZP 38 gr. 30 m. ZS 40 gr. and ZPS 31 gr. 34 m. 26 sec. to find the Angle PZS.

  1. Find the Angle PZR as before.
  2. As the Tangent of ZS 40 00 00 9.9238135
    To the Tangent of ZP 38 30 00 9.9006051
     
      232083
     
    So the Cosine of PZR 64 18 50 9.6369311
    To the Cosine of SZR 65 44 21 9.6137228

These 28 Cases are those which I set down in the Use of the Sector, and all that are commonly required in a Spherical Triangle. I will here add two more, to shew how that which is found before by the 22, 23, 26, and 28 Propositions may sometimes be found more easily, viz.

29. To find a Side, by knowing the other two Sides, and their opposite Angle.

As in the Triangle ZPS, having PS 70 gr. and PZS 130 gr. 3 m. 11 sec. together with ZS 40 gr. and ZPS 31 gr. 34 m. 26 sec. to find the third Side ZP.

  • As the Sine of half the Difference of the Angles given, To the Sune of half the Sum of those Angles:
  • So the Tangent of half the Difference of the Sides given, To the Tangent of half the Side required.

30. To find an Angle, by knowing the other two Angles, and their opposite Sides.

As in the Triangle ZPS, having the former parts PS, PZS, ZS, and ZPS, to find the third Angle ZSP.

  • As the Sine of half the Difference of the Sides given, To the Sine of half the Sum of those Sides:
  • So the Tangent of half the Difference of the Angles given, To the Cotangent of half the Angle required.
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