The Construction of the Wonderful Canon of Logarithms/Trigonometrical Propositions

by John Napier

SOME VERY REMARKABLE
PROPOSITIONS FOR THE
solution of spherical triangles
with wonderful ease.
To solve a spherical triangle without dividing it
into two quadrantal or rectangular triangles.

Prop. 1.

Iven three sides, to find any angle.

And conversely,
Prop. ‍2.Given three angles, to find any side.

This is best done by the three methods explained in my work on Logarithms, Book II. chap. vi. sects. 8, 9, 10.

Prop. ‍3.Given the side A D, & the angles D & B, to find the side A B.

Multiply the sine of A D by the sine of D; divide the product by the sine of B, and you will have the sine of A B,

4.Given the side A D, & the angles D & B, to find the side B D.

Multiply radius by the sine of the complement of D; divide by the tangent of the complement of A D, and you will obtain the tangent of the arc C D: then multiply the sine of C D by the tangent of D; divide the product by the tangent of B, and the sine of B C will result: add or subtract B C and C D, and you have B D.

5.Given the side A D, & the angles D & B, to find the angle A.

Multiply radius by the sine of the complement of A D; divide by the tangent of the complement of D, and the tangent of the complement of C A D will be produced; whence we have C A D itself. Similarly multiply the sine of the complement of B by the sine of C A D; divide by the sine of the complement of D, and the sine of B A C will be produced; which being added to or subtracted from C A D, you will obtain the required angle B A D.

6.Given A D, & the angle D with the side B D, to find the angle B.

Multiply radius by the sine of the complement of D; divide by the tangent of the complement of A D, and the tangent of C D will be produced; its arc C D subtract from, or add to, the side B D, and you have B C: then multiply the sine of C D by the tangent of D; divide the product by the sine of B C, and you have the tangent of the angle B.

7.Given A D, & the angle D with the side B D, to find the side A B.

Multiply radius by the sine of the complement of D; divide the product by the tangent of the complement of A D, and the tangent of C D will be produced; its arc C D subtract from, or add to, the given side B D, and you have B C. Then multiply the sine of the complement of A D by the sine of the complement of B C; divide the product by the sine of the complement of C D, and the sine of the complement of A B will be produced; hence you have A B itself.

Given A D, & the angle D with the side B D, to find the angle A.

This follows from the above, but the problem would require the “Rule of Three” to be applied thrice. Therefore substitute A for B and B for A, and the problem will be as follows:—

Given B D & D, with the side A D, to find the angle B.

This is exactly the same as the sixth problem, and is solved by the “Rule of Three” being applied twice only.

8.Given A D, & the angle D with the side A B, to find the angle B.

Multiply the sine of A D by the sine of D; divide the product by the sine of A B, and the sine of the angle B will be produced.

9.Given A D, & the angle D with the side A B, to find the side B D.

Multiply radius by the sine of the complement of D, divide the product by the tangent of the complement of A D, and the tangent of the arc C D will be produced. Then multiply the sine of the complement of C D by the sine of the complement of A B, divide the product by the sine of the complement of A D, and you have the sine of the complement of BC, Whence the sum or the difference of the arcs B C and C D will be the required side B D.

10.Given A D, & the angle D with the side A B, to find the angle A.

Multiply radius by the sine of the complement of A D, divide the product by the tangent of the complement of D, and the tangent of the complement of C A D will be produced, giving us C A D. Again, multiply the tangent of A D by the sine of the complement of C A D, divide the product by the tangent of A B, and the sine of the complement of B A C will be produced, giving B A C. Then the sum or difference of the ares B A C and C A D will be the required angle B A D.

11.Given A D, & the angle D with the angle A, to find the side A B.

Multiply radius by the sine of the complement of A D, divide the product by the tangent of the complement of D, and you have the tangent of the complement of C A D; C A D being thus known, the difference or sum of the same and the whole angle A is the angle B A C. Multiply the tangent of A D by the sine of the complement of C A D; divide the product by the sine of the complement of B A C, and you will have the tangent of A B.

12.Given A D, & the angle D with the angle A, to find the third angle B.

Multiply radius by the sine of the complement of A D, divide the product by the tangent of the complement of D, and the sine of the complement of B will be produced, from which we have the angle required.

Given A D, & the angle D with the angle A, to find the side B D.

This follows from the above, but in this form the problem would require the: “Rule of Three” be three times applied. Therefore substitute A for D and D for A, and the problem will be as follows:—

Given A D & the angle A with the angle D, to find the side B A.

This is the same throughout as problem 11, and is solved by applying the “Rule of Three” twice only.

The use and importance of half-versed
Sines.

1.GIven two sides & the contained angle, to find the third side.

From the half-versed sine of the sum of the sides subtract the half-versed sine of their difference; multiply the remainder by the half-versed sine of the contained angle; divide the product by radius; to this add the half-versed sine of the difference of the sides, and you have the half-versed sine of the required base.

Given the base and the adjacent angles, the vertical angle will be found by similar reasoning.

2.Conversely, given the three sides, to find any angle.

From the half-versed sine of the base subtract the half-versed sine of the difference of the sides-multiplied by radius; divide the remainder by the half-versed sine of the sum of the sides diminished by the half-versed sine of their difference, and the half-versed sine of the vertical angle will be produced.

Given the three angles, the sides will be found by similar reasoning.

3.Given two arcs, to find a third, whose sine shall be equal to the difference of the sines of the given arcs.

Let the arcs be 38° 1’ and 77°. Their complements are 51° 59’ and 13°. The half sum of the complements is 32° 29’, the half difference 19° 29’, and the logarithms are 621656 and 1098014 respectively. Adding these, you have 1719670, from which, subtracting 693147, the logarithm of half radius, there will remain 1026523, the logarithm of 21°, or thereabout. Whence the sine of 21°, namely 358368, is equal to the difference of the sines of the arcs 77° and 38° 1’, which sines are 974370 and 615891, more or less.

4.Given an arc, to find the Logarithm of its versed sine.[a]

Let the arc be 13°; its half is 6° 30’, of which the logarithm is 2178570. From double this, namely 4357140, subtract 693147, and there will remain 3663993. The arc corresponding to this is 1° 28’, and the number put for the sine is 25595; but this is also the versed sine of 13°.⁂

5.Given two arcs, to find a third whose sine shall be equal to the sum of the sines of the given arcs.

Let the arcs be 38° 1’ and 1° 28’; their sum is 39° 29’ and their difference 36° 33’, also the half sum is 19° 44’ and the half difference 18° 16’. Wherefore add the logarithm of the half sum, viz. 1085655, to the logarithm of the difference, viz. 518313, and you have 1603968; from this subtract the logarithm of the half difference, namely 1160177, and there will remain the logarithm 443791, to which correspond the are 39° 56’ and sine 641896. But this sine is equal, or nearly so, to the sum of the sines of 38° 1’ and 1° 28’, namely 615661 and 25595 respectively.

6.Given an arc & the Logarithm of its sine, to find the arc whose versed sine shall be equal to the sine of the given arc.

Let the arc be 39° 56’, to which corresponds the logarithm 443791, the sine being unknown. To the logarithm 443791 add 693147, the logarithm of half radius, and you have 1136938. Halve this logarithm and you have 568469. To this corresponds the arc 34° 30’, which being doubled gives 69° for the arc which was sought. This is the case since the sine of 39° 56’ and the versed sine of 69° are each equal, or nearly so, to 641800.

[b] Of the spherical triangle A B D, given the sides & the contained angle, to find the base.

LEt the sides be 34° and 47°, and the contained angle 120° 24’ 49″. Half the contained angle is 60° 12’ 24½″, and its logarithm 141766. To the double of the latter, namely 283533, add the logarithms of the sides, namely 581260 and 312858, and the sum is 1177651. This sum is the logarithm of half the difference between the versed sine of the base and the versed sine of the difference of the sides; it is also the logarithm of the sine of the are 17° 56’, which are we call the “second found,” for that which follows is first found,

Halve the difference of the sides, namely 13°, and you have 6° 30’, the logarithm of which is 2178570. Double the latter and you have 4357140 for the logarithm of the half-versed sine of 13°; it is also the logarithm of the sine of the arc 0° 44’, which are we call the “first found.”

The sum of the two arcs is 18° 40’, the half sum 9° 20’, and their logarithms 1139241 and 1819061 respectively. Also the difference of the two arcs is 17° 12’, the half difference 8° 36’, and their logarithms 1218382 and 1900221 respectively.

Now add the logarithm of the half sum, namely 1819061,

either	or
to the logarithm 1218382, and the sum will be 3037443; from this subtract the logarithm 1900221 and there will remain 1137222.	to the logarithm of the complement of the half difference, namely 11307, and the sum will be 1830368; from this subtract 693147 and there will remain 1137221.

Halve the latter and you have the logarithm 568611, to which corresponds the arc 34° 30’, and double this arc is the base required, namely 69°.

Conversely, given the three sides, to find any angle. The solution of this problem ts given in my work on. Logarithms, Book II. chap. vi. sect. 8, but partly by logarithms and partly by prosthapharesis of arcs.

It ts to be observed that in the preceding and following problems there ts no need to discriminate between the different cases, since the form and magnitude of the several parts appear in the course of the calculation.

Another direct converse of the preceding problem follows.—

[Given the sides and the base, to find the vertical angle.]

HAlve the given base, namely 69°, and you have 34° 30’, the logarithm of which is 568611. Double the latter and you have 1137222; corresponding to this is the arc 18° 42’, which note as the second found.

As before, take for the first found the arc 0° 44’, corresponding to the logarithm 4357140.

The complements of the two arcs are 89° 16’ and 71° 18’; their half sum is 80° 17’, and its logarithm 14449; their half difference is 8° 59’, and its logarithm 1856956. Add these logarithms and you have 1871405; subtract 693147 and there remains 1178258. The arc corresponding to this logarithm is 17° 56’, which arc we call the third found.

From the logarithm of the third found, subtract the logarithms of the given sides, namely 581260 and 312858, and there remains 283533; halve this and you have 141766 for the logarithm of the half vertical angle 60° 12’ 24½”. The whole vertical angle sought is therefore 120° 24’ 49”.

Another rule for finding the base by prosthapharesis.—

[Given the sides and vertical angle, to find the base.]

NOte the half difference between the versed sines of the sum and difference of the sides, and also the half-versed sine of the vertical angle. Look among the common sines for the values noted, and find the arcs corresponding to them in the table. Then write for the second found the half difference of the versed sines of the sum and difference of these arcs.

Also, as before, take for the first found the half-versed sine of the difference of the sides.

Add the first and second found, and you will obtain the half-versed sine of the base sought for.

Conversely—[given the sides and the base, to find the vertical angle.]

The first found will be, as before, the half-versed sine of the difference of the sides.

From the half-versed sine of the base subtract the first found and you will have the second found.

Multiply the latter by the square of radius; divide by the half difference between the versed sines of the sum and difference of the sides, and you have as quotient the half-versed sine of the vertical angle sought for.

[c]Of five parts of a spherical triangle, given the three intermediate, to find the two extremes by a single operation. Or otherwise, given the base and adjacent angles, to find the two sides.

(*)

OF the angles at the base, write down the sum, half sum, difference and half difference, along with their logarithms.

Add together the logarithm of the half sum, the logarithm of the difference, and the logarithm of the tangent of half the base; subtract the logarithm of the sum and the logarithm of the half difference, and you will have the first found.

Then to the logarithm of the half difference add the logarithm of the tangent of half the base; subtract the logarithm of the half sum, and you will have the second found.

Look for the first and second found among the logarithms of tangents, since they are such, then add their arcs and you will have the greater side; again subtract the less arc from the greater and you will have the less side.

Another way of finding the sides.

ADd together the logarithm of the half sum of the angles at the base, the logarithm of the complement of the half difference, and the logarithm of the tangent of half the base; subtract the logarithm of the sum and the logarithm of half radius, and you will have the first found.

Again, add together the logarithm of the half difference, the logarithm of the complement of the half sum, and the logarithm of the tangent of half the base; subtract the logarithm of the sum and the logarithm of half radius, and you will have the second found.

Proceed as above with the first and second found, and you will obtain the sides.

Another way of the same.

MUltiply the secant of the complement of the sum of the angles at the base by the tangent of half the base.

Multiply the product by the sine of the greater angle at the base, and you will have the first found.

Multiply the same product by the sine of the less angle, and you will have the second found.

[d]

Then divide the sum of the first and second found by the square of radius, and you will have the tangent of half the sum of the sides.

Also subtract the less from the greater and you will have the tangent of half the difference of the sides.

Whence add the arcs corresponding to these two tangents, and the greater side will be obtained; subtract the less arc from the greater and you have the less side.

Of the five consecutive parts of a spherical triangle, given the three intermediate, to find both extremes by one operation and without the need of discriminating between the several cases.

(*)

OF the angles at the base, the sine of the half difference is to the sine of the half sum, as the sine of the difference is to a fourth which is the sum of the sines.

And the sine of the sum is to the sum of the sines as the tangent of half the base is to the tangent of half the sum of the sides.

Whence the sine of the half sum is to the sine of the half difference of the angles as the tangent of half the base is to the tangent of half the difference of the sides,

Add the arcs of these known tangents, taking them from the table of tangents, and you will have the greater side; in like manner subtract the less from the greater and the less side will be obtained.

FINIS