As radius

90°,00' 10,00000

To the tangent of BCA = 35°,00′ = 9,84523

So is the base AC

= 440

= 2,64345

To the perpendicular AB 308 = 2,48868

Again, if the hypothenuse be made radius, and the angle at B be made the centre, the given base AC will become the sine of the angle at B; from which statement is obtained the following proportion;

As the sine of the angle at B 55°,00' 9,91336

The hypothenuse may also be discovered arithmetically, as was already shown, by adding together the squares of the given base, and the perpendicular when found, and extracting the square root of the sum, which will be the hypothenuse required.

CASE IV. Fig. 9, Plate 3.

Given the hypothenuse and the angles. Required the base and the perpendicular.

In the plane triangle ABC, right-angled at A, are given the hypothenuse BC= 537, and the angle at B 55°,00′, consequently the angle at C35°,00'; required the two sides AB and AC.

Supposing an arch of a circle to be described from the point C as a centre, with the radius CB, the perpendicular AB will become the sine of the angle BCA; whence arises the proportion :

As

Again, to find the base AC, with the hypothenuse for

Or, when the perpendicular AB was found as above, another proportion might have been employed, in which AB being radius, the base AC would have become the tangent of the angle at B;

As radius

90°,00' 10,00000

To the tangent of the angle at B = 55°,00' 10,15477 So is the perpend. AB

To the base AC

308

= 2,48868

be

The hypothenuse being given, and one of the sides forming the right angle being found, the other side may discovered as before, arithmetically, by subtracting the square of the side found from that of the hypothenuse, and extracting the square root of the remainder, which will be the side required.

It was already said, that proportion in logarithms is performed

formed by adding together those of the 2d and 3d terms, and subtracting from their sum the logarithm of the 1st term: but the same effect will be produced if, instead of subtracting the log. of the 1st term, we add what is called its arithmetical complement to the sum of the 2d and 3d terms. This arithmetical complement is obtained by subtracting the given logarithm from 10,00000, and thus procuring the reciprocal logarithm of the given number: for example, the logarithm of the natural number 248 being 2,39445, its arithmetical complement, or the difference between this logarithm and 10,00000, will be 7,60555. Were it, therefore, required to find a fourth proportional to any three numbers, as 248, 350, and 482, we would to the logarithms of the second and third terms add the arithmetical complement of the logarithm of the first term, when the sum (subtracting from it 10,0000) would be the logarithm of the fourth term required.

The preceding cases comprehend all the varieties of rightangled plane trigonometry: the following are those belonging to oblique-angled triangles, or such as have no right angle.

CASE I. Fig. 10, Plate 3.

Two angles, and a side opposite to one of them, being given, to find the other angle and sides.

In the oblique-angled plane triangle ABC, are given the angle BAC 42°, 30, the angle ACB=56°, 30, and the side AB 144; required the remaining angle ABC, and the sides BC and AC.

=

By the 7th Prop. of Geometry it was shown, that all the angles of any plane triangle are equal to two right angles, or equal to 180 degrees: if, therefore, from 180 we subtract the sum of the two given angles 42°, 30' + 56, 50=99 degrees, the remainder, 81°, 00', is the measure of the required angle at B.

It was shown in Prop. 2 of Trigonometry, that the sides of any plane triangle are one to another in the proportion of the sines of the respectively opposite angles: hence we have this proportion, as the sine of the angle at C to the sine of the angle at A, so is the log. of the side AB, opposite to the angle at C, to the log. of the side BC, opposite to the angle at A.

Sine of ACB 56,30 9,92111

Or, by taking the arithmetical complement of 9,92111,

the sine of 56°, 30' = 0,07889.

Arith. comp. of sine of 56, 30' 0,07859'
Log of sine of 42,30 = 9,82968 add

Again, to find the remaining side AC, we may employ either of the other sides with its opposite angle, thus;

Two sides, and an angle opposite to one of them, being given, to find the remaining side and angles.

In the oblique-angled triangle ABC, are given the side AB 144, the side BC posite to AB56o, 30';

and the angles at A and B.

116,6, and the angle at C, oprequired the remaining side AC,

This case being the converse of the preceding, we have this proportion; as the side AB to the side BC, so is the sine of the angle at C, opposite to AB, to the sine of the angle at A, opposite to BC.

Log. of AB 144 2,15836