If the hypothenuse be assumed equal to the radius, the sides, that is, the base and the perpendicular, will be the sines of their opposite angles. And, if either of the sides be considered as the radius, the other side will be the tangent of its opposite angle, and the hypothenuse the secant of the same angle.

Thus. Let A B C be a right angled plane triangle; if the hypothenuse AC be made radius, the side BC will be the sine of the angle A, and AB the sine of the angle C.-If the side A B be made radius, BC will be the tangent, and AC the secant, of the angle A:-And, if BC be the radius, AB will be the tangent, and AC the secant of the angle C.

For, if we make the hypothenuse AC radius (Fig. 1.), and upon A, as a centre, describe the arch CD to meet A B produced to D; then it is evident that B C is the sine of the arch DC, which is the measure of the angle BAC; and that A B is the co-sine of the same arch :-and if the arch AE be described about the centre C, to meet C B produced to E, then will A B be the sine of the arch A E, or the sine of the angle AC B, and B C its co-sine.

Again, with the extent A B as a radius (Fig. 2.), describe the circle BD; then BC is the tangent of the arch B D, which is evidently the measure of the angle BAC; and AC is the secant of the same arch, or angle.

Lastly, with CB as a radius (Fig. 3.), describe the arch BD; then AB is the tangent of the arch BD, the measure of the angle A CB, and A C the secant of the same arch or angle.

In the computation of right angled triangles, any side, whether given or required, may be made radius to find a side; but a given side must be made radius to find an angle: thus,

To find a Side:

Call any one of the sides of the triangle radius, and write upon it the word radius:-observe whether the other sides become sines, tangents, or secants, and write these words on them accordingly, as in the three preceding figures: then say, as the name of the given side, is to the given side; so is the name of the side required, to the side required.

And, to find an Angle :

Call one of the given sides the radius, and write upon it the word radius: observe whether the other sides become sines, tangents, or secants, and write these words on them accordingly, as in the three foregoing figures; then say, as the side made radius, is to radius; so is the other given side to its name: that is, to the sine, tangent, or secant by it represented.

Now, since in plane trigonometry the sides of a triangle may be considered, without much impropriety, as being in a direct ratio to the sines of their opposite angles, and conversely; the proportion may, therefore, be stated agreeably to the established principles of the Rule of Three Direct, by saying

As the name of a given angle, is to its opposite given side; so is the name of any other given angle to its opposite side.-And, as a given side, is to the name of its opposite given angle; so is any other given side to the name of its opposite angle.

The proportion, thus stated, is to be worked by logarithms, in the following manner; viz.,

To the arithmetical complement of the first term, add the logs. of the second and third terms, and the sum (rejecting 20, or 10 from the index, according as the required term may be a side or an angle,) will be the logarithm of the required, or fourth term.

Remarks.-1. The arithmetical complement of a logarithm is what that logarithm wants of the radius of the Table; viz., what it is short of 10.000000; and the arithmetical complement of a log. sine, tangent, or secant, is what such logarithmic sine, &c. &c. wants of twice the radius of the Tables, viz., 20. 000000.

2. The arithmetical complement of a log. is most readily found by beginning at the left hand and subtracting each figure from 9 except the last significant one, which is to be taken from 10, as thus ;—if the given log. be 2.376843, its arithmetical complement will be 7.623157:-if a given log. sine be 9. 476284, its arithmetical complement will be 10. 523716, and

so on.

3. The arithmetical complement of the log. sine of an arch, is the log. co-secant of that arch ;-the arithmetical complement of the log. tangent of an arch, is the log. co-tangent of that arch; and conversely, in both

cases.

Solution of Right-angled Plane Triangles, by Logarithms.

PROBLEM I.

Given the Angles and the Hypothenuse, to find the Base and the
Perpendicular.

Example.

Let the hypothenuse A C, of the annexed triangle ABC, be 246. 5, and the angle A 53:7:48"; required the base A B, and the perpendicular B C ?

Note. Since there is no more intended, in this place, than merely to show the use of the Tables; the geometrical construction of the diagrams is, therefore, purposely omitted.

By making the hypothenuse AC radius; BC becomes the sine of the angle A, and A B the co-sine of the same angle.-Hence,

Making the base A B radius; B C becomes the tangent of the angle A,

and AC the secant of the same angle.-Hence,

The perpendicular BC being made radius ; the base AB becomes the tangent of the angle C, or co-tangent of the angle A, and the hypothenuse A C the secant of the angle C, or co-secant of the angle A.-Hence,

To find the Perpendicular BC:

As the angle A= 53:7:48 Log. co-secant Ar. compt.

9.903090 2.391817 10.000000

As the angle A53:7:48" Log. co-secant Ar. compt. 9.903090

Is to hypothenuse A C = 246. 5 Log. =
So is the angle A= 53:7:48

To the base AB=147.9=

Log. co-tangent

Log.=

2.391817

9.875063

PROBLEM II.

Given the Angles and One Side, to find the Hypothenuse and the other

Example.

Side.

Let the base AB of the annexed triangle ABC, be 300.5, and the angle A 40:54:40%; required the hypothenuse A C, and the perpendicular B C ?

The hypothenuse A C being made radius; the perpendicular B C will be the sine of the angle A, and the base AB the co-sine of the same angle.

To find the Perpendicular B C :

As the angle A= 40:54:40" Log. co-sine Ar. compt.

Is to the base AB

So is the angle A

300. 5 Log. =

40:54:40" Log. sine.

To the perpendicular BC= 260.4 Log..

The base A B being made radius; the perpendicular B C will be the tangent of the angle A, and the hypothenuse A C the secant thereof.-Hence,

The perpendicular B C being made radius; the base A B will be the tangent of the angle C, or co-tangent of the angle A, and the hypothenuse the secant of the angle C, or co-secant of A.-Hence,

To find the Hypothenuse AC:

As the angle A40:54:40 Log. co-tang. Ar. compt. = 9.937802 Is to the base A B = 300. 5 Log. =

So is the angle A= 40:54:40" Log. co-secant =

2.477845 10. 183833

To the hypothenuse AC = 397.6 = Log. =

2.599480