The geometrical square, with the use thereof in plain and spherical trigonometrie chiefly intended for the more easie finding of the hour and azimuth / by Samuel Foster ...

Of the use of the Square, in Right-lined Triangles.

IF the Proportions be between Tangents and equal parts, then are we to use the equal parts on the sides AB, AG, as also the larger Tangents upon the two other sides of the Square, and then the work will be the same, for form, that was before in Tangents and sines, for the lines on the superficies will carry the parts of either of these Scales to and fro, as they did before the parts of the Scales of the lesser Tangents.

If the Proportions be between sines and equal parts, then are we to make use of the sines inscribed upon the Scales BI, CG, together with the former equal parts, the lines upon the super∣ficies still acting their former parts of carrying from the one to the other.

Examples in these kinds, And first of sines, and equal parts, or Numbers.

SUppose at the two stations DC, I had observed the an∣gles BCA, 30 gr. BDA, 50 gr. and CD the difference of Stations 40 feet, and by these observations, I require to know the altitude AB.

First, I must find the length of the lines CB, or DB, in this Ex∣ample of CB, after this manner, be∣cause BCA is 30 gr. and BD A 50 grad. therefore their difference CBD is 20 gr. Now then, As the sine of CB D 20 gr.

Is to CD 40 feet, So is the sine of BDC or BDA, 50 gr.

[illustration]

To the length of CB required.

To resolve this upon the Square, from C, I count the line of 20 gr. to a, and observe the line there meeting me, then upon the side AC, I count A d 40 equal parts or feet, and thirdly, I reckon C c the third term, which is the sine of 50 gr. and follow the line there meeting me, till it crosse the threed (which was to be applied to b, the intersection of the lines a b,

coming from the first term, and d b rising from the second) at e, and there I find another line concurring, namely, e h, which I follow down to h, and there it shewes in the equal parts A h 89 feet, and 58 centesmes or hundredth parts of a foot, And this is the length of CB, now to get BA, by a second work, I say,

As CB the radius, to BA the sine of BCA 30 gr.

So BC 89 58/100 feet, to BA the altitude in feet.

[illustration]

To perform this Proportion, Upon the Square I take AH, equal to BI the radius, and upon HE, I count H o, equal to C r the sine of 30 gr. thereto applying the threed; Then from A to h, I count the length of CB, that is 89, 58, and so follow the line there arising, up to the threed to s, where I find the line s u, limiting out A u, 44, 79, that is 44 feet, and 79 cen∣tesmes of a foot, and such is the altitude of AB required.

Thus by having the three Angles of a plain Triangle, and one side you may find the two other sides; And by having two sides and an Angle opposite to one of them, you may find the other two Angles and third side, in any Right-lined Trian∣gles whatsoever.

Examples of equal parts, and Tangents.

This kind of work may sufficiently be explained in the solution of this Probleme.

¶By having an Angle and the two sides comprehending it, to find the other Angles.

First, if the Angle comprehended be a Right-angle the work is easie.

ANd here we are to use the Scales of equal parts, with the larger Tangents onely. Suppose then in the Rect-angle Triangle ABC, By having the two sides including the right angle AB 30, BC 20 parts, I would find the angles at A and C, because this Proportion holds

[illustration]

As AB 30, to BC 20

So AB the radius to BC,

The Tangent of BAC.

Therefore upon the Square I count A w 30 equal parts, and follow w f, till it stand even with 20 equal parts counted on the side AB, and laying the threed at f, I find it to cut in the limb of the greater Tangent C y, which is 33 gr. 41 min. And such is the quantity of the angle CAB. And the complement of it 56 gr. 19 min. is the quantity of the angle ACB.

Further more it is to be noted, That if by having the right angle with the two including sides, you would find the sub tending side AC. In this case one of the acute angles must first be sought, and then by the Proportions of sines and equal parts, the side AC may be had.

So also, If by having the distance AB 30 foot, and the angle CAB 33 gr. 41 min. I would know the Height BC, Upon the Square I lay the threed from C to y the Tangent of 33 gr. 41 min. then upon the equal parts I count A w 30, & follow the line rising at w, till it meet with the threed at f, and at f, I find the line f q crossing also, which followed to q, shewes in the limb 20 equal parts for the altitude BC.

By these mixt Proportions of equal parts with sines and Tan∣gents, may all mensurations be performed, as also all conclu∣sions upon the Common Sea-chart, with Mr. Gunters corre∣ctions of it, to make it sufficient for Sea-mens use.

Secondly, in any plain Triangle whatsoever.

The former Probleme may be resolved in general by this Pro∣portion, As the sum of the two sides, is to their difference, So is the Tangent of half the sum of their opposite Angles, to the Tangent of the half difference of those Angles.

AS here, the two given parts, are AB 40 parts, & BC 20, the sum of them is 60, the difference is 20, the angle at B 120 gr. & therefore the sum of the two other 60 gr. the half sum of 30 gr. Now,

[illustration]

As 60 the sum of the sides, is to 20 their difference,

So is the Tangent of 30 gr. the half sum of the angles at A and C, to the Tangent of their half difference.

The best way for the solution of Proportions in this kind is first (as was before admonished in the joynt use of Sines and Tangents) to seek out a Tangent whereon the Radius is in Pro∣portion as the sum of the legs is to the difference of them, which Tangent is ever lesse then the Tangent of 45 gr. or radius, because the difference of the legs is alway lesse then the sum of them. And when the radius is brought in, the Proportion may be absolved upon the lesser Square, which is fitted for the Pro∣portions of Sines and Tangents, in the same manner as was shewed in the like Examples before. And the Proportion will then stand thus.

As the radius, is to this new found Tangent, So is the Tan∣gent of half the sum of the angles, to the Tangent of half their difference.

To make it plain by the former instance, As 60 parts, are to 20; So is the radius, to what Tangent?

Upon the Scale of equal parts, I account A k 60 and on the line there arising I account k x 20, thereto applying the threed, and then I see it cut off in the greater Tangents C 18 gr, 26 min. which is the Tangent sought. And now that the radius is brought in, the next Proportion will be thus,

As the radius, is to the Tangent of 18 gr. 26 min.

So is the Tangent of 30 gr. half the sum of the angles, to what Tangent?

Upon the lesser Square, I take EG for the radius, and count G ζ the tangent of 18 gr. 26 min. thereto applying the threed, then upon ID, I count I x the Tangent of 30 gr. and follow the line thence passing to the threed at σ, where the line σ φ shewes, in the limb the Tangent G φ, which is the Tangent of 10 gr. 54 min. the half difference of the angles required, which added to 30 gr. the half sum, makes the greater angle 40 gr. 54 min. And taken from the same 30 gr. leaveth 19 gr. 6 min. for the lesser angle.

¶By having the three sides of a plain Triangle, to find the Angles.

THe first work here will be to let fall a perpendicular, and to know where it will fall, and so reducing the Triangle to two Rectangles, you may resolve them as Rectangles, either by sines and equal parts, or Tangents and equal parts.

The manner of dividing a Triangle, into two Rectangles, as also to find the place where the perpendicular falls, is shewed by Mr. Gunter in the first Book of his Cross-staff, and the Proportion for the solution of it, is a proportion of equal parts or numbers onely, the manner of which is hereafter shewed in the next use of the Square in numbers or equal parts alone.

Thus farre of the use in Right-lined Triangles.