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THE Triangle which I shall make use of in the Solution of the severall Cases appertaining to an Oblique-angled plain Triangle shall be this following, C D B, in which
| parts | |||
| D B, the Base, | contains | 335 | |
| C B, the longer Side, | 271 | ||
| D C, the shorter Side, | 100 | ||
| And | deg. | m. | |
| C, the obtuse Angle, | contain | 122 | 00 |
| D, the 2 acute Angles, | 43 | 20 | |
| B, the 2 acute Angles, | 14 | 40 |
DRaw a right Line D B representing the Base of your Triangle, which, by help of your Scale of equal parts, make to contain 335. Then upon the Point D, with the distance of 60 degr. of your Line of Chords, describe the Arch k l, and from your Chords take 43 degr. 20 min. the quantity of the Angle at D, and set it upon the Arch∣line from l to k, drawing the Line C D. And because your other given Side B C contains 271 parts, take 271 out of your Line of equal parts, and setting one foot in B, with the other describe the Arch m n, crossing the for∣mer Arch k l in the Point C: then draw the Line C B. So shall you have constituted the Triangle C D B. Lastly, be∣cause it is the Angle at C that is required, take 60 degr. of your Chords, and upon C describe the Arch g h, and taking the distance between g and h, apply it to your Line of Chords,
and you shall finde it to reach from the beginning thereof be∣yond the end of the Line; wherefore take 90 degr. the whole Line, and set that distance from g to o; then take the remainder of the Arch o h, and measure that upon your Chord, and you shall finde it to contain 32 degr. which added to 90 degr. make 122 degr. and that is the quantity of the An∣gle at C, which was required.
The Analogie or Proportion is,
As the Logarithm of B C is to the Sine of D,
So is the Logarithm of D B to the Sine of C.
DRaw a right Line, as D B, containing 335 of your Scale of equal parts, which shall be the Base of your Triangle. Then with 60 degr. of your Line of Chords, upon the Point D describe the Arch k l; and because the given Angle at D contains 43 degr. 20 min. take 43 degr. 20 min. from your Line of Chords, and set it from l to k, drawing the Line D k. Again, because the given Side D C contains 100, set 100 of your Line of equal parts from D to C; then drawing a right Line from C to B, you shall by that means find the oblique-angled Triangle C D B. Lastly, being the other two Angles at B and C are to be found, with 60 degr. of your Chord on the Point B describe the Arch e f; also upon the Point C describe the Arch g o h. Then if you take the distance be∣tween e and f in your Compasses, and measure it upon your Line of Chords, you shall finde it to contain 14 degr. 40 min. And that is the quantity of the Angle at B. Then being the Angle at C, which is also required, is obtuse, and contains a∣bove 90 degr. take 90 degr. out of your Line of Chords,
The Analogie or Proportion is,
As the Log. of the Sum of the two Sides given, C D and C B, is to the difference of those Sides,
So is the Tang. of half the Sum of the two unknown Angles, C and B, to the Tangent of half their difference.
DRaw a right Line C D, containing 100 of your Line of equal parts. Then the other side of the Triangle being 271, take 271 out of your Scale of equal parts, and setting one foot of the Compasses in C, with the other describe the Arch r s. Also take the length of your Base 335 out of your Line of equal parts, and setting one foot of the Compasses in
D, with the other describe the Arch p q, crossing the former Arch r s in the Point B. Then draw the Lines C B and D B, and they shall form your Triangle C D B. Lastly, because the Angle at B is sought, take 60 degr. of your Line of Chords, and setting one foot of the Compasses in B, with the other describe the Arch e f. Then take the distance between e and f, and measure it upon the Line of Chords, and you shall find it to contain 14 degr. 40 min. which is the quantity of the Angle at B. And in the same manner might any of the other An∣gles at C or D have been found.
The Analogie or Proportion is,
As the Log. of the greater side D B is to the Sum of the other two Sides, D C and C B,
So is the difference of the two Sides to a fourth Sum.
Which fourth Sum being taken from the Base, will leave another number, the half whereof will be the place in the Base where a Perpendicular let fall from the obtuse Angle would fall upon the Base: and so the oblique Triangle is re∣duced into two right-angled, and may be resolved by the Precepts of right-angled Triangles.
IN this Case, where the three Angles are given, and none of the Sides, you are to take notice, that the Sides cannot be absolutely found themselves, but the Proportionality of them. Wherefore
Draw a Line, as D B, of any length, and taking 60 degr. of your Line of Chords, set one foot of the Compasses upon D, and with the other describe the Arch l k; also set one foot in B, and with the other describe the Arch e f. Then, be∣cause the Angle at D contains 43 degr. 20 min. take 43 degr.
DRaw a right Line, as C B, containing 271 of your Line of equal parts, and on the Point C, with 60 degr. of your Line of Chords, describe the Arch g o h. Then, because the given Angle C contains 122 degr. it being 32 degr. above 90 degr. first take 90 degr. from off your Line of Chords, and set it upon the Arch from g to o; and then take 32 degr. from your Chord also, and set them upon the same Arch from o to h, and draw the Line C h: then take 100, the length of the other given Side, and set it from C to D, and draw the Line D B,
which will contain 335 of your Scale of equal parts. And that is the length of the Base D B, which was required.
The Analogie or Proportion.
You must first finde the two Angles at D and B. Then make choice of one of the Sides, as C D, to work your Pro∣portion by. Then
As the Sine of B is to the Sine's Complement of C,
So is the Log. of C D to the Log. of D B.