Side {BD05 Yards given:
Angle BCD 31d. 49m..

This Triangle is made by Prob. 15. of Geometry, in Page 19. 1. For the Angle BDC the Proportion is,

As the Side BD, is to the Sine of the Angle BCD; fo is the Side BC, to the Sine of the Angle BDC required. Or thus, Side BD ..S. BCD: Side BC ..S. BDC. 65 Yards. S. 31d. 49m.:: 106 Yards.. S. 59d. 17m. which Subtract from 18od. com.

:

120d. 43.

Remainder is the Angle BDC Note; The Proportion produceth 59. 17m. for the required Angle: But being Obtufe, you must take it's Supplement to 18od. viz. 120d. 43m. as above is done.

2. Find the third Angle by the 9th of Section I. of this Chapter, in Page 35. then you may find the Side CD by the first Cafe.

This Cafe hath been omitted by moft, the Reafon (I fuppofe) is the doubtfulness of the Required Angle; but if determined (before) to be either Acute or Obtufe, the third Side is limited, and then may be a Cafe as well as any other; and the Proportions may be,

..Side BD: S. CBD

· Side CD.

S. BCD S. 31d. 49m...65Yards: S. 27d. 28m... Yds. 56.88. Or, S. BDC. Side BC :: S. CBD S. 120d 43m... 106Yds.:: S. 27d. 28m... Yds. 56.88 Tenths as before.

..Side CD.

Axiom 3. IN all Plain Triangles; as the Sum of two Sides, is to their Difference; fo is the Tangent of the Half-Sum of their two oppofite Angles, to the Tangent of the half-Difference of the faid two oppofite and unknown Angles. Then,

Add the half Difference of the Angles to their half Sum, finds the greater Angle; and fubtract the half Difference from the half Sum, finds the lesser Angle.

Prob. VII. Cafe 4 and 5. Two Sides and their contained Angle given; to find either of the other Angles, and the third Side.

Example. In the Oblique Triangle BCD. Plate 2. Fig. 7.

BC 109

Sde { B

BD

76 Leag.

Angle CBD iord. 30m.

Angle { BDC or

{ Angle {

} given: {

BCD and req. Side CD

This Triangle is made by Problem 16. of Geometry, in Page 19.

1. For the Angle BDC, and BCD, the Operation is, 109 The three Angles

Side BD
BC

180d. oom.

76 Subtract the given Angle CBD-- 101d. 30m. Sum is 78d. 30m. The two oppofite Angles { Sum is 39d. 15m.

Sum of Sides 185
Their Diff.

33

Then, As the Sum of the Sides BC and BD, is to their Difference; fo is the Tangent of half the Sum of the Angles BDC and BCD, to the Tangent of half their Difference. Or thus, Sum BC & BD.. Diff. BC & BD:: T. Sum Angle. T. Diff. : 33 Leagues T. 39d. 15m... T.8d.17m. 185 Leagues

The half Diff. of the AnglesAdded, is the greater Angle Subtract, is the leffer Angle

:

08d. 17m.

47d. 32m. BDC2
30d. 58m. BCDS

req.

2. The Proportion for the Side CD, (by the firft Cafe of Oblique Triangles) may be this:

S. BCD

Side B D :: S. CB D

.. Side CD.

S. 30d. 58m... 76 Leag. S. 101d. 30m... Leag 144.7 tenths,

Axiom 4. FRom the half Sum of the three Sides, fubtract each Side (but firft that Side oppofite to the Angle required, then the reft) feverally, noting the Remainders. Then, As the Product of the half Sum of the Sides, and firft Remainder, is to the Product of the other two Remainders; fo is the Square of Radius, to the Square of the Tangent of half the Angle oppofite to that firft Remainder.

Prob. VIII. Cafe 6. Three Sides given, to find an Angle.

Example. In the Triangle BCD. Plate 2. Fig. 8.

The Side

BC- - 105

SBDC)

BD- 85 Feet given: Angle BCD req.

CD-- 500

CBDS This Triangle is made by Prob. 17. of Geometry, in P. 19 and 20. The Operation for the Angle CBD. Feet.

This Axiom finds an Angle at one Operation, yet not being applicable to the inftrumental way of working Proportions, you have this fourth Axiom in other Terms; which finds an Angle at two Proportions, and may be wrought both Inftrumentally and Logarithmically.

Axiom 4. Ufeful when three Sides of a Triangle are given; to find an Angle.

As the longeft Side, is to the Sum of the two fhorteft; fo is the Difference of the two fhorteft, to the Difference of the Segments of the Bafe or longest Side.

Note; Let fall a Perpendicular (from the Angle oppofite) to the longest Side, which divideth it into two Segments; and the Oblique Triangle into two Right-Angled-Triangles.

As in the aforefaid Triangle BCD. Plate 2. Fig. 8.

Let fall the Perpendicular DA, which makes the Segments of the Bafe to be BA and AC, and the two Right-Angled-Triangles BAD and CAD, and the Difference of the Segments BE.

1. To find BE the Difference of the Segments of the Bafe.

Shorteft Sides

{

BD

CD

Added, is the Sum of the two shortest Sides
Subtracted, is their Difference

Then, as the Side BC, is. to the Sum of BD and CD; fo is the Difference of BD and CD, to BE the Difference of the Segments BA and AC. Or thus.

Side BC. Sum BD& CD:: Diff. BD & CD .. BE the Diff. of Seg. 135 Feet: 35 Feet

105 Ft.

The Side BC 105 Feet. Diff.Segments BE 45

Feet

.45 Feet.

30 AC the leffer Sment.

2. The Angles BCD or CBD, may be found by the 4th Cafe of Right-Angled-Triangles, in Page 39. Thus.

Hypot. BD. Radius: Leg AB. S. ADB.
85 Feet

.. S. 9od.:: 75 Feet S. 61d. 56m.
god. oom.

Which fubtracted from

Remainder is the Angle CBD --. 28d. 04m. as before.

Thus much for Plane Triangles; and to compleat Trigon metry, Spheric fhould be next. But I think the Application of this before the Doctrine of that, moft conducible to the

Learner's Advantage: Therefore will defcend to the neceffary Ufes of Plane Trigonometry in Plane and Mercator's Sailing, which will make way for Spheric Trigonometry.

CHAP. III. Plane Trigonometry applied in Problems of Sailing by the Plane Sea-Chart, commonly called PlaneSailing.

AND that nothing may be wanting for the Accomplishment of Navigation, we will begin now with the Gregorian Calendar, and then the ufe of the Plane Chart, before we apply Plane Trigonometry to Plane-Sailing.

Section I. The common Notes of the Gregorian Calendar, or New-Stile, to find the Prime, Epact, Dominical Letter,Eafter Day, the Moon's Age, Southing, and Time of High-water.

Problem I. To find the Golden Number, Cycle of the Sun, and Roman Idiction.

Definition 1. THE Golden Number or Prime, is a Circular Revolution of 19 Years; in which Space of Time (it has been fupppofed) the Sun and Moon finish all their Variety of Afpects; by this we find the Epact, and confequently whatever thereon depends.

2. The Cycle, or Circle of the Sun, maketh its Revolution in 28 Years; in which Time all the Variety of Dominical Letters, and Leap-Years expire, and the 29th Year the Circle begins again; which Number affifts in finding the Dominical Letter for any Year, paft, prefent, or to come.

3. Roman Indiction confifteth of 15. Years; for once in 15 Years the fubdu'd Nations were to Pay Tribute to the Romans; a thing now out of Ufe with us.

The Rule out of Mr. Street's Memorial Verfes on the Eclefiaftic and Civil Calendar.

When 1, 9, 3, to the Year hath added been;
Divide by 19, 28, 15.

Example. I would know the Golden Number, Cycle of the

Problem II. To find the Epact until the Year 1799 inclufive. Definition. The Epact is 11 Days the Year of the Moon lacketh of the Sun's Year: the Lunar being 354 Days, and the Solar Year, 365 Days.

Note 1. The Epact never exceedeth 29; alters every Year 11, and is used to find the Moon's Age, and Eafter-Day.

Note 2. When the Golden Number is 1, the Epact is o, conftantly; when 2, then 11, &c.

The RULE is,

1. Find the Golden Number, by Prob. I.

2. Subtract from the Golden Number, what refts divide by 3, and note the Remainder.

3. Multiply the Remainder by 10, and note the Product.

4. To that Product add the Golden Number lefs I; the Sum (if it exceeds not 30) is the Epact; but if it doth, fubtract 30 therefrom, and the Remainder is the Epact.

Example. For the Year 1757, I demand the Epact?

To the Year 1757 add 1, and the Sum is 1758, which divide by 19, the Quotient is 92, and the Remainder is 10; fo that the Golden Number is 10, fubtract 1, there refts 9, which divide by 3, the Quotient is 3, and the Remainder o; which multiply by 10, is 0, to which add 9, equal to the Golden Number lefs 1, the Sum is 9, the Epact for the Year 1757, the fame as the Golden Number lefs 1.

Problem III. To find the Biffextiles, or Leap Years.

The Old or Julian Leap-Years, is every fourth Year, and so called from its leaping a Day more that Year, than in a common Year; for in the common Year any fixed Day of the Month changeth fucceffively the Day of the Week, but in the Leap-Year, it fkips, or leaps over one Day: Thefe Leap-Years (in the New-Stile) are continued exactly in the fame Order and Succeffion as heretofore; the Centuries in the next Paragraph taken Notice of, only excepted.