. BC 106 :{ 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. Remainder is the Angle BDC 120d. 43m. Note; The Proportion produceth 59d. 17m. for the required Angle: But being Obtufe, you must take it's Supplement to 180d. viz. 120d. 43m. as above is done. 2. Find the third Angle by the 9th of Sect. I. of this Chapter, in Page 34. 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 CD. S. BCD Side BD :: S. CBD S. 31d. 49m, 65 Yards: S. 27d. 28m... Yds. 56.88. Or, S. BDC. Side BC: S. CBD S. 120d. 43m... 106 Yds. :: S. 27d. 28m. as before. .. Side ČD. Axiom 3.1 N all Plane 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 leffer 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. Side {BC 199} Leag. } given: { Angle { BD 76 Angle CBD roid. 30m. BCD and req. Side CD This Triangle is made by Prob. 16, of Geometry, in Page 19. 1. For the Angle BDC, and BCD, the Operation is, 109 The three Angles 76 Subtract the given Angle CBD Side BD 185 The two oppofite Angles { 180d. oom. 101d. 30m. Sum is 39d. 15m. Sum is 78d. 30m. 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. SumAngle.. T. Diff. T. 39d. 15m... T.88.17m. 08d. 17m. 33 Leagues 185 Leagues The half Diff. of the Angles Added, is the greater Angle Subtract, is the lesser Angle 47d. 32m. BDC 30d. 58m. BCD req. 2. The Proportion for the Side CD, (by the first Case of Oblique Triangles) may be this: S. BCD Side BD:: S. CBD .. Side CD. S. 30d. 58m... 76 Leag. :: S. 1oid. 30m... Leag.144.7 Tenths. Axiom 4. Rom 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 first 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 BD 85 Feet given: Angle BCD CD 50 CBD. req. This Triangle is made by Prob. 17. of Geometry, in P. 19 and 20. BC105The half Sum This A iom 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 longest Side, is to the Sum of the two shortest; fo is the Difference of the two fhorteft, to the Difference of the Segments of the Base 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. BD { 85 Feet CD 50 Feet 135 Feet 35 Feet Added, is the Sum of the two shortest Sides 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. 105 Ft... 135 Feet:: 35 Fect The Side BC 105 Feet. Diff. Segments BE45 .45 Feet. Feet Subtracted is 60 30 AC the leffer ment. 2. The Angles BCD or CBD, may be found by the 4th Case of Right-Angled-Triangles, in Page 38. Thus. Hypot. BD.. Radius:: Leg AB. S. ADB. Remainder is the Angle CBD - 28d. Thus much for Plane Triangles; and to compleat Trigonometry, Spheric fhould be next. But I think the Application of this before the Doctrine of Spheric, 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 Indiction. Definition 1. THE Golden Number or Prime, is a Cycle or Revolution of 19 Years; in which Space of Time (it has been supposed) the Sun and Moon finish all their Variety of Afpects; by this we find the Epact, and confequently whatever thereon depends. 2. The Cycle of the Sun, maketh its Revolution in 28 Years; in which Time all the Variety of Dominical Letters, and LeapYears expire, and the 29th Year this Cycle 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 Ecclefiaftic and Civil Calendar. When 1, 9, 3, to the Year hath added been; Example. I would know the Golden Number, Cycle of the Sun, and Roman Indiction for the Year 1772. 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. 1. 2. Subtract 1 from the Golden Number, what refts multiply by 11; reject thirties, the Remainder is the Epact. Example. For the Year 1772, I demand the Epact? To the Year 1772 add I, and the Sum is 1773; which divide by 19, the Quotient is 93, and the Remainder is 6; fo that the Golden Number is 6; fubtract 1, there refts 5, which multiply by 11, gives 55; from this Product fubtract 30, the Remainder 25 is the Epact. Problem II. To find the Biffextiles, or Leap Years. The Old or Julian Leap-Years, is every fourth Year, and fo called from its leaping a Day more that Year, than in the common Year; for in the common Year any fixed Day of the Month changeth fucceffively the Day of the Week, but in the LeapYear, it leaps over one Day. Thefe Leap-Years (in the NewStile) are continued exactly in the fame Order and Succeffion as heretofore; the Centuries in the next Paragraph taken Notice of, only excepted. |