322 Fig. First, Draw two Lines at pleasure, as A B and A C, making the LXVII. Angle at A, equal to 107. co de. Secondly, Take 356. 00 Miles, in your Compaffes, and enter them between the two Lines at pleasure, as at B and C: conftituting the Triangle ABC. Thirdly, Continue the Side B A to D, making A D equal to A C, and joyn CD, conftituting another Oblique-angled Triangle DCA, in which there is given. (1) The Angle at CAB, 73.00 deg. (the Complement of BAC to 180 deg.) And (2) The Angles ACD and ADC, each of them equal to half the Angle BAC, viz. to 53.50 deg Then in the Oblique-Angled Triangle DCD, fay, As the given Side B C, 356.00 M. Is to the Sine of BD C, 53.50 de. To the Sine of 73.27 deg. 2.551450 9.905178 2.627488 12.532666 9.981216 Which is the quantity of the Angle DCB: From which fubftract the Angle ACD, 53.50 de. there will remain 19.77 d. for the Angle ACB. Then in the Triangle B AC, As the Sine B A C, 107.00 d. (73.00) Is to the Side B C, 356.00. So is the Sine of A CB, 19.77 de. To B A, 125.90 M. 9.980619 2.551450 9.529231 12.080681 2.100062 Which fubftracted from BD, 424.12, there remains 298 M, for the Side A C. the Distance that the other Ship failed. So that the Port A, is diftant from the Head-Land C 298.22 M. And from that at B, 125.90.. PROB. XI. Fig. Two Ships fet fail from two Head-lands at B and C, diftant LXVIII. from each other 356. 00 Miles; and meeting together at A, they obferve that the two Head Lands bear fo from their Ship, that they make an Angle of 107. 00 deg. Alfo, the way that both the Ships have made from B and C to A, doth amount inte 424. 12 Miles: I demand how much of this each Ship failed. The First, The Geometrical Conftruction. DRAW 107.00 deg. upon Secondly, The Distance of the two Head-lands being 356. co Miles enter them between the two fides, at B and C, and joyn BC, making the Triangle B A C Thirdly, Make A D equal to A C, and joyn CD, so have you conftituted another Triangle A C D. Now in the Triangle A B C, is Given. (1) The Side BC, 356 co. (2) The Angle B A C 107. oo deg: And in the Triangle A CD, the Angle B A C being 107. 00 de. the Angle DAC must be 73.00 de. And the Sides AC and A D, being equal (by conftruction) the Angles ADC and ACD, muft be alfo equal, viz. (each of them) 53. 50 deg. And being thus far prepared, you may proceed to find the Angle ACB. By Trigonometrical Calculation. Fig. LXVIII. As the Side B C, 356. 00 2.551449 To the Sine of the Angle ADC, 53. 50 d. 9.905178 So is the whole Side B AD, 424. 12 M. 2.627000 12.532178 To a fourth Sine, 73 06 d. 9.980729 This 73. 06 d. fhould be the quantity of the Angle DCB, but (by the conftruction) the Angle you fee is Obtufe; and therefore the Complement of 73. 06 deg. to 180, viz. 106. 94 deg. is the quantity of the whole Angle DCB. From which if you fubftract the Angle A CD, 53. 50 deg. there will remain 53. 44 deg. for the Angle AC B. To the Side B A, 299. 00. 9.980619 2.551449 9.904842 12.456291 2.475672 And fo much did the Ship from B, fail to A, and that fubftracted from 424. 12 there remains 175. 12, and fo much did the Ship from C to A Sail. II. of MERCATO R's Sailing, by the True Sea-Chart. LL the foregoing Problems are performed by that kind of Navigation, commonly called Plain Sailing, or Sailing by the Plain Sea-Chart; in which Chart, the Degrees of Longitude and La-titude, in all Places, are fuppofed to be equal; which is Erroneous, though moft practifed: But there are two other ways of Sailing, both more exact than the former, the one called Mercator's, the other, Sailing by the Middle Latitude. That of Mercator's requires that the Degrees of Latitude in that Chart be inlarged as they go farther from the Equinoctial towards either of the Poles, which is done by Reducing them into Meridional Parts: but the Degrees of Longitudes into Miles and Centefms, as in Plain Sailing: And for the Reducing of Degrees and Minutes of Latitude into Meridional Parts, I have here inferted a Table for the ready performance thereof, to every Degree and Quarter, or 25 Centefms of a Degree. A |