CASE V. Two Sides, and the Angle comprehended given, the other Side required. Example. cb=64 In the Obl. a dcb, Scd=129 given, A {id=121-15, } requ.b.d. 24d=101°15'. Preparation. Find the Angles b d, by Cafe the fourth. S,4b:cd::S Lc: b d. S56°15': 139: S, 101, 15: 164. The Operation is the fame with Cafe the fecond. Fig. 62. The three Sides given; the Angles required. Preparation by Axiom 4. From the Vertical Angle, upon the Bafe--bd Let fall the Perpendicular ca. Then the whole, is divided { Las cad, ca b Fale into 2 Segments ad, ab. c d Proportion per Ax. 4. db: cdcb: cd-cb: df = da-ba 105: 120 :: 20: 22, 8. Operation. To Ar. co. Log. bd = 105. Add { Log. cd— cb = 20 8. 9788.11 1.079181 0.301030 Sum- Radius Log. df 22; 8. 10.359022. By Gunters Scale.. Το 22.8 bd=105=Bafe. fd=22.8 Differ. Segments, Sum 127.8 Diff. 82.15 its 63.9=d=<>Segment: 41.7=ba=> Then the Argles are found by Cafe 4th of Right Angled Triangles.. Proportion. In the cabcb: Rad.: ab: S. L.c 50: S, 90 :: 41.7: S,56. 30.. (od: Rad. :: ad : SLC In the acad 70: S, 90:: 639:5,465° 369. From 90° OCH Suh, 56 30, Sum = 121. 36=lc in a b c d. Practical Trigonometry. Wherein the Do&t. ine of Pline Triargl . s are af plied to Practice. 1 N this Section, I shall treat only of such Practical parts thereof, as the Doctrine of Plane Right - lined Triangles becomes fulfirvient rc: As, 1. In ALTIMETRIA; By which the Height of any Object acceslible, inacceflible, may be found; As of Trees, Steeples, Towers, Etc. 'or 2. In 2. In LONGIMETRIA: By which the Diftance of one Obj et from any place, or of many Objects one from another, whether approachable, or in-approachable may be known, their Pofitions laid down, and a Map made of them. Of Altimetria. Prob. 1. Of an Altitude that is Acceffible. Fig. 63. Let AB be a Tower, whofe Height you would know. First, At any convenient distance, as at C, place your Quadrant, or any other Inftrument you make ufe of, and there obferve the Angle ACB, which let be 58, fo much is your Angle of Altitude. Meature next the di ftance between your Inftrument and the Foot of the Tower, viz. The Line CD which let be 25 Yards; then have you in a right Angled Triangle, one Angle c given, and one Leg CB to find the other AB; which you may do as you were taught in Cafe 1. of Trigonometry: For if you take 58 from-90, there remains 32 for the Angle at A. Then fay, As the Sine of the LA 32 9.724210 I. 39740 Add the two tc>gether and from Bafe CD 25 So is the Sine of the LC 58 9.928420) the Sum Sub ftract the first. 11.326360 To the Log. Height of the 1. 602150. To this 40 Yards, you mnft add the heighth of your Inftrument from the Ground. In this way of taking Heights, the Ground ought to be very Level, or you may make great mistakes: Alfo the Tower or Tree fhould ftand Perpendicular. Prob. 2. Of an Altitude inacceffible. In the the foregoing Figure, let AB be the Tower or Steeple, and fuppofe CB to be a Mote, or fome other hindrance, that you cannot come nearer then C; plant your Inftrument, and take the Angle ACB 58 deg. Then go backward any convenient ditance, as to G, there alfo take the Angle AGB 38 deg. This done Subftract 58 from 180, to have you 122 deg. the A gle ACG, then 122 and 38 being taken from 180, remains 20 for the Angle GC, the diftance GC meafured is 26. Trigonometry fay, As the Sine of the 4 A 20 is the Log. of the diftance GC26 So is the Sine of the Angle G 38 Now by 9. 534052 1. 414972 9.789242 I 1.204314 to the Log. of the Line AC 47 1. 670269 Again, |