Fig. XXXVI. Fig. XXXVII. Then, Ass DCB: Log. DB :: SCBD: Log. CD 116.20d. 1270 And, 37.43 d.. 865 Ass DCB: Log. DB:: SCDB: Log. CB 1270 26.37 d. PROB. III. 632 In the Right-angled Plain Triangle ABC, there is given, the Base N the given Triangle ABC, extend the Perpendicular ID, taking into A C to CB; so shall you have constituted a new Triangle DAB; in which there is given, (1.) The Right Angle at A. (2.) The Base AB 28 Foot. (3.) The Side A D, equal to the Sum of the Sides A C and CB, by which you may find the Angles A DB, and ABD, (by Cafe I. of R. Ang. Tri.) As Log. AB: Rad. :: Log. DA:DBA 56 63.43 d. Whose Complement 26.57 d. is the Angle ADB, to which the Angle CBD is equal, because the Sides C B and CD, fubtending them, are equal by Conftruction. Then from the Angle DBA 63.43 d. before found, you substract the Angle DB C, 26.57 d. laft found, the Remainder 36.86 d. is equal to the Angle CBA, and the Complement thereof, 53.14 d. to the Angle C A B. And then, As SACB: Log. AB:: SCBA: Log. CA 53.14 d. : Otherwife, by this Theorem. Fig. If from the Square of the Sum of the Base and Perpendicular D A XXXVII. PROB. IV. The Three Sides of a Right-lined Triangle, being given; To find A RULE. DD the Three Sides of the given Triangle together, and take the half thereof: From which half Sum, fubtract each Side of the Triangle severally, and note the several Differences: Then multiply the Half Sum, by any one of the Differences, and that Produit multiply by another of the Differences, and that Produt multiply by the third Difference: The Square Root of this third Product, shall be the Area of the given Triangle. Example. In the Triangle ABC, whose Three Sides are A B 20, AC Fig. 34, and BC 42: Their Sum is 96, the half whereof is 48: XXXVIII. From which fubtract the several Sides, 20, 34, and 42, and there will remain these Differences, 28, 14, 6. Then, multiply 48 (the half Sum) by 28 (the first Difference) the Product will be 1344: Which multiplied by 14 (the second Difference (the Product will be 18816: And this Product multiplied by 6 (the third Difference) produceth 112896: The Square Root whereof is 336, for the Area of the Triangle ABC. By Logarithms. To the Logarithm of half the Sum of the Sides, add the Logarithms of the several Differences of the Sides from the half Sum: Half the Sum of those Logarithms shall be the Logarithm of the Area of the Triangle. 2.5263392 Which is the Logarithm of 336, the Area of the Triangle. PROB. V. By the Three Sides given; To find the Point in the longer Side, where a Perpendicular shall fall, and the Length of that Perpendicular. RULE. AKE the Sum and. Difference of the Two Sides containing. multiply them together, and divide the Product by the third Side, upon which the Perpendicular is to fall, the Quotient added to the third Side, or substracted from it, shall be the Double of the.. Greater or Leffer Segment on either Side of the Perpendicular. Example Fig. In the former Triangle A B C, the Sum of the Sides A B and XXXVIII. A C, is 54, and their Difference is 14, which multiplied together, produce 756; which divided by the Side BC 42, the Quotient is 18; which added to the Side B. C 42, gives 60; the halfwhereof 30, is the Greater Segment of the Side BC; or the Quotient 18, substracted from the Side BC, leaves 24; the half whereof 12, is the Leffer Segment. B D, wherethe Perpendicular is to fall. By Logarithms. To the Sum of the Logarithms of the Sum and Difference of the Sides containing the Angle from whence the Perpendicular is to fall, substract the Logarithm of the Side upon which it is to fall. The Remainder shall be the Logarithm of a Number; which added to, or substracted from the Side on which the Perpendicu-lar is to fall, shall be the Double of the Greater or Leffer Segments of the Side on which the Perpendicular is to fall. Example. Example. The Sum of A B and AC is 54 The Difference of A B and AC is 14 Their Sum The Logarithm of the third Side B C 42, substract Remains 1.7323937 1.1461281 2.8785218 1.6232493 1.2552725 Which is the Logarithm of 18. Which 18 added to B C 42, makes 60; the half whereof 30, is the Greater Segment CD, or 18 substracted from B C 42, leaves 24; the half whereof 12, is the Leffer Segment B D. For the Length of the Perpendicular. RULE. Multiply the Sum of A B and BD 32, by the Difference of A B and BD, 8; the Product will be 256, whose Square Root is 16; for the Length of the Perpendicular A D. By Logarithms. Half the Sum of the Logarithms of 32 and 8, the Sum and Difference of A B and BD is the Logarithm of the Perpendicular.. The Logar. of the Sum of A B and BD 32, 1.50515 The Logar. of the Difference of A B and A D 8. 090309 1.20412 Fig. XXXVIII. Which is Logarithm of 16, the Length of the Perpendicular AD.. PROB. VI. The Base BA (or longest Side) of a Plain Triangle BAD, being 47.8 P. and a Perpendicular DC, let fall from the Angle oppofite to that Side 17.33 P. being given; To find the Area of that Triangle. Fig. O the Logarithm of half the given Side, add the Logarithm Tof the the Logarithm of the Area of that Triangle. The The Logarithm of 414.2 2.6171965 Fig. XL. Which is the Area of the Triangle. After the fame manner, if the Base of a Triangle were 42, and the Perpendicular 16, the Area will be found to be 336. PROB. VII. There is an Oblique-angled Plain Triangle A B C, one of whose Angles at the Base BC is 53.48 d. And the Sum of the Two Sides opposite to the Base B C, viz. A B and AC, is 129, and the Base CB, is longer than the longer fide A C, by 16.8. The reSpective Sides and Angles of the Triangle are required. D CONSTRUCTION. Raw a Right Line BC, at Liberty, for the Base; and upon one end thereof, at B, make an Angle ABC, to contain 53.84 d. Then the Sum of the Two Sides A B and AC being together 120, break it into any Two equal or unequal Parts, as into 43.2 and 76.8, then (by the Propofition) must the Base B C, be 93.6, which is greater than A C by 16.8. So in the Obliqueangled Triangle ABC, you have given, (1.) The Side AB 43.2. (2.) The Side A C 76.8. (3) The Angle opposite thereto, ABC 53 d. 50 m. by which you may find the Angle ACB. For, As A C 76.8: Istos ABC 53.84 d. Of the Menfuration of the Area of a Spherical Triangle. 1 Lemma I. The Lunary Superficies of the Hemisphericks, are as the Fig. XLI. T HE Proof of this Lemma (amongst many other Ways) may be this: Let the Meridian Semicircle AED be imagined to be equally moved over the Longitude of the Equator BEC, upon the Poles A and D. Therefore the Angles on the other fide of A and D (to |