Geometrically. 1. Draw AB equal to 345 by a scale of equal parts.2. With the centers A and B, and radii 174°07 and 232, taken from the fame scale, describe arcs interfecting in c. 3. Draw AC and BC, and it is done. I Then by measuring the angles, they appear to be nearly of the following dimentions, viz. LA B=27°, and c = 115° 4. Arithmetically. 37° Having let fall the perpendicular CP, it will be AS AB 345 BCCA 406.07 CA=406·07 :: BO CA -- 2 AS AC To AP So s. LP To s. LACP and 345-68-18 Then in the triangle APC, right-angled at P, BP, 138.41 AP. 174 07 2.2407239 138.41 90° 00′ 2.1411675 10 0000000 Alfo 52° 40′ LACP added to 62 56 LBCP makes 115 36 LACB Whence the A 37° 20′, the LB = 27° 04', and the Lc 115° 36′. Inftrumentally. In the first proportion.-Extend from 345 to 406, on the line of numbers; that extent will reach, upon the fame line, from 58 to 68.2, the difference of the fegments of the bafe. In the fecond proportion.-Extend from 174 to 138 on the numbers; that will reach, on the fines, from 90° to 52°3. In the third proportion.-Extend from 232 to 206, and that extent will reach from 90 to 630. These three problems include all the cafes of plane triangles, as well right-angled as oblique; befides which there are fome other theorems, fuited to fome particular forms of triangles, which are often more expeditious in ufe than the general ones; one of which, as the cafe for which it ferves fo often occurs, take as follows. Given the Angles and a Leg, of a Right-angled Triangle; to find the Other Leg and the Hypotenuse. With the center A and any radius AD, defcribe an arc DE, and erect the perpendicular DF; which, it is evident, will be the tangent, and AF the fecant of the arc DE, or angle A, to the radius AD. And in the fimilar triangles ADF, ABC, it will be AD: AB:: DF: BC: AF: AC. 2. E. D. GENERAL SCHOLIUM. Befides thefe rules, I fall here fet down fome new theorems concerning the relations of the fides and angles of triangles, independent of any tables. PRO EXAMPLE. In the plane triangle ABC, right-angled at B, AB 162 Given { LA 53° 07′ 48′′ E 3 PROPOSITION. If 2 a denote a fide of any triangle, a the number of degrees contained in its oppofite angle, and r the radius of the circle circumfcribing the triangle: Then I fay that A is equal to 57°2957795 × ; a a3 + &c. For fince 2a is the chord of the arc upon which the angle, whose measure is A, infists; a will be the fine of half that arc, or the fine of the angle to the radius r, fince an angle in the cir cumference of a circle is measured by half the arc upon which it stands: now it appears, from rule 3. prob. 6. fect. 1. part 2. of this work, that the faid half arc z is equal to half the circumference of the fame circle, or the arc of 180 degrees, we shall have 3.14159r: 180° :: z: 180x 3.14159r T 2.373 2.4.575 2.4.6.7r1` the degrees in the angle or half arc. Corollary 1. By reverting the above feries we obtain Corollary 2. If 2a be the hypotenufe of a right-angled triangle, a will be = r, and then the general feries will become n × : a+ Geometrically. So Make AB 162, and the angle A = 53° 07′ 48′′; then raise the perpendicular вc meeting AC in c. fhall AC measure 270, and BC 216. Calcu Corollary 3. Since the chord of 60 degrees is fine of 30 degrees = half the radius, putting a for feries, will given X:- + + + the radius, or the r in the general 3.5 2 2.3.23 2.4.5.252.4.6.7.27 30; and hence the fum of the infinite feries &c= one-fixth of the circumference of the circle whofe diameter is 1. Corollary 4. It might eafily be fhewn, from the principles of common geometry, that the fine of 60 degrees is to the radius, as 3 is to 1; substituting, then, r3 for a in the general fe I + ries, we fhall have 3 X 3.5.33 + 3 3.32 + 2 2.3.23 2.4.5.25 2.4.6.7.27 &c= 60; and hence the fum of the infinite feries to the infinite feries in the third corollary, as 2 is to 3. Corollary 5. If b, c be the halves of the other two fides of the triangle, and B, c the degrees contained in their oppofite angles; b 63 fince BnX: + + 2.373 2.4.5rs and c = n x :- + &c, and the 3 angles of any triangle r 2.373 are equal to 180 degrees; we fhall have 180=A+B+C= I a3+b3+c3 3 a5+65+c5 3.5 a+b+c7 + 2.4.5 1.5 2.4.6.7 &c |