THEOREM III. In any plane triangle, if a line is drawn from the vertical angle perpendicular to the base, dividing it into two segments: then, the whole base, or sum of the segments, is to the sum of the other two sides, as the difference of those sides, to the difference of the segments. 65. Let BAC be a triangle, and AD perpendicular to the base BC; then will BC: СА + АВ :: СА — АВ For, ᎪᎬ = BD+AD (Bk. IV., AC DO2 + AD2 CD-DB. A Prop. XI.); and by subtraction, AC2 - ABCD - BD. B D But, since the difference of the squares of two lines is equal to the rectangle contained by their sum and difference (Leg., Bk. IV., Prop. X.), we have, and AC-AB2 = (AC + AB). (AC – AB) CD2 - DB2 = (CD + DB) . (CD — DB) therefore, (CD+DB) . (CD — DB) = (AC+ AB). (AC — AB) hence, CD + DB: AC + AB :: AC – AB : CD DB. THEOREM IV. In any right-angled plane triangle, radius is to the tangent of either of the acute angles, as the side adjacent to the side opposite. 66. Let CAB be the proposed triangle, and denote the radius by R then will R: tan C:: AC : AB. For, with any radius as CD, describe the arc DH, and draw the tangent DG. From the similar triangles CDG and CAB, we have, CD : DG :: CA AB; hence, : R: tan C:: CA : АВ. By describing an arc, with B as a centre, we could show in the same manner that, R tan B :: AB : A C. THEOREM V. In every right-angled plane triangle, radius is to the cosine of either of the acute angles, as the hypothenuse to the side adjacent. 67. Let ABC be a triangle, right-angled at B: then will R : cos 1 :: AC : AB. For, from the point 4 as a centre, with any radius as AD, describe the arc DF, which will measure the angle A; and draw DE perpendicular to AB: then will AE be the cosine of A. EF The triangles ADE and ACB, being similar, we have, 68. The relations between the sides and angles of plane triangles, demonstrated in these five theorems, are sufficient to solve all the cases of Plane Trigonometry. Of the six parts, which make up a plane triangle, three must be given, and at least one of these a side, before the others can be determined. If the three angles only are given, it is plain, that an in angles of which may be respectively equal to the angles that are given, and therefore, the sides could not be determined. Assuming, with this restriction, any three parts of a triangle, as given, one of the four following cases, will always be presented. I. When two angles and a side are given. II. When two sides and an opposite angle are given. A CASE I. 69. When two angles and a side are given. 1. In a plane triangle, ABC, there are given the angle = 58° 07', the angle B = 22° 37', and the side AB = c = 408 yards; to final C, a and b.* GEOMETRICALLY. Draw an indefinite straight line, AB, and from the scale of equal parts lay off B AB equal to 408. Then at A, lay off an angle equal to 58° 07', and at B an angle equal to 22° 37', and draw the lines AC and BC: then will ABC be the triangle. The angle C may be measured either with the protractor or the scale of chords, and will be equal to 99° 16'. The sides AC and BC may be measured by referring them to the scale of equal parts. We shall find AC = 158.9 and BC=351 yards. TRIGONOMETRICALLY. Add the given angles together, and subtract their sum from 180 degrees. The remaining parts of the triangle can then be found by Theorem I. The sides lying opposite the angles A, B and C, are denoted by a, b and c. as it exceeds 90°, we use the supplement 80° 44'. Ans. C 98° 16', a = 351.024, and b = 158.976. = NOTE. The logarithm of the fourth term of a proportion is obtained by adding the logarithm of the second term to that of the third, and subtracting from their sum the logarithm of the first term. But to subtract the first term is the same sum (Sec. I., Art. 17): hence, the arithmetical complement of the first term added to the logarithms of the second and third terms, minus ten, will give the logarithm of the fourth term. 2. In a triangle ABC, there are given A 38° 25', B = 57° 42', and c = 400: required the remaining parts. Ans. C = 83° 53', a = 249.974, b = 340.04. CASE II. 70. When two sides and an opposite angle are given. = 216, In a plane triangle, ABC, there are given AC = b CB =α = 117, the angle ▲ 22° 37′, to find the other parts. ཡ GEOMETRICALLY. Draw an indefinite right line ABB': from any point, as A, draw AC, making BAC 22° 37', and make AC = 216. With C as a centre, and a radius equal Br d to 117, the other given side, describe the arc B'B; draw CB and CB'; then will either of the triangles ACB or ACB', answer all the conditions of the question. The ambiguity in this, and similar examples, arises in consequence of the first proportion being true for either of the angles ABC, or AB'C, which are supplements of each other, |