(K) Twice the rectangle of any two sides of a plane triangle, is to the sum of the squares of the same two sides diminished by the square of the third side; as radius, is to the cosine of the angle contained between those sides. Let ABC be any plane triangle and cd a perpendicular from the vertical angle upon the base or upon the base produced. I. If the perpendicular falls within the triangle Aco = AB" + BC” - 2AB x BD. (EUCLID 13 of II.) 2. If the perpendicular falls without the triangle AC” – AB* + BC” + 2AB x D [. BD. (EUCLID 13 of II.) AB* +BC?– Aco 2AB Aco-AB"-Bc. 2AB But BC : rad.::BD : cosine of the angle B (Z. 34.) COS, # **, here Z1 is acute in the first triangle, and obtuse in the second. By substitution, cos. Z. BxBc_AB*-i-Bco-Aco From the first BD= From the second BB = NotE. If the sides of the triangle be large, the rule given above is inconvenient for arithmetical calculation. SCHOLIUM, (L). Various other propositions might be given, but the preceding are sufficient for solving every case of plane trigonometry. The analogies of plane trigonometry are only particular cases of spherical trigonometry, and may in general be deduced from thence by simple inferences only. VIDE Book III. Chap. V. Prop. xxviii. and the Scholium. SOLUTIONS OF THE DIFFERENT CASES OF RIGHT ANGLED PLANE TRIANGLES. Rules for solving the different cases both of right angled plane triangles, and oblique C angled plane triangles, have already been investigated. The solutions here given are adapted to the table of natural sines”; and Z, a; the method of notation is that used by Lagrange and Legendre, the three angles of the triangle being represented by A, B, C, cT B and their opposite sides by a, b, c, as in the figure annexed; raradius=sine of 90° CASE II. Given the angles and the base, to find the hypothenuse and the perpendicular. * Logarithmical formulac arc easily supplied, thus in the first case (log sine A { log. b)-10–log. a. The introduction of such formulae would only increase the size of the book, without any real advantage to the student, CASE III. Given the angles and the perpendicular, to find the base and the hypothenuse. CASE IV. Given the hypothenuse and the base, to find the angles and the perpendicular. CASE W. Given the hypothenuse and the perpendicular, to find the angles and the base. S CASE VI. Given the base and the perpendicular, to find the angles and the hypothenuse. Or, having found the angles, the hypothenuse may be found by Case II. or by Case III. SoLUTIONs of THE DIFFERENT CASES OF OBLIQUE ANGLED - PLANE TRIANGLES. CASE I. Given the angles and one C side, to find the other two sides. CASE II. Given two sides and an angle opposite to one of them, to find the other angles, and the third side. NotE. If the given Z be obtuse; or if the side opposite to the givenz be the greater of the two given sides, then the z sought is always acute: in every other case it is ambiguous. wo angles being now given, their sum deducted from 180°, leaves the third Z; hence the remaining side may be found by Case I. CASE III. Given two sides and the angle contained between them, to find the other angles; and the third side. The remaining side may be found from the angles, by Case I. Case IV. Given the three sides, to find the angles. I. PRACTICAL RULEs Fort THE solution of ALL THE DIF- THEIR APPLICATION BY LOGARITHMS. (M) In every right angled plane triangle there must be two given quantities, one of which must be a side, (together with the right angle,) to find the rest. I. To FIND A SIDE. Call any one of the sides of the triangle radius, and write upon it the word radius; observe whether the other sides be come sines, tangents, or secants, and write these words on ) |