P, then will AB.AC:({P-AC). (İPAB):: rada : sin? ABAC. For, the same construction being made as in the two preceding propositions, in the right angled triangles AHK, ADG, AK: HK:: rad: (sin HAK=) sin }BAC, AG: DG:: rad : (sin DAG=) sin BAC; therefore, AK. AG: HK. DG :: rad : sin BAC; but AK.AG was shown to be equal to AB.AC, and DG.HK to be equal to DB.BH; wherefore, AB.AC: DB.BH:: rad : sina BAC; that is, AB.AC:({P-AC). (AP-AB):: rad? : sino {BAC. Q. E. D. PROP. X. Fig. 12. 13. In a plane triangle, the base is to the sum of the sides as the difference of the sides is to the sum or dif. ference of the segments of the base made by the perpendicular upon it from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the less side and the base. Let ABC be a plane triangle; if from A the vertex be drawn a straight line AD perpendicular to the base BC, the base BC will be to the sum of the sides BA, AC, as the difference of the same sides is to the sum or difference of the segments CD, BD, according as the square of AC the greater side is greater or less than the sum of the squares of the less side AB, and the base BC. About A as a centre, with AC the greater side as a distance, let a circle be described meeting AB produced in E, F, and CB in G: It is manifest, that FB is the sum, and BE the difference of the sides : And since AD is perpendicular to GC; GD is equal to CD; consequently GB is equal to the sum or difference of the segments CD, BD, according as the perpendicular AD meets the base, or the base produced ; that is, (by Conv. 12. 13. 2.) according as the square of AC is greater or less than the sum of the squares of AB, BC: But (by 35. 3.) the rectangle CBG is equal to the rectangle EBF; that is, (16. 6.) BC is to BF, as BE is to BG; that is, the base is to the sum of the sides, as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the less side and the base. Q. E. D. * SECT. II. RULES FOR TRIGONOMETRICAL CALCULATION. The general problem that Plane Trigonometry proposes to resolve is, of the three sides, and three angles of a plane triangle, the numerical values of any three being given, and one of these three being a side, to find any of the other three. This general problem is usually divided into the two following, according as the triangle has or has not one of its angles a right angle. PROBLEM I. Fig. 15. In a right angled plane triangle, of the three sides and either of the acute angles, any two being given, to find the other two, and the remaining acute angle. The several cases of this problem may be resolved by Proposition 1, as in the following table ; observing, that when one of the acute angles of a right angled triangle is given, the other is also given, for it is the complement of the former; and that therefore the sine of either of the acute angles is the cosine of the other. When the two acute angles only are given, the sides cannot be found from them; but their ratios may, which are the same with those of the sines of their opposite angles. Cases. Given. Sought. . Solution. 1. The hypotenuse AC, The legs AB Rad: sin C:: AC: AB, and an acute angle C. and BC. Rad: cos C:: AC:BC. 2. A leg AB, and an The hyp. AC Cos A : rad:: AB: AC, acute angle A. and perp.BC. Rad : tan A :: AB: BC. 3. The hypotenuse AC, The angle A, AC: BC :: rad: sin A, and a leg BC. and the base Rad: cos A:: AC: AB; AB. or AB=~[(AC+ BC). (AC-BC)]by 47.1. and Cor. 5. 2. 4. The two legs AB, The angle A, AB: BC:: rad: tan A, BC. and the hyp. Cos A: rad :: AB: AC; AC. or AC= V(AB* + BC) |(47. I.) PROB. II. Fig. 16. 17. In an oblique angled plane triangle, of the three sides, and any two of the angles, any three being given, to find the other two and the remaining angle. The four cases of this problem may be resolved by means of the foregoing propositions, as in the following table; observing, that when two angles of any plane triangle are given, the third is also given, being the supplement of the sum of the two given angles; and if one of the angles is given, the sum of the other two is also given. When the three angles only are given, the sides cannot be found from them, but their ratios may, being the same with those of the sines of their opposite angles. 2. 3. AB C AC: AB :: sin B : sin C (2.) A=180°—(B+C) BC sin B:sin A:: AC: BC. B AB+AC: AB-AC :: tan с *(C+B): tan {(C-B) (3.) BC then by Lem. 3. C=} (C+B) AB+AC:BC (4.) :: AB-AC: BC. |