is a right angle, BA the less side is to AD or AC'the greater as the radius is to the tangent of the angle ABD; and because BGD is a right angle, BG is to GD or GF as the radius is to the tangent of GBD, which is the excess of the angle ABD above ABF half a right angle. But because EB is parallel to GD, BG is to GF as ED is to DF, (2. 6.) that is, (since ED is the sum of the sides BA, AC, and FD their difference, (3. of this,)) as the tangent of half the sum of the angles B, C, at the base to the tangent of half their difference. Therefore, in any plane triangle, &c. Q. E. D. PROP. VI. FIG. 9. & 10. In any triangle, twice the rectangle contained by any two adjacent sides is to the difference between the sum of the squares of these two sides, and the square of the base, as the radius is to the cosine of the angle included by the two sides. Let ABC be a plane triangle, twice the rectangle contained by AB, BC, is to the difference between the sum of the squares of AB, BC, and the square of the base AC, as the radius to the cosine of the angle ABC. From A draw AD perpendicular to BC, then (by 12. and 13. 2. El.) the difference between the sum of the squares of AB, BC, and the square of AC, is equal to twice the rectangle BC.BD; but twice the rectangle BC. BA is to twice the rectangle BC.BD; [that is, to the difference between the sum of the squares of AB, BC, and the square of AC]; as AB to BD; that is, by Prop. 1. as radius to the sine of BAD, (which is the complement of the angle ABC,) that is, as radius to the cosine of ABC. COR. If I be assumed as radius, then cos ABC = AB+BC-AC2 2AB.BC * PROP. VII. FIG. 11. In any plane triangle, the rectangle contained by half the perimeter, and its excess above the base, is to the rectangle contained by the excesses of half the perimeter above the other two adjacent sides, as the square of the radius is to the square of the tangent of half the contained angle. Let ABC be a plane triangle, of which the base is BC, and let P denote the perimeter of the triangle; then P(P-BC): (P—ÂB). (¿P—AC) :: rad: tan2 BAC. Bisect the angles ABC, BAC by the straight lines BG, AG, meeting each other in G; join GC; and having produced the sides AB, AC, bisect the exterior angles CBH, BCL, by the straight lines BK, CK, meeting AG in K; and from the points G, K, draw GD, GE, GF; KH, KL, KM, perpendiculars to the sides AB, AC, BC. Then, because G is the centre of the circle inscribed in the triangle ABC, (4. 4.) GD, GE, GF are equal, and AD is equal to AE, and BD to BF; also, since the angles at E and Fare right angles, the square of GC is equal to the squares of GE and EC together, or to the squares of GF and FC together; but the square of GE is equal to the square of GF; therefore, the square of EC is equal to the square of FC; and EC is equal to FC: and (8. 1.) the angle ECG is equal to FCG. For a like reason, KH, KL, KM are equal, and BH is equal to BM, and AH to AL, because the angles HBM, HAL are bisected by the straight lines BK, KA; and because in the two triangles KCL, KCM, the sides LK, KM are equal, KC is common to both, and KLC, KMC are right angles, CL is equal to CM. Since, therefore, BM is equal to BH, and CM to CL, BC is equal to BH and CL together: to these equals add A B and AC together, then AB, AC and BC together are equal to AH and AL together, but AH is equal to AL; wherefore each of them is equal to half the perimeter of the triangle ABC. But since AD is equal to AE, and BD to BF, and also CE to CF; AB and FC together are equal to half the perimeter of the triangle, to which AH or AL was shown to be equal; from these equals take away AB, which is common, and the remainder FC is equal to the remainder BH. In the same manner it may be demonstrated, that BF is equal to CL; and since the points B, D, G, F are in a circle, the angle DGF is equal to the exterior and opposite angle FBH, (22. 3); wherefore their halves BGD, HBK are also equal to one another. The right angled triangles BGD, HBK are therefore equiangular, and DG: DB:: BH: HK; therefore the rectangle DG.HK is equal to DB.BH. BAC. But AH: HK:: rad: (tan HAK=) tan. And AD: DG:: rad: (tan DAG=) tan. BAC; therefore, AH.AD: HK.DG :: rad2 : tan2 BAC, or by substitution, AH.AD: DB.BH:: rad: tan2 BAC; that is, P(P-BC): (P—AC). (¿P—AB):: rad: tan2 BAC. Q. E. D. * PROP. VIII. FIG. 11. In a plane triangle, the rectangle contained by any two adjacent sides, is to the rectangle contained by half the perimeter, and its excess above the base, as the square of the radius to the square of the cosine of half the contained angle. In the plane triangle ABC, if the perimeter be denoted by P, then will AB.AC: P(P-BC):: rad: cos BAC. For, the same construction being made as in the preceding proposition, in the right angled triangles AHK, ADG, AK: AH:: rad: (cos HAK=) cos BAC, And AG: AD:: rad: (cos DAG=) cos BAC; therefore, AK.AG: AH.AD:: rad2: cos2 ¿BAC. But since the straight lines BG, BK bisect the adjacent angles ABC, CBH, the angle GBK is a right angle, and for a like reason, GCK is a right angle; the points B, G, C, K are therefore in a circle, and the angles BKG, (or BKA,) and BCG in the same segment are equal; but BCG has been shown to be equal to ACG, and BAK is, by construction, equal to GAC; therefore the triangles ABK, AGC are equiangular; consequently AK: AB:: AC: AG, and the rectangle AK.AG is equal to AB.AC. Wherefore, AB.AC: AH.AD: : rad: cos2 BAC; that is, AB.AC: PP-BC): : rad2: cos2 BAC. Q. E. D. * PROP. IX. FIG. 11. In a plane triangle, the rectangle contained by any two adjacent sides is to the rectangle contained by the excesses of half the perimeter above those sides, as the square of the radius to the square of the sine of half the contained angle. In the plane triangle ABC, if the perimeter be denoted by P, then will AB.AC: (¿P—AC). (†P—AB) : : rad2 : sin2 BAC. 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 : : rad2 : sin2 & 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: sin2 BAC; that is, 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 difference 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 posite angles. Cases. Given. Sought. Solution, op 1. The hypotenuse AC, The legs AB Rad: sin C:: AC: AB, and an acute angle C. 2. A leg AB, and an acute angle A. and BC. Rad: cos C:: AC: BC. The hyp. AC Cos A: rad: : AB: AC, and perp.BC. Rad: tan A:: AB: BC. 3. The hypotenuse AC, The angle A, AC: BC :: rad: sin A, and a leg BC. 4. The two legs AB, BC. and the base Rad: cos A:: AC: AB; |