This proposition is demonstrated by the writers on Algebra, and may be proved geometrically as follows.

A.

B.

-C

B

Let AB, BC be two unequal magnitudes, of which AB is the greater. Place AB, BC in the same straight line AC; bisect AC in D, and take AE-BC. Then AC is the sum, and EB is the difference of the two magnitudes AB, BC. Because AD is equal to DC, and AE to BC, AD-AE is equal to DC-BC, that is, DE=DB; thereA E D B C fore DE or DB is half the difference of the magnitudes AB, BC. But AB=BD+DA, that is, to half the difference added to half the sum; and BC= DC-DB, that is, to the excess of half the sum above half the difference.

58. Cor. If the semi-sum be subtracted from the greater quantity, the remainder will be the semi-difference. ABAD=BD.

PROP. V.

59. In any triangle, if a perpendicular from the vertex to the base fall within the triangle, the base is to the sum of the other two sides, as their difference is to the difference of the segments of the base made by the perpendicular.

Let ABC be the proposed triangle, C the vertex, AB the base, CD the perpendicular dividing the base into the segments AD, DB. About the centre C, with the radius CB, the less of the two sides, describe a circle cutting the base in G, the side AC in F, and AC produced in H; then will A

F

G

D B

H

AH AC+CH-AC+CB, be the sum of the sides, and AF= AC-CF-AC-CB, be the difference of the sides. Because DB=DG (3. 3), AG=AD—DB is the difference of the segments of the base. Hence ABXAG=AHXAF (Cor. 36. 3), therefore AB: AH: AF: AG (16. 6); that is,

AB: AC+CB:: AC-CB: AD-DB.

60. Scholium. In any triangle ABC, if a perpendicular be let fall from the greatest angle C, it will fall within the triangle, and will divide it into two right-angled triangles ACD, BCD, whose hypothenuses AC, BC are the sides of the proposed triangle, and whose bases AD, DB are the segments of its base. Now if all the three sides of the triangle ABC be given, then the base AB, or the sum of the segments AD, DB, is given, and the difference of the segments may be found by the proposition. But if the sum and difference of the segments be known, the segments may be found by article 57. Thus, in the two right-angled triangles ACD, BCD, the hypothenuses. AC, BC are given, and the bases AD, BD are found by the proposition; whence the angles at the base CAD, CBD may be found by art. 49, and consequently the angle ACB, which is the supplement of the angles at A and B, may be found by Cor. 32. 1.

PROP VI.

61. In any triangle, the sum of any two sides is to their difference, as the tangent of half the sum of the opposite angles is to the tangent of half their difference.

F

H

Let ABC be the proposed triangle, whose sides are AC, BC, and base AB. About the centre C, with the radius CB, the less of the two sides, describe a circle cutting the longer side AC in F, and AC produced in H; then is AH the sum of the sides AC and CB, and AF their difference. Draw FB and HB, then is the angle HCB the sum of the angles at the base (32. 1), and the angle HFB=half the angle HCB (20. 3), half the sum of the angles at the base. The triangle FCB is isosceles, therefore the angle CFB CBF (5. 1), therefore the angle CBF is half the sum of the angles at the base. But ABF= CBA-CBF, that is, the angle ABF is the difference between the greater of the angles at the base and their semi-sum, therefore ABF is the semi-difference of the angles at the base (58). Draw FE parallel to BH, then the angle EFB=ÈBF (29. 1) a right angle (31. 3). Therefore if F be made the centre, and FB the radius, HB will be the tangent of the angle HFB, the semi-sum of the angles at the base; and if B

=

A

C

E

B

be the centre, with the same radius FB, then FE will be the tangent of the angle EBF, the semi-difference of the angles at the base.

The triangles AFE, AHB, having the angles AFE and AHB equal (29. 1), and the angle at A common, are similar. Therefore AH: AF :: HB: FE, or AC + CB : AC—CB :: HB: FE; that is, the sum of the sides: difference of the sides :: tan. of half the sum of the angles at the base : tan. of half their difference.

62. Scholium. If two sides AC, BC of any triangle ABC, and the angle ACB included between them, be given, the other side and angles may be found.

For, from the angle ACB we can find its supplement to 180° (18. 1), equal to the angle HCB, half of which is the semi-sum of the angles at the base; and their semi-difference is found by this proposition. Then, the semi-sum and semidifference of the angles being known, the angles themselves, CBA and CAB, may be found by art. 57. Two sides and all the angles of the proposed triangle being now known, the third side AB may be found by art. 54, as follows. Sine ZA : BC :: S. LACB: AB.

Note. The preceding propositions are sufficient for the resolution of all the cases of rectilinear triangles. The following proposition is added, because it is useful in many geometrical investigations, and in the solution of triangles in certain cir

cumstances.

PROP. VII.

65. In any rectilinear triangle, twice the rectangle contained by any two sides is to the difference between the sum of the squares of those sides and the square of the base, as the radius to the cosine of the angie contained by the two sides.

Let ABC be any triangle; 2 AB. BC: difference between AB+BC and AC:: R: cos. B.

From A draw AD perpendicular to BC, then the difference between AB2+BC2 and AC2 is equal to 2 BC. BD (13. 2). In the right-angled triangle ABD, BA: BD :: R: cos. B (49), therefore 2 BC. BA: 2 BC. BD :: R: cos. B (15. 5), that is, 2 AB. BC: difference between AB + BC and AC2 :: R: cos. B.

64. Cor. 1. If radius=1, then BD=BÁxcos. B (49), therefore 2 BC. BA x cos. B=2 BC. BD; therefore when B is acute, 2 BC. BA x cos. B=BC2+BA2-AC (13. 2), therefore AC2+2 cos. BxBC, BA=BC2 + BA2, therefore AC2= BC2-2 cos. BxBC. BA+BA2, therefore AC = √(BC2 — 2 cos. B x BC. BA+BA2).

If B be an obtuse angle, it may be proved in the same manner, that AC=√(BC2 +2 cos. BxBC. BA + BA3).

65. Cor. 2. Because AC2-AB2+BC2-2 cos. BxAB. BC, 2 cos. BxAB. BC=AB2 + BC2—AC2=AB2+(BC+AC)× AB2+(BC+AC)×(BC—AC) (BC-AC), therefore cos. B=2 AB.BC

66. Scholium. When two sides and the included angle of any plane triangle are given, the third side may be found from art. 64. This method of finding the third side of a triangle is very useful in many geometrical investigations, but is not adapted to computation by logarithms, because the quantity under the radical sign cannot be resolved into simple multipliers.

When the three sides of any plane triangle are given, the angles may be found immediately from art. 65. When the sides are expressed by great numbers it will be more convenient to find the difference of the segments of the base, by prop. 5, and then the angles at the base, by prop. 1.

SECTION II.

RULES OF TRIGONOMETRICAL CALCULATION.

67. THE general problem which trigonometry proposes to resolve is this. In any plane triangle, of the three sides and the three angles, three parts being given, one of which is a side, to find any of the other three parts.

The parts of a triangle which are said to be given, are understood to be expressed by numbers, or by their numerical

values; the angles in degrees, minutes, &c.; and the sides in feet, or any other known measure.

The reader must remember, that the sines, cosines, &c. which are used in the calculations in this section, are the numerical measures of the geometrical lines bearing the same names, the measure of the radius of the circle being supposed unity.

The problem is restricted to those cases in which one side at least is given, because, if the three angles only be given, the magnitudes of the sides cannot be determined, but only the ratios of their magnitudes. Innumerable triangles, equiangular to one another, may exist; but the sides of none of them may be equal to the sides of another, though the ratios of the sides to one another will be the same in all the triangles (4. 6). Therefore, if the three angles only of a triangle be given, we can determine nothing but the ratios of the sides, which are the same as the ratios of the sines of the opposite angle (56).All the varieties which can happen in the solution of plane triangles are comprised in three cases, as follows.

1. When two of the three given things are a side and its opposite angle.

2. When two sides and their included angle are given. 3. When the three sides are given.

For the conveniency of calculation it is usual to divide the general problem into two problems, according as the proposed triangle is right-angled or oblique-angled.

A right-angled triangle consists of three variable parts, the three sides and the two acute angles. The right angle being a constant quantity is not reckoned. An oblique-angled triangle consists of six variable parts, the three sides and the three angles.

PROBLEM I.

68. In a right-angled triangle, of the three sides and the three angles, two being given, besides the right angle, and one of them being a side, it is required to find the other three parts of the triangle.

If one angle of a triangle be right, the other two angles are acute, and together make a right angle (32. 1). Therefore when one of the acute angles of a right-angled triangle is given, the other also is given, being its complement to a right angle, or 90 degrees.

The sine of either of the acute angles is the cosine of the other (23), and therefore may be used instead of the other, whenever it renders the operation more simple.