As the sum of the two given sides,
Is to their difference;
So is the tangent of half the sum of the two unknown

angles,
To the tangent' of half their difference.*

This half difference of the two unknown angles, added to their half sum, will give the angle opposite the greater

for a

a

* DEMONSTRATION. Let ABC, Fig. 45, be the proposed triangle, having the two given sides AB, AC, including the given angle A. About A as a centre, with AC the greater of the given sides,. distance, describe a circle meeting AB

produced in E and F, and BC in D; join DA, EC, and FC, and draw FG parellel to BC, meeting EC produced, in G.

The angle EAC (32.1.) is equal to the sum of the unknown angles ABC, ACB, and the angle EFC at the circumference is equal to the half of EAC at the centre (20.3.); therefore EFC is half the sum of the unknown angles; but (32.1.) the angle ABC is equal to the sum of the angles BAD and ADB or BAD and ACB; therefore FAD is the difference of the unknown angles ABC, ACB, and FCD, at the circumference is the half of that difference; but because of the parellels DC, FG, the angles GFC, FCD are equal; therefore GFC is equal to half the difference of the unknown angles ABC, ACB; but since the angle ECF in a semicircle, is a right angle, EG is perpendicular to CF, and therefore CF being radius, EC, CG are the tangents of the angles EFC, CFG; it is also evident that EB is the sum of the sides BA, AC, and that BF is the difference; therefore since BC, FG are parellel EB : BF: : EC: CG (2.6.): that is, the sum of the sides AC, AB, is to their difference, as the tangent of half the sum of the angles ABC, ACB, is to the tangent of half their difference.

To demonstrate the latter part of the rule, let AC and AB, Fig. 46, represent any two magnitudes whatever; in AB produced, take BD equal to AC the less, and bisect AD in E.

Then because AE is equal to ED and AC to BD, CE is equal to EB; therefore AE or ED is half the sum of the given magnitudes AB, AC, and CE or EB is half their difference; but AB the greater is equal to AE, EB, that is to half the sum added to half the difference, and AC the less, is equal to the excess of AE, half the sum, above CE, half the difference.

of the two given sides, and being subtracted from the half sum, will give the angle opposite the less given side.

1

After finding the angles, the other side may be found by Case 1.

EXAMPLES.

1. In the triangle ABC, there are given AB = 128, AC = 90, and the angle A = 48° 12', to find the angles B and C, and the side BC.

By Construction, Fig. 47. Draw AB = 128, and make the angle A = 48° 12'; draw AC = 90, and join BC. The angle B will measure 44° 37', the angle C 87° 11', and the side BC 95.5.

As the sum of the sides AB, AC, 218

2.33846 Is to their difference, 38

- 1.57978 So is the tang.of half sum of angles B & C,65° 54' 10.34938

11.92916 • 9.59070

To tang. of half their difference, 21° 17'

Half sum of the angles B and C. 65° 54'
Add and subtract half their difference 21 17

Angle C

87 11

By Gunter's Scale. Extend the compasses from 218, the sum of the sides, to 38, their difference, on the line of numbers, and apply this extent to the line of tangents from 45° to the left hand; then keeping the left leg of the compasses fixed, move the other leg to 65° 54' the half sum of the angles; that distance will reach from 45° on the same line, to 21° 17', the half difference of the required angles. Whence the angles are obtained as before.

To extend the second proportion, proceed as directed in Case 1st.

2. In a triangle ABC, are given AB = 109, BC = 76, and the contained angle B = 101° 30', to find the other angles and side. Ans. The angle A = 30° 57', C = 470 33, and the side AC = 144.8.

3. Given, in a right angled triangle, the base AB = 890 and the perpendicular BC = 787, to find the angles and hypothenuse. Ans. The angle A = 41° 29' C = 48 31' and the hypothenuse AC = 1188.

CASE 4.

Given the three sides, to find the angles.

RULE 1.

Consider the longest side of the triangle as the base, and on it let fall a perpendicular from the opposite angle. This perpendicular will divide the base into two parts, called segments, and the whole triangle into two right angled triangles. Then,

As the base, or sum of the segments,
Is to the sum of the other two sides;
So is the difference of those sides,
To the difference of the segments of the base.*

* DEMONSTRATION. Let ABC, Fig. 48, be a triangle, and CD be perpendicular upon AB. About Casa centre with the less side BC for a radius, describe a circle, meeting AC produced, in G and E, and AB in F. Then it is evident that AE is equal to the sum of the sides AC, BC, and that AG is equal to their difference; also because CD bisects FB (3.3), it is plain that AF is the difference of the segments of the base; but AB X AF = AEX AG (36.3. cor.); therefore AB: AE :: AG:AF (16.6); that is, the base, is to the sum of the sides, as the difference of the sides, is to the difference of the segments of the base.

Cor. If AF be considered the base of the triangle AFC, then will CD be a perpendicular on the base produced; AE will be equal to the sum of the sides AC, FC, and AG will be equal to their difference; also AB will be equal to the sum of the segments AD, FD. But by the preceding demonstration and (16.5), AF: AE :: AG:AB; hence when the perpendicular falls without the triangle, the base, is to the sum of the sides, as the difference of the sides, is to the sum of the segments of the base.

A rule might, therefore, be given, making either side of a triangle, the base; and such a rule would be rather more convenient, in some cases, than the one above: but then, on account of the perpendicular, sometimes fallinga Within and sometimes without thì triangle, it To half the base, add half the difference of the segments and the sum will be the greater segment; also from half the base, subtract half the difference of the segments, and the remainder will be the less segment.

Then, in each of the two right angled triangles, there will be known two sides, and an angle opposite to one of them; consequently the other angles may be found by Case 2d.

EXAMPLES

1. In the triangle ABC, are given AB = 426, AC = 365, and BC = 230; required the angles.

By Construction, Fig. 49. Draw AB = 426; with AC = 365 in the dividers, and one foot in A, describe an arc, and with BC = 230, and one foot in B describe another arc, cutting the former in C; join (AC, BC, and ABC will be the triangle required. The angles measured by a scale of chords, will be A = 32° 39, B = 58° 56' and C = 88° 25'.