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 B101° 30', to find the other angles and side. Ans. The angle A = 30° 57′, 30° 57′, C = 47° 33', and the side AC = 144.8.

6

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

CASE 4.

Given the three sides, to find the angles.

RULE.

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 C as a 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 × AF= AE X 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 falling within, and sometimes without the triangle, it would require two cases, and consequently would be less simple.

The following rule, by which the necessity of letting fall a perpendicular is obviated, is deduced from a proposition, demonstrated by most writers on tri

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 2nd.

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, 32° 39', B 58° 56′, and C

will be A

By Calculation.

= 88° 25′.

To the arithmetical complements of the logarithms of the sides containing the required angle, add the logarithms of half the sum of the three sides, and of the difference between this half sum and the side opposite the required angle; then half the sum of these four logarithms will be the logarithmic cosine

To the diff. of the segments AD, DB 188.6

2.27544

From 180° subtract the sum of the angles A, and B, 91° 35', and the remainder 88° 25' is the angle C.

H

By Gunter's Scale.

Extend the compasses from 426 to 595 on the line of numbers, that extent will reach on the same line from 135 to 188.6 the difference of the segments of the base. Whence the segments of the base are found as before. To extend the other proportions, proceed as directed in case 2nd.

2. In a triangle ABC, there are given AB = 64, AC 47, and BC 34; required the angles. Ans. Angle A 31° 9', B 45° 38', and C = 1039 13'. =

=

3. In a triangle ABC, are given AC = 88, AB = 108, and BC 54, to find the angles. Ans. Angle A = 29° 49′, B = 54° 7', and C = 96° 4.

The four preceding rules solve all the cases of plane triangles, both right-angled and oblique. There are however other rules, suited to right-angled triangles, which are sometimes more convenient than the general ones. Previous to giving these rules, it will be necessary to make the following,

Remarks on right-angled triangles.

· 1. ABC, Fig. 50, being a right-angled triangle, make one leg AB radius, that is, with the centre A, and distance AB, describe an arc, BF. Then it is evident that the other leg BC represents the tangent, and the hypo