From 180° subtract the sum of the angles A, and B,

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 2d.

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 = 103° 13'.

3. In a triangle ABC, are given AC = 88, AB = 108, and BC = 54, to find the angles. Ans. Angle A = 290 49', B = 54° 7', and C = 96° 4'.

RULE 2.

Add together the arithmetical complements of the los garithms of the two sides containing the required angle, the logarithm of the half sum of the three sides, and the logarithm of the difference between the half sum and the side opposite the required angle. Then half the sum of these four logarithms, will be the logarithmic cosine of half the required angle.*

* DEMONSTRATION. Let ABC, be a triangle of which the side AB, is greater than AC: make AD = AC, join DC, bisect it in E, and join AE; draw EH parallel and equal to CB; join HB and produce it to meet AE produced in G.

Now in the triangles AED, AEC, all the sides of the one are equal to the sides of the other, each to each; therefore (8.1) the angle EAD = EAC, and AED = AEC; consequently AED is a right angle.

D

EXAMPLES.

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

EFD =

S

Because EH is equal and parallel to BC, BH is also equal and parallel to EC (33.1); now in the triangles EDF and HBF, the angle

BFH, the angle FED FHB (29.1) and ED = EC = BH; therefore (26.1) EF=FH, and FD = FB. Again the angle HGE

DEA = a right angle; if therefore with the centre F, and radius FE = FH, a circle be described, it will pass through the point G (31.3).

Now 2 AF 2 AD + 2 DF = AD + AD + DB AD + AB AC + AB; therefore AF = 1 AC + AB; Also FK-; IK = 1 EH = BC; hence by adding equals to equals, AF + FK = } AC + 1 AB + 1 BC, or AK = } (AC + AB + BC); again AI = AK

1 = IK = } (AC + AB + BC) - BC.

But (Dem. to rule, case 1st.) AD: AE :: sin. AED : sin. ADE:: rad. : cos. EAD (cos. Į BAC). Also, AB: AG:: sin. AGB: sin. ABG:: rad. : cos. BAG (cos. ¿ BAC).

Hence (C. 6.) AB X AD :: AG X AE:: rad. :(cos. { BAC).

:

If the other angles are required, they may be found by Case 1st.

2. In a triangle ABC, are given AB = 64, AC = 47, and BC = 34, to find the angle B. Ans. Angle B= 45° 38.

a

3. In a triangle ABC are given AC = 88, AB = 108, and BC = 54, to find the angle C. Ans. C = 96° 4'.

=

The 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 of the arc BF, or of the angle A, and the hypothenuse AC the secant of it.

2. In like manner, if the leg BC, Fig. 51, be made radius; then the other leg AB will represent the tangent, of the arc BG, or angle C, and the hypothenuse AC the secant of it.

=(AC + AB + BC) x (1 (AC + AB + BC) – BC); therefore AB X AC:1 (AC + AB + BC) (3 (AC + AB + BC) - BC) :: rad.2 : (cos. Į BAC).

Now it is evident, that in working this proportion by logarithms, and taking the arithmetical complements of the logarithms of the first term, viz. of the two sides including the required angle, if we omit the logarithm of the square of radius, which is 20, it is just equivalent to rejecting 20 from the sum of the logarithms; which would otherwise have to be done.

3. But if the hypothenuse be made radius; then each leg will represent the sine of its opposite angle; namely, the leg AB, Fig. 52, the sine of the arc AE or angle C, and the leg BC the sine of the arc CD, or angle A.

The angles and one side of a right-angled triangle being

given to find the other sides.

RULE.

Call any one of the sides radius, and write upon it the word radius; observe whether the other sides become sines, tangents, or secants, and write these words on them accordingly. Call the word written upon each side the name of that side. Then,

As the name of the side given,
Is to the name of the side required;
So is the side given,
To the side required. *

Two sides of a right-angled triangle being given, to find

the angles and other side.

RULE.

Call either of the given sides radius, and write on them as before. Then,

* DEMONSTRATION. Let ABC, Fig. 53, be a right-angled triangle; then it is evident that BC is the tangent, and AC the secant of the angle A, to the radius AB. Let AD represent the radius of the tables, and draw DE perpendicular to AD meeting AC produced in E; then DE is the tangent, and AE the secant of the angle A, to the radius AD. But because of the similar triangles ADE, ABC, AD:DE:: AB : BC; that is the tabular radius: tabular tangent :: AB : BC. AlSO AD: AE :: AB: AC; that is, the tabular radius: tabular secant : : AB : AC. These proportions correspond with the rule. When either