The angle C and B being found; BC is had as before, by theo. 1. of this sect. Thus,

Because the two first terms are of the same kind, extend from 420 to 60 on the line of numbers; lay that extent from 45° on the line of tangents, and keeping the left leg of your compasses fixed, move the right leg to 716 40'; that distance laid from 45° on the same line will reach to 23o 30', the half difference of the required angles. Whence the angles are obtained, as before.

The second proportion may be easily extended, from what has been already said.

CASE IV.

PL. 5. fig. 17.

The three sides given, to find the angles.

In the triangle ABC, there is given, AB 64, AC 47, BC 34: the angles A, B, and C, are required.

1st. By Construction.

The construction of this triangle must be manifest, from prob. 1. sect. 4.

2d. By Calculation.

From the point C, let fall the perpendicular CD on the base AB; and it will divide the triangle into two right-angled ones, ADC and CBD; as well as the base AB, into the two segments AD and DB.

AC 47
AB 33

Sum 81

By theo. 3. of this sect.

As the base or the longest side, AB

is to the sum of the other sides, AC and BC

So is the difference of those sides

to the difference of the segments of the base, AD, DB

64

81

13

16.46

By theo. 4. of this sect.

To half the base, or to half the sum of the segments AD

32

and DB

Add half their difference, now found,

8.23

Their sum will be the greatest segment, AD

40.23

Subtract, and their difference will be the least segment, DB 23.77

In the right-angled triangle ADC, there is AC 47, and AD 40.23, given, to find the angle A.

This is resolved by case 4, of right-angled plane trigonometry, thus,

AD: R: AC: Sec. A

40.23 900 :: 47: 310 08'

Or it may be had by finding the angle ACD, the complement of the angle A, without a secant; thus,

AC: R:: AD: S. ACD.

44: 90° 40.23 : 58° 52'

90-58° 52' 31° 08', the angle A.

Then by theo. 1, of this sect.

BC : S. A :: AC : S. B.

34: 310 08′ : : 47: 45° 37/

By cor. 1, theo. 5, sect. 4, 1800-the sum of A and B=C.
A 31° 08′

B 45 37

1800-76 45-1030 15', the angle C.

3d. By Gunter's Scale.

The first proportion is extended on the line of numbers; and it is no matter whether you extend from the first to the third, or to the

second term, since they are all of the same kind: if you extend to the second, that distance applied to the third will give the fourth; but if you extend from the first to the third, that extent will reach from the second to the fourth.*

The methods of extending the other proportions have been already fully treated of.

RULE 2.

Either of the angles, as A, may be found by adding together the arithmetical complements of the Logarithms of the two sides AB, AC, containing the required angle, the Log. of the half sum of the three sides, and the Log. 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 co-sine, of half the required angle. It is required to find the angle A, in the last problem, by this rule, the sides remaining the same.

If the other angles were required, they can be found by Case 1, or by Theo. 1, of this Sect.

RULE 3.‡

Add the three sides together, and take half the sum, and the differences betwixt the half sum and each side: then add the com

The reader is referred to Hutton's Math. Vol. 2. N. Y. edition, for the method of investigating Plane Trigonometry analytically.

The demonstration of this rule is evident from Theo. 5, and the nature of Logarithms; but in working the proportion by Logarithms, we omit the Log. of the square of Radius or 20, which is just equivalent to rejecting 20 from the sum of the four Logarithms, which should be done, because, for every arithmetical complement that is taken, 10 must be rejected: but the Ar. Co. of the two sides, containing the required angle, is taken; consequently 20 should be rejected, which is equal to the Log. of the square of radius.

For the demonstration of this rule the reader is referred to Leslie's Geometry,

plements of the Logarithms of the half sum, and of the difference between the half sum and the side opposite to the angle sought, to the Logarithms of the differences of the half sum and the other sides: half their sum will be the tangent of the angle required.

Example. In the triangle ABC, having the side AB 562, AC 800, and BC 320, to find the angle ABC.

Whose double 128° 4' is the angle ABC. Whence the other angles can be easily found by theo. 1. of this section.

* An example in each case of oblique-angled triangles.

1. In the triangle ABC, having AB 106, AC 65, and the angle B 31° 49', to find the 4s A and C, and the side BC.

Ans. The C=59° 17′ or 120° 43', the LA 27° 28′ or 88° 54', and the side BC=43.2 or 123.2.

2. In the triangle ABC, having the side AB 2200, the 4 A 35°, and the LB 47° 24', to find the sides AC and BC, and the C. Ans. The C 97° 36', the side AC 1636, and the side BC 1272.

3. In the triangle ABC, having the side AB 240, AC 263.7, and the angle A 46° 30', to find the other angles and the side BC.

Ans. The LC 60° 31', the ▲ B 72° 59′, and the side BC 200.

4. In the triangle ABC, having the sides given, viz. AB=144.8, BC=109, and AC=76, it is required to find the angles by each of the three rules given to case 4.

Ans. The least angle 29° 49′, next greater 54° 07′, and the greatest 96° 04'.

Additional Exercises with their Answers.

QUESTIONS FOR EXERCISE.

1. Given the hypothenuse 108, and the angle opposite the perpendicular 25° 36'; required the base and perpendicular. Ans. The base is 97.4, and the perpendicular 46.66.

2. Given the base 96, and its opposite angle 71° 45'; required the perpendicular and the hypothenuse.

Ans. The perpendicular is 31.66, and the hypothenuse 101.1.

3. Given the perpendicular 360, and its opposite angle 58° 20; required the base and the hypothenuse.

Ans. The base is 222, and the hypothenuse 423.

4. Given the base 720, and the hypothenuse 980; required the angles and the perpendicular.

Ans. The angles are 47° 17′ and 42° 43', and the perpendicular 664.8.

5. Given the perpendicular 110.3, and the hypothenuse 176.5; required the angles and the base.

Ans. The angles are 38° 41′ and 51° 19′, and the base 137.8.

6. Given the base 360, and the perpendicular 480; required the angels and the hypothenuse.

Ans. The angles are 53° 8' and 36° 52', and the hypothenuse 600.

7. Given one side 129, an adjacent angle 56° 30', and the opposite angle 81° 36'; required the third angle and the remaining sides. Ans. The third angle is 41° 54', and the remaining sides are 108.7 and 87.08.

8. Given one side 96.5, another side 59.7, and the angle opposite the latter side 310 30'; required the remaining angles and the third side.

Ans. This question is ambiguous; the given side opposite the given angle being less than the other given side (see Rule 1.); hence, if the angle opposite the side 96.5 be acute, it will be 57° 38', the remaining angle 90° 52', and the third side 114.2; but if the angle opposite the side 96.5 be obtuse, it will be 122° 22', the remaining angle 26° 8', and the third side 50.32.

9. Given one side 110, another side 102, and the contained angle 113° 36'; required the remaining angles and the third side. Ans. The remaining angles are 34° 37′ and 31° 47', and the third side is 177.5.

10. Given the three sides respectively, 120.6, 125.5, and 146.7 ; required the angles.

Ans. The angles are 51° 53′, 54° 56′, and 73° 9'.

The student, who has advanced thus far in this work with diligence and active curiosity, is now prepared to study, with ease and pleasure, the following part; which comprehends all the necessary