By theo. 3. of this sect.

As the base or the longest side, AB

64

is to the sum of the other sides, AC and BC, 81

To half the base, or to half the sum
of the segments AD and DB.
Add half their difference, now found,

32

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 AC47, 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 90° : 47: 31° 08′

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

AC: R:: ADS. 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 31° 08′ 47: 45° 37'.

By cor. 1. theo. 5. sect. 4. 180°-the sum of A and B=C.

A 31°.08'
B 45. 37

180°-76.45-103°. 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 reader is referred to Hutton's Math. Vol. 2. N. Y. Edition, for the me thod 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.

the angle A, in the last problem, by this rule, the sides remaining the same.

Cos. A. 15° 34'
14

2

9.983760

Whose double 31° 08′ is the angle A.

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 complements 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.

* For the demonstration of this rule the reader is referred to Leslie's Geometry, prop. 12. page 372.

1

2

Tang of 64° 2' = sum 10.312431 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 angles LSA and C, and the side BC.

A

Ans. The C = 59° 17′ or 120° 43′, the 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 A 35°, and the ▲ B 47° 24'; to find the sides AC and BC, and the 4 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 C 60° 31', the ▲ B 72° 59′, and the side BC 200.

4. In the triange 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.

Answer. The Base is 97.4, and the Perpendi cular 46,66.

2. Given the Base 96 and its opposite Angle 71° 45'; required the Perpendicular and the Hypothenuse.

Answer. 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.

Answer. The Base is 222, and the Hypothenuse 423.

4. Given the Base 720 and the Hypothenuse 980; required the Angles and the Perpendicular. Answer. 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.

Answer. The Angles are 38° 41′ and 51° 19′, and the Base 137.8.

6. Given the Base 360 and the Perpendicular 480; required the Angles and the Hypothenuse. Answer. The Angles are 53° 8' and 36° 52′, and the Hypothenuse 600.