CASE V

The perpendicular and hypothenuse given, to find the angles and base.

PL. 5. fig. 8.

In the triangle ABC there is BC 306, and AC 370 given; to find the angles A and C; and the base AB.

1st. By Construction.

Draw a blank line from any point, in which, at B, erect a perpendicular, on which lay BC 306, from a scale of equal parts: from the same scale, with AC 370, in the compasses, cross the first drawn blank line in A, and the triangle ABC is constructed.

Measure the angle A (by prob. 17. sect. 4.); and also AB, from the same scale of equal parts the other sides were taken from, and the answers are now found.

The operations by calculation, the square root, Gunter's scale, and Natural sines, are here omitted, as they have been heretofore fully explained: the statings, or proportions must also be obvious, from what has already been said.

Answers; The angle A 55° 48'; therefore the angle C 34° 12', and AB 208.

CASE VI.

The base and perpendicular given; to find the angles and hypothenuse.

PL. 5. fig. 9.

In the triangle ABC, there is AB 225, and BC 272, given; to find the angles A and C, and the hypothenuse AC.

1st. By Construction.

Draw a blank line, on which lay AB 225, from a scale of equal parts; at B, erect a perpendicular; on which lay BC, 272, from the same scale: join A and C, and the triangle is constructed.

As before, let the angle A, and the hypothenuse AC be measured; in order to find the answers.

2d. By Calculation.

1. Making AB the radius.

AB: R:: BC: T. A.

R.: AB:: sec. A: AC.

2. Making BC the radius.

BC: R::AB: T. C.

R.: BC:: Sec. C: AC.

By calculation; the answers from the foregoing proportions are easily obtained, as before.

But because AC, by either of the said proportions is found by means of a secant; and since there is no line of secants on Gunter's scale; after having

found the angles, as before, let us suppose AC the radius, and then

1. S. A: BC : : R. : AC.

or 2. S. C: AB:: R: AC.

These proportions may be easily resolved, either by calculation or Gunter's scale, as before; and thus the hypothenuse AC may be found without a

secant.

From the two given sides, the hypothenuse may be easily obtained, from cor. 1. theo. 14. sect. 4.

=

Thus the square of AB 50625
Add the square of BC= 73984

124609 (353 = AC

9

65)346

325

703)2109
2109

From what has been said on logarithms, it is plain,

1. That half the logarithm of the sum of the squares of the two sides, will be the logarithm of the hypothenuse. Thus,

The sum of squares, as before, is 124609; its log. is 5.095549, the half of which is 2.547774;

and the corresponding number to this, in the tables, will be 353, for AC.

2. And that half of the logarithm of the difference of the squares of AC and AB, or of AC and BC, will be the logarithm of BC, or of AB. The following examples are inserted for the exercise of the learner.

the angle C'64° 40′ SAB
AC 3876 required.

1. Given, }

2. Given, }

the angle C 47° 20'

AB 17

{ A Crequired.

3 Given, }

the angle C 28° 30' AB
BC 27187 Acrequired.

The answers are omitted, that the learner may resolve them for himself by the foregoing methods; by which means he will find and see more distinctly their mutual agreements: and become more expert, and better acquainted with the subject.

OBLIQUE ANGLED

PLANE TRIGONOMETRY.

BEFORE

EFORE we proceed to the solution of the four cases of Oblique angled triangles, it is necessary to premise the following theorems.

THEO. I.

PL. 5. fig. 10.

In any plane triangle ABC, the sides are proportional to the sines of their opposite angles; that is, S. C : AB : : S. A: BC, and S. C: AB :: S. B: AC; also S. B: AC:: S. A: BC.

By. theo. 10. sect. 4. the half of each side is the sine of its opposite angle; but the sines of those angles, in tabular parts, are proportional to the sines of the same in any other measure; and therefore the sines of the angles will be as the halves of their opposite sides; and since the halves are as the wholes, it follows, that the sines of their angles are as their opposite sides; that is, S. C: AB:: S. A: BC, &c. 2. E. D.

THEO. II.

Fig. 11.

In any plane triangle ABC, the sum of the two given sides AB and BC, including a given angle ABC, is to their difference, as the tangent of half the sum of the two unknown angles A and C is to the tangent of half their difference.

Produce AB and make HB-BC, and join HC: let fall the perpendicular BE, and that will bisect