Page images
PDF
EPUB

REMARKS on the Solutions in the TABLE.

In the fecond cafe, when AC and C are given to find the hypotenufe BC, a folution may also be obtained by help of the fecant, for CA: CB:: 1: Sec. C.; if, therefore, this proportion be made 1: Sec. C:: AC: CB, CB will be found.

In the third cafe, when the hypotenufe BC and the fide AB are given to find AC, this may be done either as direced in the Table, or by the 47th of the 1ft; for fince AC= BCBA, AC/BCBA. This value of AC will be easy to calculate by logarithms, if the quantity BC2-BA be feparated into two multipliers, which may be done; becaufe (Cor., 5. 2.), BC2 — BA2= (BC + BĂ) (BC — BA), V. Therefore AC(BC+BA) (BC-BAC.).

When AC and AB are given, BC may be found from the 47th, as in the preceding inftance, for BCBA2 + AC2. But BA2 + AC cannot be feparated into two multipliers; and therefore, when BA and AC are large numbers, this rule is inconvenient for computation by logarithms. It is best in such cafes to feek firft for the tangent of C, by the analogy in the Table, AC: AB:: R: tan C; but if C itfelf is not required, it is fufficient, having found tan C by this proportion, to take from the Trigonometric Tables the co-fine that corresponds to tan C, and then to compute CB from the proportion cof C: R :: AC : CB.

CASE

IN

PROBLEM II.

N an oblique angled triangle, of the three fides and three angles, any three being given, and one of these three being a fide, it is required to find the other three.

This problem has four cafes, in each of which the folution depends on fome of the foregoing propofitions.

CASE I.

Two angles A and B, and one fide AB, of a triangle ABC, being given, to find the other fides.

SOLUTION.

Because the angles A and B are given, C is alfo given, being the fupplement of A+ B; and, (2.)

Sin C: fin A :: AB: BC; alfo

Sin C fin B:: AB: AC.

A

[ocr errors][merged small][merged small][merged small]

CASE II.

Two fides AB and AC, and the angle B oppofite to one of them being given, to find the other angles A and C, and also the other fide BC.

SOLUTION.

The angle C is found from this proportion, AC : AB : : fin B: fin C. Alfo, A180°-B-C; and then, fin B: fin A: AC: CB, by Cafe 1.

In this cafe, the angle C may have two values; for its fine being found by the proportion above, the angle belonging to that fine, may either be that which is found in the tables, or it may be the fupplement of it, (Cor. def. 4.). This ambi guity, however, does not arife from any defect in the folution, but from a circumftance effential to the problem, viz. that whenever AC is lefs than AB, there are two triangles which have the fides AB, AC, and the angle at B of the fame magnitude in each, but which are nevertheless unequal, the angle oppofite to AB in the one, being the fupplement of that which is oppofite to

it in the other. The truth of this appears by defcribing from the centre A with the radius AC, an arch interfecting BC in C and C'; then, if AC and AC' be drawn, it is evident that the triangles ABC, ABC' have

B

A

the fide AB and the angle at B common, and the fides AC and AC' equal, but have not the remaining fide of the one equal to the remaining fide of the other, that is, BC to BC', nor their other angles equal, viz. BC ́A to BCA, nor BAC' to BAC. But in these triangles the angles ACB, AC ́B are the fupplements of one another. For the triangle CAC' is ifofceles, and the angle ACC'the AC'C, and therefore, AC'B, which is the fupplement of AC'C, is alfo the fupple

ment

ment of ACC' or ACB; and these two angles, ACB, AC'B are the angles found by the computation above.

From these two angles, the two angles BAC, BAC' will be found the angle BAC is the fupplement of the two angles ACB, ABC, (31. 1.), and therefore its fine is the fame with the fine of the fum of ABC and ACB. But BAC' is the difference of the angles ACB, ABC; for it is the difference of the angles ACC and ABC, (because AC'C, that is, ACC' is equal to the fum of the angles ABC, BAC, (32.1.)). Therefore to find BC, having found C, make fin C: fin (C+B)::AB: BC; and again, fin C : fin (C—B) :: AB: BC'.

Thus, when AB is greater than AC, and C confequently greater than B, there are two triangles which fatisfy the conditions of the question. But when AC is greater than AB, the interfections C and C' fall on oppofite fides of B, fo that the two triangles have not the fame angle at B common to them, and the solution ceases to be ambiguous, the angle required being neceffarily less than B, and therefore an acute angle.

CASE III.

Two fides AB and AC, and the angle A, between them, being given, to find the other angles B and C, and also the side BC.

SOLUTION.

Firft, make AB+ AC: AB-AC :: tan (B+C):

tan (B-C.). Then, fince (B+C) and (B-C) are both given, B and C may be found. For B—1⁄2 (B+C) + 1⁄2 (B—C), and C (B+C) — ÷ (B—C.)

V.222

Το

To find BC.

Having found B, make fin B: fin A:: AC: BC. But BC may also be found without seeking for the angles B and C; for BCAB-2 cof A X AB.AC + AC2, Prop. 6.

This method of finding BC is extremely useful in many geometrical investigations, but it is not very well adapted for computation by logarithms, because the quantity under the radical fign cannot be feparated into fimple multipliers. Therefore, when AB and AC are expreffed by large numbers, the other folution, by finding the angles, and then computing BC is preferable.

CASE IV,

The three fides AB, BC, AC, being given, to find the angles A, B, C.

SOLUTION I.

Take F fuch that BC: BA+AC: BA-AC: F, then Fis either the fum or the difference of BD, DC, the fegments of the base, (5.) If F be greater than BC, F is the fum, and BC the difference of BD, DC; but, if F be less than BC, BC is the fum, and F the difference of BD and DC. In either cafe, the fum of BD and DC, and their difference being given, BD and DC are found.

Then,

« PreviousContinue »