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REMARKS on the Solutions in the TABLE.

In the second case, when AC and C are given to find the hypotenuse BC, a solution may also be obtained by help of the secant, for CA: CB : :1: Sec. C.; if, therefore, this proportion be made 1 : Sec.C:: AC : CB, CB- will be found.

In the third case, when the hypotenuse BC and the fide AB are given to find AC, this may be done either as dire&ed in the Table, or by the 47th of the Ift; for fince AC BC? - BA, AC EN BC? — BA?. This value of AC will be easy to calculate by logarithms, if the quantity BC?-BA? be separated into two multipliers, which may be done; be.

cause (Cor., 5. 2.), BC? — BA? =(BC + B) (BC-BA). Vism. Therefore AC=N(BC + BA) (BC – BAС.).

When AC and AB are given, BC may be found from the 47th, as in the preceding instance, for BC=NBA? + AC. But BA? + AC? cannot be separated 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 cases to seek first 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 sufficient, 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:: AG:CB.

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PROBLEM II.

IN

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 cases, in each of which the folution depends on some of the foregoing propofitions,

CA SE I.

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

SOLUTION.

Because the angles A and B are given, C is also given, being the supplement of A +B; and, (2.) Sin C : fin A :: AB: BC; also

А
Sin C:fio B:: AB: AC.

B

CASE

CASE II.

Two sides AB and AC, and the angle B opposite 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 :: sin B : sin C. Also, A=180° — B-C; and then, sin B : fin A :: AC: CB, by Case 1.

In this case, the angle C may have two values ; for its fine being found by the proportion above, the angle belonging to that line, may either be that which is found in the tables, or it may be the supplement of it, (Cor. def. 4.). This ambi. guity, however, does not arise from any defect in the solution, but from a circumstance effential to the problem, viz. that whenever AC is less than AB, there are two triangles which have the sides AB, AC, and the angle at B of the same magnitude in each, but which are nevertheless unequal, the angle opposite to AB in the one, being the supplement of that which is opposite to it in the other. The

A truth of this appears by describing from the centre A with the radius AC, an arch intersecting BC in Cand C'; then, if AC and AC' be drawn, it is evident that the tri

B angles ABC, ABC'have 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 supplements of one another. For the triangle CAC' is isosceles, and the angle ACC = the ACC, and therefore, AC'B, which is the supplement of ACʻC, is also 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 supplement of the two angles ACB, ABC, (31. 1.), and therefore its fine is the fame with the fine of the sum of ABC and ACB. But BAC' is the difference of the angles ACB, ABC; for it is the difference of the angles AC'C and ABC, (because AC'C, that is, ACC' is equal to the sum of the angles ABC, BAC, (32.1.)). Therefore to find BC, having found C, make sin C: fin (C+B):: AB: BC; and again, sin C:fin (C-B):: AB : BC'.

Thus, when AB is greater than AC, and C confequently greater than B, there are two triangles

which satisfy the conditions of the question. But when AC is greater than AB, the intersections C and C fall on opposite sides of B, so that the two triangles have not the same angle at B common to them, and the folution ceases to be ambiguous, the angle required being necessarily 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 Band C, and also the fide BC,

SOLUTION.

First, make AB+AC: AB-AC :: tan (B+C): tan Ź (B-C.). Then, fince (B+C) and (B-C) are both given, B and C may be found. For B= (B+C)+(B-C), and C= (B+C) - (B-C.)

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Having found B, make sin B : fin A:: AC : BC.

But BC may also be found without seeking for the angles B and C; for BC =N AB* — 2 cof AX AB.AC + AC, 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 separated into simple multipliers. Therefore, when AB and AC are expressed by large numbers, the other solution, by finding the angles, and then computing BC is preferable.

CASE IV,

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

8.

SOLUTION I.

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

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