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The reason of the restriction in this problem to those cases in which at least one side is given, is evident from this, that by the angles alone being given, the magnitudes of the sides are not determined. Innumerable triangles, equiangular to one another, may exist, without the sides of any one of them being equal to those of any other; though the ratios of their sides to one another will be the same in them all, (4. 6.) If, therefore, only the three angles are given, nothing can be determined of the triangle but the ratios of the sides, which may be found by trigonometry, as being the same with the ratios of the sines of the opposite angles.

For the conveniency of calculation, it is usual to divide the general problem into two; according as the triangle has, or has not, one of its angles a right angle.


In a right angled triangle, of the three sides and three angles, any two being given, besides the right angle, and one of those two being a side, it is required to find the other three.

It is evident, that when one of the acute angles of a right angled triangle is given, the other is given, being the complement of the former to a right angle; it is also evident that the sine of any of the acute angles is the cosine of the other.

This problem admits of several cases, and the solutions, or rules for calculation, which all depend on the first Proposition, may be conveniently exhibited in the form of a Table; where the first column contains the things given; the second, the things required; and the third, the rules or proportions by which they are found.

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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: CBR: sec. C.; if, therefore, this proportion be made R : sec. C: AC: CB, CB will be found.

In the third case, when the hypotenuse BC and the side AB are given to find AC, this may be done either as directed in the Table, or by the 47th of the first; for since AC2=BC2 —BA2, AC=

BC-BA. This value of AC will be easy to calculate by logarithms, if the quantity BC2-BA2 be separated into two multipliers, which may be done; because (Cor. 5. 2.), BC2—BA2=(BC+BA) (BC-BA). Therefore AC=√(BC+BA) (BC—BA).

When AC and AB are given, BC may be found from the 47th, as in the preceding instance, for BC=/BA+AC3. But BA2+AC2 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, ACAB:: R: tan. C; but if C itself is not required, it is sufficient, having found tan. C by this proportion, to take from the Trigonometric Tables the cosine that corresponds to tan. C, and then to compute CB from the proportion cos. CR AC: CB.


In an oblique angled triangle, of the three sides and three angles, any three being given, and one of these three being a side, it is required to find the other three.

This problem has four cases, in each of which the solution depends on some of the foregoing propositions.


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


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

Sin. C sin. A :: AB : BC; also,


Sin. C sin. B:: AB: AC.

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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 side BC.


The angle C is found from this proportion, AC: AB sin B: sin C. Also, A=180°-B-C; and then, sin B: sin A :: AC: CB, by Case 1.

In this case, the angle C may have two values; for its sine being found by the proportion above, the angle belonging to that sine may either be that which is found in the tables, or it may be the supplement of it, (Cor. def. 4.). This ambiguity, however, does not arise from any defect in the solution, but from a circumstance essential 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 truth of this appears by describing from the centre A with the radius AC, an arch intersecting BC in C



and C'; then, if AC and AC' be drawn, it is evident that the triangles ABC, ABC' have the side AB and the angle at B common, and the sides AC and AC' equal, but have not the remaining side of the one equal to the remaining side of the other, that is, BC to BC', nor their other angles équal, 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 AC'C, and therefore, AC'B, which is the supplement of ACC, is also the supplement 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, (32. 1.), and therefore its sine is the same with the sine 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: sin (C+B): AB BC; and again, sin C: sin (C-B): AB: BC'.


Thus, when AB is greater than AC, and C consequently 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 solution ceases to be ambiguous, the angle required being necessarily less than B, and therefore an acute angle.


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


First, make AB+AC: AB-AC :: tan (C+B): tan (C-B.). Then, since (C+B) and (C-B) are both given, B and C may be found. For B1 (C+B)+} (CB), and C=1(C+B)−1(C—B.). (Lem. 2.).

To find BC.

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

But BC may also be found without seeking for the angles B and C ; for BC AB2-2 cos AXAB.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 sign 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.


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


Take F such that BC: BA+AC: BA-AC: F, then F is either the sum or the difference of BD, DC, the segments of the base, (5.). If F be greater than BC, F is the sum, and BC the difference of BD, DC; but, if F be less than BC, BC is the sum, and F the difference of BD and DC. In either case, the sum of BD and DG, and their difference being given, BD and DC are found. (Lem. 2.)

Then, (1.) ČA: CD :: R: cos. C; and BA: BD :: Ŕ: cos. B; wherefore C and B are given, and consequently A,



Let D be the difference of the sides AB, AC. Then (Cor. 7.) 2✔✅AB.AC: ✔ (BC+D)(BC—D) :: R: sin↓ BAC.


Let S be the sum of the sides BA and AC. Then (1 Cor. 8.) 2 ✔ AB.AC : ✔ (S+BC) (S—BC) : : R : cos ¦ BAC.


S and D retaining the significations above, (2 Cor. 8.) ✓ (S+BC) (S—BC): ✔ (BC+D)(BC—D) :: R : tan BAC.

It may be observed of these four solutions, that the first has the advantage of being easily remembered, but that the others are rather more expeditious in calculation. The second solution is preferable to the third, when the angle sought is less than a right angle; on the other hand, the third is preferable to the second, when the angle sought is greater than a right angle; and in extreme cases, that is, when the angle sought is very acute or very obtuse, this distinction

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