cosine that corresponds to tan C, and then to compute CB from the proportion cos C:R:: AC: CB.

PROB. II.

Solution of oblique-angled triangles.

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

CASE I.

Two angles A and B, and one side 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 (Prop. 2),

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

Sin C: sin B::AB: AC,

B

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 side 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: sin A:: AC: CB, by Case 1.

A

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 (Art. 9). 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, and B is an acute angle, 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 arc 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 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 angle AC'C, and, therefore, AC'B, which is the supplement of AC'C, 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 (I. 32), 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 ACC and ABC, because AC'C, that is, ACC' is equal to the sum of the angles ABC, BAC (I. 32). 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, the side adjacent to the given acute angle, is greater than AC, the side opposite to it, there are two triangles which satisfy the conditions of the question. But when AČ 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.

CASE III.

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.

SOLUTION.

First, By Prop. V., AB+AC: AB-AC:: tan (C+B) : tan (C-B). Then, since (C+B) is known being the complement of ≥ A; } (C—B), and thence B and C may be found. For C = (C+B) + (C—B), and B=3 (C+B)—1⁄2 (C—B) (Lemma).

To find BC.

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

But BC may also be found without seeking for the angles B and C; for, when the radius is unity, BC=√[AB2-2 cos Ax AB.AC+AC2] (Prop. 3, Cor. 1).

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.

CASE IV.

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

SOLUTION I.

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 (Prop. 4). 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 DC, and their difference being given, BD and DC are found. (Lemma).

Then (Prop. 1) CA: CD::R: cos C; and BA: BD::R: cos B; wherefore C and B are found, and, consequently, A.

SOLUTION 11.

Let D be the difference of the sides AB, AC. Then (Prop. 6, Cor.) 2 (AB.BC): √[(BC+D) (BC-D)]:: R : sin BAC.

SOLUTION 111.

Let S be the sum of the sides BA and AC. Then (Prop. 7, Cor. 1) 2√(AB.AC): √[(S+BC) (S-BC)]:: R: cosBAC.

SOLUTION IV.

S and D retaining the same significations as above (Prop. 7, Cor. 2), [(S+BC) (S-BC)]: [(BC+D) (BC-D)]: R tan BAC; or, p(p—a): √(p—b) (pc): R: tan Ã.

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 is very material to be considered. The reason is, that the sines of angles, which are nearly =0°, or the cosines of angles, which are nearly 90°, vary very little for a considerable variation in the corresponding angles, as may be seen from a table of sines and cosines. The consequence of this is, that when the sine or cosine of such an angle is given (that is, a sine or cosine nearly equal to the radius), the angle itself cannot be very accurately found. If, for instance, the natural sine .9998500 is given, it will be immediately perceived from the tables, that the arc corresponding is between 89° and 89° 1'; but it cannot be found true to seconds, because the sines of 89° and of 89° 1' differ only by 50 (in the two last places), whereas the arcs themselves differ by 60 seconds. Two arcs, therefore, that differ by 1", or even by more than 1", have the same sine in the tables, if they fall in the last degree of the quadrant.

The fourth solution, which finds the angle from its tangent, is not liable to this objection; nevertheless, when an arc approaches near to 90°, the variations of the tangents become very great, and too irregular to allow the proportional parts to be found with exactness, so that when the angle sought is extremely obtuse, and its half of consequence very near to 90°, the third solution is the

best.

It may always be known, whether the angle sought is greater or less than a right angle by the square of the side opposite to it being greater or less than the squares of the other two sides.

It may be useful to have all the solutions of the oblique-angled triangle reduced to a form purely arithmetical, not requiring the inspection of a diagram, and brought together in one table.

Let A, B, C be the angles of the triangle, and a, b, c the sides respectively opposite to them.

SECTION V.

PROBLEMS.

I. Problems on angles.

1. The distance of the sun from the earth is 95 millions of miles, and his mean apparent diameter seen from the earth, and considered as a circular arc, is 32'. What is the sun's diameter in miles?

By Art. 31, the circumference of the circle, of which the radius is the distance of the sun from the earth, is 3.1416×2×95,000,000= 596,904,000,000, and we have 360×60: 32:: 596,904,000,000 : 884,302 miles for the diameter of the sun, regarded as a circular The linear diameter is actually about 883,000 miles.

arc.

2. How many miles is the earth carried round the sun in an hour. Ans. 68,000.

3. The moon's diameter is 2160 miles, and her apparent size is, on the average, just the same as that of the sun; what is her distance from the earth?

By Art. 3, we have 883,000: 2160: 95,000,000: 232,396 miles. The mean distance of the moon from the earth is actually about 237,000 miles, or 60 radii of the earth.

4. The parallax of the nearest fixed star a Centauri, or the angle which the earth's distance from the sun subtends at it, is very nearly 1". Required the distance of this star from the earth.

We have, 1: 360 × 60 × 60 :: 95,000,000: circumference of circle, of which the radius is the distance of the star= 123,120,000,000,000, which, divided by 2 and by 3.1416, gives nearly 19,600,000,000,000, or about 20 millions of millions of miles as the distance of the nearest fixed star from the earth or sun. We conclude that there is no star so near to the sun as within 6 or 7000 times the distance of the planet Neptune.

5. Civil engineers allow 8 inches of depression to the mile, on account of the curvature of the earth. What do they suppose the earth's diameter to be?

If, in the figure (III. 32) CF be drawn perpendicular to BF; CF will be the depression, provided BC be a mile. Now, the triangles CBF, CAB will be similar; hence,

CF: BC BC : BA, i. e.

:

8 in. 63360 in. : : 1 mile: 7920 miles, the earth's diameter required.