Examples of the solution of right-angled triangles.

1. In the triangle BAC (fig. 13), the hypothenuse a and a Fig. 13. side c being given, to find the angle C, opposite to this side; let the hypothenuse a = 13chains,178, and the side c=7ch,357. We have, in order to determine sin C, the proportion

To render the calculation more simple we almost always make radius equal to unity; its logarithm is then zero, and no account need be taken of it; and, instead of performing the subtractions, we employ the arithmetical complements, the theory of which has already been explained (Alg. 248). Thus

1 c = 17,357 =

..0,8667008 arith. comp. la arith. comp. 1 13,178 = 8,8801505

sum or 1 sin C=...

I

9,7468513

which in the tables answers to 33° 56′ 13′′ C.

2. The angle C 52° 21′ 59′′, the hypothenuse a = 33ch,253, being given, to find the side b. We have

whence

R: sin B or cos C::a: b, (31),

b =

a X cos C
R

lbla+1 cos CIR la + 1 cos C.

which answers in the tables to 20ch,,228 b to within one thousandth.

3. The side c5ch,391, the angle B 31° 30' 44", being given, to find the side b. We have

Fig. 16.

which answers in the tables to 3ch,305 = c.

Examples of the solution of oblique-angled triangles.

1. In the triangle ABC (fig. 16), the side c, and theangles A and B, being given, to find the side b.

Let A

112° 30′ 21′′, B = 52° 54′ 40′′, c = 27ch,348; the angle C will be

180° — (A + B) = 180° 165° 25′ 01′′ 14° 34′ 59′′, and we have

arith. comp. I sin Carith. comp. I sin 14° 34′ 59′′ = 0,599633

which answers in the tables to 86ch,642 = b.

2. In the triangle ABC the two sides a, b, and the included angle C, being given, to find the third side c.

Let a = 28ch,442, b = 17ch,803, C = 75° 50′. We begin with finding the other angles by means of the proportion (35)

In order to determine the side c, we have the proportion

arith. comp. I sin Barith. comp. sin 35° 37′ 45′′ = 0,2346766

sum or 1 c=

which answers in the tables to 29ch,632 = c.

3. In the triangle ABC, the three sides a, b, c, being known, to find the angle A.

Let a = 29ch,037, b = 18ch,743, c = 13ch,782.

According to art. 38, we add the three sides a, b, c, together, which gives 61,562; and from half of the sum 30,781, we subtract successively b, c; the remainders are 12,038 and 16,999, we have then

sum, or 1 sin A =

9,9494132

the half which answers in the table to 62° 52′ 55′′ = A, therefore A 125° 45′ 30′′.

40. A work of this nature does not admit of a particular ac

count of the applications of plane trigonometry. I shall confine myself to the solution of three questions, which may be regarded as the basis of the art of drawing plans.

The first is, having given in magnitude and position upon a Fig. 17. plane a right line AB (fig. 17), to determine the position of a point C, situated in the same plane, or which is the same thing, to find the distances AC and BC.

In order to resolve this question, the side AB, which is the base of the operation, must be measured, as also the angles CAB, CBA, comprehended between this base and the lines, which connect the extremities with the point C; the distances sought, AC and BC, may be calculated by the rule laid down in art. 32; and these being known, the triangle ABC may be constructed by means of a scale of equal parts, and the relative position of the three points A, B, C, may be calculated.*

We can then, by the resolution of the right-angled triangle ACP, in which the side AC and the angle CAP are known, find the length of the perpendicular CP let fall upon AB, or of the shortest distance of the point C from AB, and the length of the segment AP. By means of these the position of the point C with respect to the line AB is determined. The situation of the point D may also be found, if it can be perceived from any two of the points A, B, and C.

41. When we have determined immediately the point D with respect to the line AB, by measuring the angles DAB, DBA, we have every thing, which is necessary in order to compute the distance of the points C and D with respect to each other; for, having resolved the triangle DAB, as also the triangle CAB, by subtracting the angle DAB from the angle CAB, we have, in the triangle CAD, the two sides AC and AD, and the included angle CAD; by applying the rules of art. 35, we shall obtain the two other angles DCA, CDA, and the third side CD, which

* I do not insist upon the angles being measured, since more might be learned by a sight of the instruments, which are employed for this purpose, than by any thing which I can say on the subject. To conceive of the possibility of performing this operation, it is sufficient to imagine, that there is placed at the centre C, a sector of a circle, the radii of which correspond to the direction of the sides AB and AC, which contain the angle to be measured.

is the distance sought. The angle DCA gives the position of the right line CD; and if we consider AC, as secant, by comparing the angles DCA and CAB, we shall be able to find the inclination of CD with respect to AB.

If we set out from the points C and D, taking CD as a new base, we may determine other points not visible from the first two, A and B, and by proceeding in this manner we can determine the relative position of all the points of a country. It is in this manner that the map of France was constructed under the direction of Cassini.

42. The second question, which I am to consider, is merely the first rendered more general by supposing the point to be determined, situated without the plane, in which the line AB is found. Let C(fig. 18) be this point, and ABC' the plane in which Fig. 18. AB is situated. The position of the point C will be known, if we have that of the foot C' of the perpendicular let fall from this point upon the plane ABC', and the length of the perpendicular CC', which shows how much the point C is elevated above C', its projection. In this case the angles C'AB and C'BA are not the angles to be measured, but we take, instead of them, the angles CAB and CBA, situated in the plane CAB passing through the lines AC and BC, which are drawn from the given points A and B to the point required; and in order to fix the position of this plane, we measure also the angle DBC, which the line CB makes with the line BD perpendicular to the plane ABC', and consequently parallel to the right line CC.* We resolve the triangle CBA, as in the preceding article, the same things being given; then in the triangle C'BC right-angled at C', knowing the hypothenuse CB and the angle CBC, which is the difference between the right angle DBC' and the measured angle DBC, we calculate the sides C'C', C'B. The first is the height of the point C above the plane C'AB, and is used in connexion

* When the question relates to points situated on the earth's surface, we take the plane of the horizon for the plane ABC', the lines CC' and BD are then vertical, and their direction is given by the plumb-line: the plane C'CB, which passes through these lines, is vertical, and is determined by the line DB and the point C, which is seen from the point B. The line C'B is horizontal and comprehended in the same plane.