This is one equation between the unknown quantities x, x'. Another is easily obtained; for since the four angles of the quadrilateral ABCD make up four right angles or 360°, we have x + x' + a + a' + ACD + BCD = 360°; the sum of the two latter angles may become known, since in the triangle ABC the angle c is determinable from the three given sides; therefore all the terms in the first member of this equation are known except x and x'. Call the sum of the known quantities B, and we shall thus have x'=ß-x, and, consequently by substitution, equation (2) becomes

The first term of this second member may be easily calculated by logarithms, and this added to the natural cotangent of ẞ gives the nat. cot. of x, and thence x' is known from the equation x' = ẞx, and CD from either of the equations (1).

This problem has a useful application in the survey of harbors.

Let the angles be taken with a sextant, from a boat, at a point where a sounding is made, to three stations on the shore. After having drawn upon a map the triangle, of which these three stations are the vertices, the following simple and elegant construction will determine the point where the sounding was made.

Upon the line joining two of the stations, on the map, make a segment, capable of containing the angle observed from the place of sounding, and subtended by this line (Plane Geom., Prob. 21); upon a line joining one of these two stations and the

third, make another segment that will contain the angle observed to be subtended by this last line, and the intersection of the arcs of these two segments will determine the point on the map, corresponding to that at which the sounding was made.

PROBLEM IV.

B

Given the angles of elevation of an object taken at three places on the same horizontal straight line, together with the distances between the stations; to find the height of the object and its distance from either station. Let AB be the object, and c, c', c'', the three stations, then the triangles BCA, BCA, BC''A, will all be right angled at a; and, therefore, to radius BA, AC, AC', AC'', will be the tangents of the angles at B, or the cotangents of the angles of elevation; hence, putting a, a', a'', for the angles of elevation, z for the height of the object, and a, b, for the distances cc', c'c'', we shall have

in order to eliminate c'r, multiply the first by b, the second by a, and add, and we shall have

x2 (b cot? a+ a cot3 a'') = (a + b) x2 cot2 a' + nb (a + b)

.. I=

ab (a + b)

b cota + a cot2 a'' —(a + b) cot2 a'

If the three stations are equidistant, then a = b, and the expression becomes

The height A B being thus determined, the distances of the stations from the object are found by multiplying this height by the cotangents of the angles of elevation.

EXAMPLE.

22 Given the hypothenuse a = 6512•4 yards, b= 6510.6, to find c

Upon inspecting the tables that are calculated to seven places of decimals only, it will be seen that, when the angles become very small, the cosines differ very little from each other. The same remark applies, of course, to the sines of angles nearly 90°. In cases, therefore, where great accuracy is required, we may commit an important error by calculating a small angie from its cosine, or a large one from its sine. We must consequently endeavor to avoid this, by transforming the expression employed.

In the example before us, c is a small angle which has been calculated from its cosine; we must, therefore, if possible, calculate this angle by means of its sine, or some other trigonometrical function.

Now, by formula (8), Art. 72, we have generally

Instead of 1° 20' 50", as obtained by the former process.

Or c might first be calculated from a and b, and then c by means of its sine. No angle which is nearly 90° ought to be calculated from its tangent, for the tangents of large angles increase with so much rapidity, that the results, derived from the column of proportional parts found in the tables, cannot be depended on as

accurate.

23. The following is the demonstration of formulas (3) and (4) of Art. 77. By Art. 69

c2 = a2 + b2 — Qab cos c = a2+ b2 — 2ab (2 cos2 c — 1)

These forms (1) and (2) are more suitable than (1) and (2) of Art. 77, if b be

nearly equal to a, because then tan p, which, in Art. 77, was equal to

is very large, or $ is near 90°, unless c is very small, and when such is the case the Increase of the tangent is not proportioned to the increase of the arc, so that the ordinary mode of calculating logarithms not exactly found in the tables would be inaccurate.

USE OF SUBSIDIARY ANGLES.

24. Formulas not adapted to logarithmic computation may often be rendered so by the use of subsidiary angles. Specimens have been given in the last Art. and Art. 77. The following is another example.

◆ may be computed from (1) by logarithms, and then + from (2), and from and +, becomes known.

25. To resolve a quadratic equation by the aid of Trigonometry.

The general form of such an equation is

Logarithms may be applied to the formulas (1), (2), and (3).

If p and q be negative the following forms should be used, which may easily be

For the resolution of a cubic equation by the aid of Trigonometry, see Alg. Art. 378.

26. To find the increment of the sine, tangent, &c. corresponding to a small increment of the angle.

Let a represent the angle, i its increment, and 8 sin a the corresponding increment of the sine. Then

• When the coefficients of a quadratic are large, the trigonometric mode of solution is convenient.