+BCD=360°; the two latter angles may be considered as 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 these known quantities ß, and we shall thus have x' =ß-x, and, consequently, by substitution, equation (2) becomes

a sin. a sin. (ẞ- x) = b sin. a' sin. x =a sin. a (sin. ẞ cos. x

cos. ẞ sin. x);

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 r' is known from the equation a'ẞr, and CD from either of the equations (1).

PROBLEM VII.

Given the angles of elevation of an object taken at three places o 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, BC'A, 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, a for the height of the object, and a, b, for the distances C C',

B

A

C"C"", we shall have

AC=x cot. a, AC′ = x cot. a', AC"=r cot. a".

Now if a perpendicular AP be drawn from A to C C", we shall have (Geom. p. 35,) from the triangle ACC

n order to eliminate C'P, multiply the first by b, the second by a, and add, and we shall have

x2 (b cot.2 a + a cot.2 a")= (a + b) x2 cot.2 a' + ab (a + b)

If the three stations are equidistant, then a= becomes

x=

a

√ cot.2 a + cot.2 a".

b, and the expression

cot.2 a'

The height AB 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.

PROBLEM VIII.

Three objects A, B, and C, whose distances are AC=8 miles, BC=7} miles, and AB=12 miles, are visible from one station D, in the line joining A and B, at which point the line joining A and C subtends

an angle of 107° 56′ 13′′. Required the distances of the objects from

the station,

AD=5 miles, DC=4.892 miles, DB=7 miles.

PROBLEM IX.

Suppose the angle of elevation of the top of a steeple to be 40° when the observer's eye is level with the bottom, and that from a window 18 feet, directly above the first station, the angle of elevation is found to be 37° 30'. Required the height and distance of the steeple.

Height =210.44 feet. Distance 250.79 feet.

PROBLEM X.

In order to determine the horizontal distance between two remote objects A, B, a base line A'B' of 536 yards was measured, and then a flagstaff being set up at each extremity, these four angles were taken from them, viz. at A′ the angular distance between A and B, 57° 40′, and the angular distance between B and B′, 40° 16′, also at B' the angular distance between A and B, 71° 7′, and the angular distance between A' and A, 42° 22′. Required the distance between the objects.

939.52 yards.

PROBLEM XI.

Three objects A, B, C, are in the same straight line, and of known distances from each other, viz. AB = 3·626 yards, and BC = 8·374 yards, the angular distance of A, B, from a station D, where all the objects are visible, is 19°, and the angular distance of B, C, is 25°. Required the distance of each object from the place of observation.

DA = 9.471 yards, DB = 10·861, DC = 16·848.

PROBLEM XII.

24′ and 14°, the middle Required the height of

At three points in the same horizontal straight line the angles of elevation of an object was found to be 36° 50′, 21° station being 84 feet from each of the others. the object.

PROBLEM XIII.

53.964 feet.

There are three towns A, B, and C, whose distances apart are as follow: from A to B, 6 miles; from A to C, 22 miles; and from B to C, 20 miles. A messenger is despatched from B to A, and has to call at a town D in a direct line between A and C. Now in travelling from B to D, he walks uniformly at the rate of 4 miles an hour, and from D to A at the rate of 3 miles an hour. Supposing him to perform his journey in 3 hours, it is required to determine the position of the town D.

The distance of D from A is 4.72 miles.

The student who has the practical applications of Plane Trigonometry more immediately in view may pass over the following chapter, on the theory of the trigonometrical lines, and proceed to the first chapter of part III., which contains the application of Trigonometry to Navigation.

46

CHAPTER IV.

INVESTIGATION OF TRIGONOMETRICAL FORMULAS.

(26.) The formulas hitherto investigated are those only which are immediately connected with the business of plane trigonometry, properly so called, that is, with the solutions of the several cases of plane triangles. Having disposed of all these cases, we shall now proceed to develop the theory of the trigonometrical lines more at large, dismissing all considerations of the sides of triangles.

The following general expressions have already been established, viz.

sin. (A+B)=sin. A cos. B + sin. B cos. A sin. (A-B)=sin. A cos. B—sin. B cos. A cos. (A+B)=cos. A cos. B―sin. A sin. B cos. (A—B) = cos. A cos. B + sin. A sin. BS From these equations we get

1. By addition,

sin. (A+B) + sin. (A — B)=2 sin. A cos. B

cos. (A+B) + cos. (A — B)=2 cos. A cos. B 2. By subtraction,

sin. (A + B) — sin. (A—B)=2 cos. A sin. B
cos. (A — B) — cos. (A+B)=2 sin. A sin. BS

(1)

(2).

It is worth while to remark here that if we make A= 60°, then since cos. 60°; (p. 9,) the first of these formulas furnish the equation

sin. B sin. (60° + B) — sin. (60° — B) . . . . (V);

which is a useful expression in the work of computing tables.