The first part of Case II. is ambiguous, for an arc and its supplement have the same sine. This ambiguity, however, is removed if the character of the triangle, as acute or obtuse, be previously known.

For the solution of the second part of Case II. the first analogy is the most usual, but the double analogy is the best adapted for logarithms. The direct expression for the side subtending the given angle is very commodious, where logarithms are not employed.

PROP. XVII. PROB.

Given the horizontal distance of an object and its angle of elevation, to find its height and absolute distance.

Let the angle CAB, which an object A makes at the station B, with an horizontal line, and also the distance BC of a perpendicular AC, to find that perpendicular, and the hypotenusal or

aërial distance BA.

In the right angled triangle BCA, the radius is to the tangent of the angle at B as AB to AC, and the

radius is to the secant of the angle at B, or the cosine of the angle at B is to the radius, as AB to BC.

PROP. XVIII. PROB.

Given the acclivity of a line, to find its corresponding vertical and horizontal length.

In the preceding figure, the angle CBA and the hypotenusal distance BA being given to find the height and the horizontal distance of the extremity A.

The triangle BCA being right angled, the radius is to the sine of the angle CBA as BA to AC, and the radius is to the cosine of CBA as BA to BC.

If the acclivity be small, and A denote the measure

A

of that angle in minutes; then AC = BA × nearly. 3438

But the expression for AC, will be rendered more accu

AC3

rate, by subtracting from it, as thus found, the quantity BA

In most cases when CBA is a small angle, the horizontal distance may be computed with sufficient exactness, by or BA x A x .000,000,0423, from the

deducting

AC2 2 BA'

hypotenusal distance.

PROP. XIX.

PROB.

Given the interval between two stations, and the direction of an object viewed from them, to find its distance from each.

Let BC be given, with the angles ABC and ACB, to calculate AB and AC.

In the triangle CBA, the angles ABC and ACB being given, the remaining or supplemental angle BAC is thence given; and consequently, S,BAC: S,ACB :: BC: AB, and S,BAC :: S,ABC: BC: AC,

A

B

PROP. XX. PROB.

Given the interval between two stations, and the directions of two remote objects viewed from them in the same plane, to find the mutual distance, and relative position, of those objects.

Let the points A,B represent the two objects, and C,D the two stations from which these are observed; the interval CD being measured, and also the angles CDA,CDB at the first station, and DCB,DCA at the second; and it is thence required to determine the transverse distance AB, and its direction.

Find, by the last problem, the distances AC and BC of both objects from one of the stations C; then the contained

angle ACB, or the excess of DCA above DCB, being likewise given, the angles at the base AB of the triangle BCA, and the base itself, may be calculated, from the analogies exhibited for the solution of the

D

B

second branch of Case II. For AC + BC: AC-BC :: Cot, ACB: Cot, ACB+CAB, and thus the angle CAB

is found; or, more conveniently by two successive operaBC:: R: T,b, and R: T,45°-b :: Cot, ACB:

tions, AC

: Cot, ACB

CAB. Now, S,CAB:: S,ACB:: BC: AB,

or AB = √(AC2+BC-2AC × BCX Cos, ACB). Again, the inclination of AB to CD is evidently the supplement of the two interior angles CAB and DCA.

Cor. Hence the converse of this problem is readily solved. Suppose two remote objects A and B, whose mutual distance is already known, are observed from the stations C and D, and it were thence required to determine the interval CD. Assume unit to denote CD, and calculate AB according to the same scale of measures; the actual distance AB being then divided by that result, will give CD: For the several triangles which combine to form the quadrilateral figure CABD, are evidently given in species.

In this and the following problem, the angles on the ground are supposed to be taken by means of a theodolite. If the sextant be employed for that purpose, such angles, when oblique, must be reduced by calculation to their projection on the horizontal plane. This reduction, however, belongs properly to Spherical Trigonometry.

In surveying an extensive country, a base is first carefully measured, and the prominent distant objects are all connected with it, by a series of triangles. To avoid, in practice, the multiplication of errors, these triangles should be chosen, as nearly as possible, equilateral.

PROP. XXI. PROB.

The mutual distances of three remote objects being given, with the angles which they subtend at a station in the same plane, to find the relative place of that station.

Let the three points A, B, and C, and the angles ADB and BDC which they form at a fourth point D, be given; to determine the position of that point.

Through D and the extreme points A and C describe a circle, draw DB cutting the circumference in E, and join AE and CE.

1. In the triangle AEC, the side AC, and the angles ACE and CAE, which are equal (III. 20. El.) to ADB and BDC, being given, the side AE is found by Case III.

2. All the sides of the triangle ABC being given, the angle CAB is found by Case I.

3. In the triangle BAE, the sides AB and AE are given, and their contained angle EAB, or the difference of CAE and CAB, are given, whence, by Case II., the angle ABE or ABD is found.

B

D

4. Lastly, in the triangle DAB, the side AB and the angles ABD and ADB being given, the side AD or BD is found by Case III., and consequently the position of