2. Suppose the three objects A, B and C to lie in the same direction.

Describe a circle about the extreme objects A, C and the station D, join DA, DB and DC, produce DB to meet the circumference in E, and join AE and CE.

E

In the triangle AEC, the side AC is given, and the angles EAC and ECA, being equal (III. 16. El.) to CDE and ADE, are consequently given; wherefore the side AE is found by Case III. The triangle AEB, having thus the sides AE, AB, and their contained angle EAB or BDC given, the angle ABE, and its supplement ABD are found by Case II. Lastly, in the triangle ABD, the angles ABD and ADB, with the side AB, are given; whence BD is found by Case III. But since the angle ABD and the distance BD are

C

B

D

assigned, the position of the station D is evidently determined.

3. Let the three objects form a triangle, and the station D lie either without or within it.

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. 16. El.) to ADB and BDC, being given, the side AE is found by Case III.

1

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 the point D, with respect to A and B is determined. By a like process, the relative position of D and C is deduced; or CD may be calculated by Case II. from the sides AC, AD, and the angle ADC, which are given in the triangle CAD.

It is obvious that the calculation will fail, if the points B and E should happen to coincide. In fact, the circle then passing through B, any point D whatever in the opposite arc ADC will answer the conditions required, since the angles ADB and BDC, being now in the same segment, must remain unaltered.

PROP. XXVI. THEOR.

The mutual distances of three remote objects, two of which only are seen at once from the same station, being given, with the angles observed at

T 1

two stations in the same plane, and the intermediate direction of these stations,-to find their relative places.

Suppose the three points A, B and C are given, with the angle AEB which A and B subtend at E, and BFC, which B and C subtend at F, and likewise the angles AEF and EFC; to find the relative situation of each of those stations E and F.

B

ཡ

Produce AE and CF to meet in D, and join BD. The angle EDF, being equal to AEF+CFE-180°, is given Now in the triangle EBF, sinBFE: sinEBF:: EB: EF; and in the triangle EDF, sinEDF: sinDFE:: EF: ED; whence, (V. 23. El.) sin BFE.sinEDF: sinEBF.sinDFE :: EB: ED, and consequently the ratio of EB to ED is found. Again the angle BED, being the supplement of AEB, is given, (Prop. 10. cor.) sinBFE.sinEDF :sinEBF.sinDFE :: R: tan b, and R: tan (45—b) : : cot BED:-cot(÷BED+EBD), or cot(180°-BED— EBD), whence the angle EDB is given. The angles which all the three objects A, B, and C subtend at the point D are therefore all given, and hence the position of D is determined by the preceding proposition. But BD, being found, the several distances BE, ED, and BF, FD are thence obtained, and consequently the position of each of the stations E and F is determined.

PROP. XXVII.

Given the angles of elevation at which an object is seen from three known points in a horizontal plane, to find its position and altitude.

Let A, B, and C be the three points of observation, and D the foot of the perpendicular from the given object to the horizontal plane. It is evident from Proposition 17, that the horizontal distances AD, BD and CD are proportional to the co-tangents of the vertical angles at the sta tions A, B, C; let these co-tangents be respectively denoted by the lines L, M, and N. Divide AB, the base of the triangle ADB, internally and externally at

the points E and F, in the

ratio of L to M, and the

lines DE and

DF joining the vertex D must

(VI, 10. cor. El.) bisect in

H

L

M

N

ternally and externally the angle; whence EDF is a right angle, and (III. 19. El.) contained in a semicircle. In the same manner, divide CB internally and externally at G and H in the ratio of M to N, and on GH describe a semicircle. The point D, being common to both semicir cles, must occur in their intersection.

From this construction, the trigonometrical calculation is readily deduced. For L+M: M::AB: BE, and L-M: M:: AB: BF; whence EF and its half DE, or the radius KE, is found. In like manner, N+M: M:: CB: BG, and N-M: M:: CB: BH; consequently DI= BG+BH

2

In the triangle IBK, the sides BI and BK,

with their included angle, are given, and therefore (Prop. 10.) the angle BKI and the base IK are found. Again, all the sides of the triangle IDK being given, the angle -IKD (Prop. 14.) is found. Hence, in the triangle ADK, the whole angle AKD and its containing sides are given, and therefore the base AD, or the horizontal distance of the object from the station A is found, and consequently its altitude. The opposite semicircles will, likewise, by their intersection, give, on the other side, a second position for that point. In practice, the ambiguity would easily be removed.

If the object be seen at the same elevation from all the three points, the arcs of the circles will evidently pass into tangents which bisect at right angles the sides of the triangle ABC. The projection D of the object on the horizontal plane will then be the centre of the circle circumscribing that triangle, and therefore the radius, or the distance AD, will be found by Prop. 18. Book VI. of the Elements.

If the three points of observation should lie in the same straight line, the centres of the determining circles will likewise occur in that line or its extension, and hence the process of calculation will be greatly abridged.

General Scholium. In all the foregoing problems, the angles on the ground are supposed to be taken by means of