In the triangles EAC and A 3DBF, find by Prop. 19. the sides AC and BD; and in the triangle CAD, the sides AC, CD, with their contained angle ACD, being given, the base IE DA and the angle CDA are C D found by Case II. But the E" distances DA, DB being now given, with their contained angle ADB, the base AB is found by Prop. 20. w PROP. xxv. 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. 1. Suppose the station D to be situate in the direction of two of the objects A, C, All the sides AB, AC and BC of the triangle ABC be ing given, the angle BAC is IB found by Case I. ; and in the 2^ triangle ABD, the side AB with -the angles at A and D being .A. D H C given, the side AD is found by –A Case III. and consequently the * * C position of the point D is determined. 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. 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 s given; whence BD is found by Case III. But since the angle ABD and the distance BD are 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. z 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 CAF, which are equal (III. 16. El.) to ADB and BDC, being given, the side AE is found by Case III. A 2. All the sides of the trian- IB gle 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 differenceof CAEand CAB, * C are given, whence, by Case II., the angle ABEor ABDis found. 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 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 * C. subtend at F, and likewise the angles AEF and EFC; to find Es: the relative situation of each of S., A' those stations E and F. * in Produce AE and CF to meet in D, and join BD. The angle EDF, being equal to AEF-H CFE—180°, is given. Now in the triangle EBF, sinBFE: sin EBF :: EB : EF; and in the triangle EDF, sin EDF : sinDFE:: EF: ED; whence, (W. 23. El) sinBFE.sineDF : sin EBF.sinl)FE :: 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.) sin BFE.sin EDF : sinEBF.sins) FE : : R : tan b, and R : tan (45–b) : : cot; BED : —cot(#BED--EBD), or cot(1809—3 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 ob. ject 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 stations 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 DFjoining the vertex D must (VI. 10. cor. El) bisect in- 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 semicircles, must occur in their intersection. |