On the ground which we suppose to be horizontal or very nearly so, measure a base AD, neither very great nor very small in comparison with the altitude AB; then at D place the foot of the circle, or whatever be the instrument, with which we are to measure the angle BCE formed by the horizontal line CE parallel to AD, and by the visual ray direct it to the summit of the building. Suppose we find AD or CE 67.84 yards, and the angle BCE=41° 04' in order to find BE, we shall have to solve the right angled triangle BCE, in which the angle C and the adjacent side CE are known. Hence, EB-59.111 yards. To EB add the height of the instrument, which we will suppose to be 1.12 yards, we shall then have the required height AB=60.231 yards. If, in the same triangle BCE it were required to find the hypothenuse, form the proportion As cos C 41° 04' ar.-comp. log. 0.122660 10.000000 1.831486 1.954146 Note. If only the summit B of the building or place whose height is required were visible, we should determine the distance CE by the method shown in the following example; this distance and the given angle BCE are sufficient for solving the right angled triangle BCE, whose side, increased by the height of the instrument, will be the height required. If for another inaccessible object C, we have found the angles CAD=35° 15', ADC=119° 32', we shall in like manner find the distance AC=1201.744 yards. 3. To find the distance between two inaccessible objects B and C, we determine AB and AC as in the last example; we shall, at the same time, have the included angle BAC BADDAC. Suppose AB has been found equal to 538.818 yards, AC 1201.744 yards, and the angle BAC 68° 40′ 55′′; to get BC, we must resolve the triangle BAC, in which are known two sides and the included angle. = As AC+AB 1740.562 ar.-comp. log. 6.759311 Is to AC-AB 662.926 2.821465 Now, to find the distance BC make the proportion, 4. Wanting to know the distance between two inaccessible objects which lie in a direct line from the bottom of a tower of 120 feet in height, the angles of depression are measured, and found to be, of the nearest, 57°; of the most remote, 25° 30' required the distance between them. : Ans. 173.656 feet. 5. In order to find the distance between two trees, A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distance of a third point C from each, was measured, viz. CA=588 feet and CB 672 feet, and also the contained angle ACB-55° 40': required the distance AB. Ans. 592.967 feet. 6. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°: then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45': required the height of the tower. Ans. 83.9983 feet. 7. Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which both of them could be seen, two points C and D, were chosen, at a distance from each other equal to 200 yards, from the former of which A could be seen, and from the latter B, and at each of the points C and D a staff was set up. From C a distance CF was measured, not in the direction DC, equal to 200 yards, and from D, a distance DE equal to 200 yards, and the following angles were taken, viz. AFC-83° ACF= 54° 31', ACD=53° 30', BDC=156° 25', BDE=54° 30', and BED 88° 30': required the distance AB. Ans. 345.46 yards. 8. From a station P there can be seen three objects, A, B and C, whose distances from each other are known, viz. AB= 800, AC 600, and BC=400 yards. There are also measured the horizontal angles, APC=33° 45′, BPC=22° 30'. It is required, from these data, to determine the three distances PA, PC and PB. Ans. PA=710.193, FC=1042.522, PB=934.291 yards. SPHERICAL TRIGONOMETRY. I. It has already been shown that a spherical triangle is formed by the arcs of three great circles intersecting each other on the surface of a sphere, (Book IX. Def. 1). Hence, every spherical triangle has six parts: the sides and three angles. Spherical Trigonometry explains the methods of determining, by calculation, the unknown sides and angles of a spherical triangle when any three of the six parts are given. II. Any two parts of a spherical triangle are said to be of the same species when they are both less or both greater than 90°; and they are of different species when one is less and the other greater than 90°. III. Let ABC be a spherical triangle, and O the centre of the sphere. Let the sides of the triangle be designated by letters corresponding to their opposite angles: that is, the side opposite B the angle A by a, the side opposite B by b, and the side opposite C by c. Then the angle COB will be represented by a, the angle COA by b and the angle BOA by c. The angles of the D EC 6 H F spherical triangle will be equal to the angles included between the planes which determine its sides (Book IX. Prop. VI.). From any point A, of the edge OA, draw AD perpendicular to the plane COB. From D draw DH perpendicular to OB, and DK perpendicular to OC; and draw AH and AK: the last lines will be respectively perpendicular to OB and OC, (Book VI. Prop. VI.) The angle DHA will be equal to the angle B of the spheri cal triangle, and the angle DKA to the angle C. The two right angled triangles OKA, ADK, will give the proportions R sin AOK :: OA: AK, or, Rx AK OA sin b. AK sin C. Hence, R2 × AD=AO sin b sin C, by substituting for AK its value taken from the first equation. In like manner the triangles AHO, ADH, right angled at H and D, give R: sin c: AO: AH, or Rx AHAO sin c Equating this with the value of R2 viding by AO, we have AD, before found, and di The sines of the angles of a spherical triangle are to each other as the sines of their opposite sides. IV. From K draw KE perpendicular to OB, and from D draw DF parallel to OB. Then will the angle DKF=COB=a, since each is the complement of the angle EKO. In the right angled triangle OAH, we have R: cos c:: OA: OH; hence AO cos c=RX OH=Rx OE+R.DF. In the right-angled triangle OKE R: cos a: OK : OE, or R× OE=OK cos a. But in the right angled triangle OKA R: cos b:: OA: OK, or, Rx OK=OA cos b. In the right-angled triangle KFD R: sin a KD: DF, or Rx DF=KD sin a. But in the right angled triangles OAK, ADK, we have |