2. At a considerable distance from a hill, I took the elevation of the top of a tower built upon it, 33° 45'; and measuring on level ground 300 feet directly towards the hill, I again took the elevations of the top and the bottom of the tower 51° and 40°. Required the height of the tower.

Ans. 46.666 yards.

3. At a window on a level with the base of a steeple, I took the elevation of its top 40°; and at another window of the same house 18 feet higher, I took again the elevation of the top of the steeple 37° 30′. Required the height of the steeple. Ans. 210-44 feet.

At

4. The elevation of the top of a hill at one station was 38° 25'. Another station was taken 450 feet from the first, but neither on a level with it nor in the direction of the hill. the first station, the line from the other station to the top of the hill subtended an angle of 67° 30′; and at the second, the line from the first to the top of the hill subtended an angle of 74° 48'. Required the height of the hill. Ans. 441.25 feet.

5. I measured directly up a hill 132 yards: there I took the depression of the hill 42°, that of the bottom of a distant object 27°, and that of its top 19°. Required the height of the object. Ans. 28-637 yards.

PROB. XII. To find the distance of a place A, from an inaccessible object B.

First. Let B be visible from A. Choose a station C, from which both A and B can be seen. Measure AC 650 yards, and take the angles BAC 72° 22′, and ACB 78° 37', with the theodolite. Then ABC is 29° 1′, and sin. B: sin. C :: CA: AB 1313-67 yards.

Secondly. Let B be not visible from A. Choose a station C from which both A and B A may be seen, and their distances from it mea

B

sured. Take the angle ACB 75° 38', and measure AC 358, and CB 560 feet.

Ans. (BC+CA)918: (BC—CA) 202 :: tan. (A+B) 52° 11′ tan. (A-B) 15° 49′7′; whence BAC is 68° 0·7′, and sin. A sin. C:: CB: BA 585.041.

3. A straight line was measured along the bank of a river 528 feet, and at its extremities the angles contained by it, and straight lines directed to a tree upon the opposite bank, were 62° 40′ and 73° 26'. Required the breadth of the river.

Ans. 648-366 perp. breadth, and 676-444 feet to the nearest

station.

4. Straight lines from a station to two places measured 694

and 456 yards, and the angle contained by them was 127° 16′.
Required the distance of the one place from the other.
Ans. 1035-773 yards.
5. To find the distance between two trees, I found the angle
it subtended at a station to be 55° 40′, and measured from the
station to the trees 588 and 672 yards. Required their dis-
Ans. 592-967 yards.

tance.

PROB. XIII. To find the distance between two places A and B, both of them inaccessible. Take two stations C and D, such, that from each of them the other station and the places A and B may be seen. Measure CD 1267 links, and at C take the angles BCA 53° 38', and BCD 34° 50′, and at D take the angles ADC 43° 44′, and ADB 58° 38'.

In the triangle ADC, the angle ACD is

88° 28', and CAD 47° 48′, and sin. A: sin. C :: CD: DA 1709-69. Also, in the triangle BCD, the angle CDB is 102° 22', and CBD 42° 48', and sin. B: sin. C :: CD : DB 1065 14. Lastly, in the triangle ADB are given AD and DB, and the angle ADB . (AD+ DB) 2774-83: (AD-DB) 644-55 :: tan. (A+B) 60° 41′ : tan. (A—B) 22° 28'; whence ABD is 83° 9', and sin. ABD : sin. ADB :: ĎA : AB 1470-304 links.

2. To find the distance between two steeples A and B, I took two stations C and D, distant 428 yards from one another; and at C took the angles ACB 54° 30′, and BCD 42° 26'; and at D took the angles CDA 40° 44′, and ADB 57° 42'. Required the distance of the steeples.

Ans. 546 704 yards. 3. To find the distance between two places M and P, I took two stations A and B, distant from one another 908-36 feet; and at A took the angles PAM 14° 34', and MAB 46° 16'; and at B took the angles ABP 96° 44′, and PBM 18° 39'. Required the distance between M and P. Ans. 674.64 feet.

NOTE. If the distance between the objects be known, and the distance between the stations be required, assume 1 or 1000 for the distance between the stations, and with it find the distance between the objects. Then, as the distance found is to the given distance, so is 1000 to the true distance between the stations.

4. Suppose the distance AB 700 feet, and at the station C let ACB be 42° 45′, and BCD 54° 12′, and let the angles at D be ADB 50° 19′, and ADC 57° 33'. Required the distance CD. Ans. 330.04 feet.

5. To find the distance between two lighthouses A and B, I measured the distance between two stations M and R 3370 yards, and at M took the angles AMB 37° 52', and BMR 91° 27', and at R the angles ARM 29° 56′, and ARB 40° 27′. Required the distance AB. Ans. 7063-465 yards.

6. At a station C took the angle ACB, subtending a line AB 3291 yards, and found it 4° 35', and the angle BCD between B and another station D 86° 52'; and at D took the ingles ADB 8° 24′, and ADC 70° 23'. Required the disance of the stations from one another. Ans. 3370-425 yards.

PROB. XIV. Given the distances of three places, A, B, C, from one another, viz. AB 317, AC 308, and BC 478 feet, and the angles which these distances subend at a station D in the same plane with them, viz. ADB 24° 50′, and ADC 27° 44; to find the distance of the station D from each of the places.

Having drawn the triangle ABC, make at the point C, on the side of BC, opposite to that on which the station D lies, the angle BCd 24° 50′, and at B the angle CBd 27° 44', and about the triangle BCd describe a circle, and join Ad, meeting the circle again in D, and join BD and DC.

The three sides of the triangle ABC are given to find the angle ABC 39° 25′ 14′′;

=

B

C

then ABd ABC+dBC = 67° 9′ 14′′, when A and d are on different sides of BC, or = 11° 41′ 14′′, when, as here, A and d are on the same side of BC. Also, the angles of the triangle BCd are given, with the side BC, to find Bd 252.7 feet. Again, in the triangle ABd are given the sides AB and Bd, and the included angle ABd, to find the angles AdB 131° 53′ 53′′, and BAd 36° 24′ 53′′. Then in the triangle ABD are given the angles and the side AB, to find BD 448.065, and AD 661-738. And in the triangle DBC are given the angles and BC, to find DC 591.57 feet.

2. If A be the place nearest to D, the angle BAd is 46° 47′ 32′′: then BD is 550·154, AD 282.25, and CD 528.42 feet.

NOTE 1. If the given station be within the triangle, as at d, make the angles BCD and CBD equal to the supplements of BdA and AdC.

NOTE 2. If two of the given places, A and B, be in a straight line with the station D, the distances BC and CA

subtend the same angle BDC. After finding the angle at B, work the triangle DBC.

NOTE 3. If the three places A, B, C, be in a straight line, the first operation will not be required. The rest are the same as before.

3. The three sides of the triangle ABC are AB 280, BC 314, and AC 326 yards; and from the station D without the triangle, the angle ADB was 25° 52′, and ADC 23o 6′, the point C being the nearest to D. Required their distances from D. Ans. AD 586.154, BD 413-41, CD 308-107 yards. 4. Suppose AB 267 feet, BC 209, and AC 346, and at the point D, within the triangle, the angle ADC is 128° 40′, and ADB 91° 20′. Required the distances of D from the angles. Ans. AD 104-05, BD 189.33, and DC 178.85 feet. NOTE. When D is in one of the sides, describe a segment on BC containing the given angle.

5. Suppose AB 122.4, BC 74, and AC 82 chains, and at D in AB, produced beyond B, the angle ADC is 22° 45'. Required the distance of D from the angles.

Ans. AD 1818, BD 594, and CD 125-4 chains. 6. Suppose AB 1234, BC 873, and AC 632 yards, and at D in AB the angle ADC is 120°. Required its distance from the angles.

Ans. AD 226.12, BD 1007.88, and CD 487·84 yards. 7. Suppose AB 138, BC 224, and AC 326, and at D the angles are ADB 7° 22′, and ADC 19° 58'. Required the distance of D from the angles.

Ans. AD 510.96, BD 385-286, and DC 204-87.

PROB. XV. Given the angles of elevation of a tower PS, taken at three stations A, B, and C, on a level plane, no two of which are in the same vertical plane with the tower, viz. PAS 20° 10′, PBS 18° 50′, and PCS 34° 30, and also the distances between the sta tions AB 324, BC 568, and AC 672 yards; to find the height of the tower.

Make the triangle ABC, of which AB is 324, BC 568, and AC 672, and make BE = BC, and BD = BA, and join ED, and upon it make the triangle EDF on either side of DE, so that BE EF:: cot. PBS: cot. PAS, and BD: DF:: cot. PBS : cot. PCS; or make EF 527-494, and DF 160-79, and join BF, and

:

B

make the angle BAP = BFE. Then erect PS perpendicular to the plane ABP, and in the plane passing through AP and PS make the angle PAS 20o 10', and PS will be the tower required.

Join PC, CS, BS, the triangles APB, FBE, being similar, AP: PB:: FE: EB:: cot. SAP: cot. SBP, therefore SBP is 18° 50′; also PB : BE=BC :: BA=BD: BF, therefore the triangles PBC and FBD are similar; and BP: PC:: BD: DF:: cot. PBS : cot. PCS, therefore PCS is 34o 30'.

In each of the triangles EBD, EFD, are given the three sides, to find the angles BED 28° 45′ 30′′, and FED 6° 47′ 26′′; and their difference 21° 58′ 4′′, or their sum 35° 32′ 56", is the angle BEF, from which, with the sides BE and EF, the angle BFE or BAP is found in the first case to be 89° 48′ 7′′, and in the other 78° 48′ 22′′. Therefore AP is 866.108 or 546-676, and PS 318.094 or 200.78.

2. Let AB be 326, BC 584, and AC 683, and the angles of elevation SAP 30o, SBP 26o, and SCP 23o; to find PS. Ans. PS is 952.14 or 168.642. 3. Let AB be 80, BC 119, and AC 140 yards, and the elevation at A 50°, at B 60°, and at C 55°. Required the height of the object D. Ans. 96.4 feet. 4. Let AB be 60, BC 72, and AC 132 feet, and the elevations of S at A 30° 48', at B 40° 33', and at C 50° 23'. Required the height of S. Ans. 94.84 feet. 5. Let AB and BC be each 84 feet, and the points A, B, C, in a straight line, and the elevation at A 36° 50', at B 21° 24', and at C 14o. Required the height of the object. Ans. 53.96 feet.

OF LEVELLING.

When the altitudes of the several parts of an irregular ascent are to be determined, a spirit-level with telescopic sights is to be used.

PROB. XVI. To find the height of g above a.

Erect a pole ab at a, and another cd at a convenient distance. Place the level between them, and, directing the sights to the pole ab, cause the point b to be marked on it; then direct the sights to the pole cd, and on it

b

f

mark m. Next erect a pole nearer to g, as at e, and place the level between it and the pole cd, and mark upon them, as before, the points d and k; and proceed in this way to g.