The Theodolite is a femi-circle divided into 180°, with an index which turns about on its centre, and retains any fituation given it, on which are two fights, called the moveable fights; there are alfo two other fights fixed on the diameter of the theodolite, which are called the fixed fights. Fig. 2. plate 4.

Sights are small pieces of wood or brafs, having small holes or flits in them, to view the object through ;---They are fixed perpendicular to the plane of the theodolite, but parallel to the plane of the quadrant.

The geometrical fquare may be made of brafs, wood, or any folid body, having equal fides and angles; from one of the angles, a thread is fufpended, with a small weight at the end, fo as to point always to the centre. The two fides oppofite to the centre of fufpenfion, are divided each of them into 100 equal parts; there is also an index, which, (when occafion ferves), may be fixed to the centre of fufpenfion, and is made fo as to turn round, and retain any fituation; on this index, are two fights. See fig. 3. plate 4.

Heights and diftances are of two kinds, viz. acceffible and inacceffible acceffible objects are houfes, growing trees, &c. inacceffible ones are all mountains, celeftial bodies, alfo houfes and trees, in certain fituations.

PROBLEM I. See Plate 4. fig. 58.

To measure acceffible heights.

EXAMPLE I.

Let AB be a horizontal plane and BC a tower, whofe height is required: From B, the foot of the tower, measure any convenient distance, 80 feet upon the horizontal plane AB. Suppose the tower to fubtend an angle of 39° 49′ from A. What is its height?

As

A tower, furrounded by a ditch 40 feet broad: from the other fide of the ditch, the tower fubtends an angle of 53° 13'. Required the height of the tower, also the length of a ladder fufficient to fcale the tower. See fig. 58. plate 4.

From the top of a ship-mast 100 feet above the level of the water, I took an angle of depreflion of another fhip's hull, 74° 15′; required the distance of the other fhip.

PROBLEM II.

To measure inacceffible heights and diflances.

EXAMPLE I. Plate 4. fig. 60.

At the foot of a hill, I took an angle of elevation of its top, and found it to be 50° 42'. I then measured back 120 yards on the horizontal plane, and obferved the angle to be 40° 12'. Required the perpendicular height of the hill.

N. B. When any fide AB of the triangle ADB is produced, the exterior angle DBC is equal to both the interior and oppofite angles DAB, ADB; therefore the angle ADB will be 10° 30'

To find BD.

is to AB 120

To find DC the height.

is to BD 425

As fine ADB 10°30′ 9.26063 | As rad. 90

2.07918

2.62839

So is fine an. A 40°12' 9.80987 So is fine DBC 50° 42' 9.88865

I obferved an object on the other side of a river, on a level with the place where I ftood; behind me was a regular declivity, which I might reckon a straight line. I marked my ftation by the fide of the river, and measured back 170 yards, when I obferved I was higher than the object. I took the angle of depreffion of the mark by the river fide 42° 18', of the bottom

H

bottom of the object 72° 8', and of its top 78° 20'. Requir ed the height and distance of the object.

Here, because the angle ABC is 42° 18′ the angle BAC is 47° 42; confequently, its fupplement, the angle BAD will be 132° 18. And fince all the angles of a triangle are equal to two right angles, and that the angle DBA is 29° 50', the remaining angle BDA will be 17° 52' Again, the angle CDE is a right angle, of which the angle BDC is a part; therefore, the angle BDE is 72° 8′, and the angle at E 101° 40′; also the angle DBE will be 6° 12'.

EXAMPLE III. Plate 5. fig. 1.

Being on a horizontal plane, I took the angle of elevation of the fummit of a hill, and of the top of a tower built upon it, and found them to be 48° 20' and 61° 25'. I then measured back 150 yards, and found the angle fubtended by the height of the tower above the plane to be 38° 19'. Required the height of the tower.

The

The exterior angle CBD, is equal to both the interior and oppofite angles, CAB, ACB; but CAB is 38° 19'; therefore, ACB will be 23° 6' and fince all the angles of a triangle are equal to two right angles, angle ABC will be 118° 35'. Or it is the supplement of the angle CBD; alfo angle BCD is 28° 35', and CEB will be 138° 20'.

From a window on a level with the bottom of a steeple, I took the angle of elvation of the top of the fteeple 50°; from another window, 20 feet perpendicular above the former, I took another angle of the top of the steeple 45° 15' Required the height and diftance of the fteeple.

Because the angle ACD is a right angle, of which the angle SCD=50° is a part, the angle SCA will be 40°, consequently, the alternate angle CSD will also be 40°. And fince the angle SAB is 45° 15', and the angle BAD a right angle: therefore, the whole angle SAC 135° 15', and the angle ASC 4° 45'•

To find CS.

As fine ASC 4° 45' 8.91807 is to AC 20

1.30103

|

So is fine SAC 135° 15'9.84758 So is tan. SCD 50°

To CS 170

To find the height of the steeple. Asfec. ang. SCD50° 10.19193′ is to SC 170

2.23045 10.07619

11.14861

12.30664

2.23054 To the height SD2

2.11471

130.2 feet.