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between two objects, which has been already determined by calculation. By this means the error of the computed distance may be discovered. This measured line is called the base of verification.

93. The area or content of any quantity of land measured in this manner may be found as follows.

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Let P denote a perpendicular drawn from the angle A on the opposite side BC, then R=1: sine ABC :: AB : P AB X sine ABC; hence, by mensuration, the area of the triangle ABC AB × BC × sine ABC. In the same manner the area of BCD = 1⁄2 BC × BD × sine CBD, of CDE = 1 CD × DE x sine CDE, of EDF = ED × DF × sine EDF. The sum of the several triangles will be the area of the whole tract of land.




94. It sometimes happens, in mak-' ing a survey, that, when the distance between two objects, C, D has been determined, it is required to find the distance between two eminences A, B, which are conveniently situated for extending the series of triangles. In this case, measure the angles CAD, CAB, DBC, DBA. Now, as there are not sufficient data in any of the triangles to compute the unknown parts, we must assume a value for AB, and thence compute the value of CD, as above. Then, the computed value of CD: its true value :: the assumed value of AB its true value. For, by changing the value of AB, while the angles at A and B remain the same, the whole figure will continue similar to itself, and consequently AB will vary as CD.

95. Military sketches, or small surveys, may be made by means of a compass fixed on the top of a staff. The staff being stuck in the ground, so that the needle of the compass may play freely, the angular distances, or bearings, as they are commonly called, must be taken from the magnetic meridian. But the compass must not be used when accuracy is required.

Let NS represent the needle, or magnetic meridian, n the true north point, and E, W the east A and west points; then if the sights of the compass be directed to the object A, and the angle NOA be 40°, the object is said to bear





NW 40°. If the sights be directed to the object B, and the angle NOB be 110°, the object is said to bear NE 110°.

But the angular distances, or bearings of objects, are commonly expressed by the letters which denote the points of the compass. Thus, if an object bear in the direction of the point of the compass, which is denoted by the letters NNE, it makes an angle with the meridian of 22° 30', from the north toward the east; and if an object bear SW, it makes an angle with the meridian of 45°, from the south toward the west.

Example 1. As I was travelling in England I perceived a fort on a high hill, which bore from me NE 22 degrees; and having gone 20 miles in the direction NW 67 degrees, I saw the fort again, which now bore NE 561 degrees. Required the distance of the fort from each place of observation.

Let A be my first station, MAm the meridian, Ċ the fort, AC its direction on bearing, B my second station, BC the bearing of the fort = angle m DC.

Angle ACB = m DC-DAC = 56° 15′-22° 30′ =33° 45'.

Angle BAC = CAD + DAB = 22°30′+ 67° 30′ = 90'.



Angle ABC 180°-BAC - ACB= 180° 123° 45′ = 56° 15'.





Sine C 33° 45': sine BAC 90° :: AB 20: BC.

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Sine C 33° 45′ : sine B 56° 15′ :: AB 20: AC.

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2. At a certain place A St. Paul's church C, in London, bore from me NE 114 degrees; and when I had travelled 15 miles farther to B, in the direction NW 22 degrees, it bore NE 50° 37. Required its distance at the last place of observation. Answer. 13.14 miles.

3. From a ship at sea a point of land C bore NE by N. The ship sailed NW 23 miles, and the same point then bore E by N. Required the distance of the point from the ship at both stations.

The angle BAC = NE by N+ NW = 33° 45′ + 45° =78° 45'.


The angle Cm DC-DAC = DAC = E by N-NE by =78° 45' - 33° 45′ 45o.

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Hence the angle B = 56° 15'.

Then (54) the distance of the land from the first station of the ship is 27.045 miles, and from the second station 31.902 miles.

4. As a ship was sailing along the coast of Carolina a cape was observed to bear NW by N, and another NNE. The ship then sailing ENE E 21 miles, the first cape bore WNW, and the second N by WW. Required the distance between the two capes. Answer. 28.0847 miles.

96. Problems for the Exercise of Learners.

The following problems are intended as exercises for students, and therefore the answers of most of them have been omitted. The reader will be able to resolve them without much difficulty, by comparing them with similar questions, of which the solutions are given above. They are not however

deemed necessary, and the solution or the omission of them is submitted to the judgment of the teacher.

1. The angle of elevation of a tower is 65°, and the distance from the place of observation to the bottom of the tower is 80 feet. Required the height of the tower.

2. A ladder 40 feet long will reach the top of a house, when the foot of the ladder is placed 15 feet from the bottom of the house. Required the height of the house, and the inclination of the ladder to the horizon.

3. Required the length of a scaling ladder, to scale a tower whose angle of elevation is 67° 41′, across a ditch 16 feet wide.

4. A tower, whose height is 100 feet, subtends an angle of 25°. Required the distance of the observer from the bottom of the tower.

5. What angle will the top of the same tower (100 feet high) subtend, when the observer is twice as far from it? and what angle will it subtend when he is only half as far from it?

6. What is the altitude of the sun when a man's shadow is half his height? and what is the altitude of the sun when a man's shadow is double his height?

1: 2: R: tan. sun's altitude. Answer. 63° 26'.

2:1 R tan. sun's altitude. Answer. 26° 34'.


7. How far will the shadow of an upright pillar extend at noon, upon a level pavement, the meridian altitude of the sun being 50°, and the height of the pillar 100 inches?

Answer. 83 inches.

8. Standing on the top of a tower 136 a tree at a distance on the plain, and pression 22° 40'. Required the distance bottom of the tower.

feet high, I observed found its angle of deof the tree from the

9. From the top of a castle on the shore, 156 yards above the level of the sea, the angle of depression of a ship is 56° 40'. Required the distance of the ship both from the bottom and of the castle.

the top

10. There are three towns A, B, C; at A the towns B and C make an angle of 20° 12'; at B the towns A and C make an angle of 38° 30'; and the distance between A and B is 23 miles. Required the distance between C and A, and C and B.

11. As I was walking on the sea-shore I observed an island, which made an angle of 95° 20′ with a straight line along the shore. At the distance of 650 links farther the island made an angle of 75° 40' with my first station. Required its distance. from the first station.

12. A man travels from A to B, 3 miles and 6; then, bending a little to the right, he goes from B to C, 4 miles and; at C he observes that A and B make an angle of 29° 16′. Required his least distance from home. Answer. 7 miles.

13. A man travels from A to B, 3 miles and; returning in a mist he loses his way, and going a little too much on the right hand comes to C, which is 4 miles and from B. The weather now cleared up, and he could see both A and B, and observed that they made an angle of 22° 16'. Required his distance from home. Answer. 1.2 mile.

14. There are three towns A, B, C. The distance between A and C is 3 miles and 3 furlongs, and the distance between B and C is 4 miles and 5 furlongs. Between A and B lies a large wood, which prevents these towns from being seen from each other, and their distance from being measured. However A and B are visible from C, and there make an angle of 71°2′. What is the distance between A and B? and how can a vista

be cut through the wood, so that A and B may be seen from

each other?

15. There are three towns A, B, C. The distance of A and B is 5 miles, of B and C 9 miles, of C and A 7 miles. What are their respective bearings from each other?

Answer. Angle A = 95° 441, B = 50° 421, C = 33° 33%.

16. How shall I plant three trees, so that the angles which every two trees make with each other may be 50, 60, and 70 degrees?

The distances between the trees will be to one another respectively as the sines of the angles which are opposite to them (56), and therefore will vary indefinitely in this proportion. Now the natural sines of 50°, 60°, and 70o, to radius 1, are 76604, ·86603, 93969; therefore these numbers, or any other in the same proportion, will represent the proportional, not the absolute distances between the trees. Suppose x to be the distance between two trees opposite to the angle 50°, then s. 50° : s. 60° :: x: xX s. 60°-s. 50° distance of two trees opposite to 60°. Let x = 20 yards, then log. 20+ log. s. 60°-log. s. 50° 1.354307, the number answering to which is 22.61 yards, the side or distance opposite to the angle 60°.

17. How must three trees A, B, C be planted in a triangular form, so that the angle at A may be double of the angle at B, and the angle at B double of the angle at C; and a line of 100 yards may go round them?

It is evident that the angles are to one another as the num

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