problems are given for their general guidance in such cases. In solving these problems, it is the logarithmical mode of calculation that will be attended to, with the view of showing the direct application of the principles of plane trigonometry to such cases. To the imagination of the ingenious, however, many other modes of obtaining an approximate value for the heights and distances of remote objects will soon present themselves: such as, by means of shadows, mirrors, unequal vertical staves, &c. &c.; but, since these methods entirely depend upon the principles of similar triangles (as demonstrated in Euclid, Book VI., Prop. 4), they admit of direct solutions without the assistance of trigonometrical tables: hence, no notice can be taken of them in this work.

Let the horizontal distance B C be 250 feet, the angle of elevation ADE = 41:45, and the height of the eye CD = 5 feet; required the height of the object AB?

This comes under Problem II. of right angled plane trigonometry, page 172; and, by making D E radius, it will be

Remark. By removing either towards or from the object, until the quadrant shows the angle of altitude to be 45 degrees, the measure of the distance between the foot of the observer and that of the object, augmented by the height of the eye, will become the altitude or height of that object.

PROBLEM II.

Given the Angle of Elevation, and the Height of an Object; to find the Observer's horizontal Distance from that Object.

RULE.

At any convenient distance, as at C, let the angle of elevation ADE be taken; then, in the triangle ADE, given AE the height of the object A B, diminished by the height of the eye CD, or its equal BE, and the angle at D; to find the horizontal distance D E = CB. D

Example.

B

Let the height of the object AB be 175 feet, the angle of elevation ADE 37:20, and the height of the observer's eye CD = 5 feet; required the horizontal distance B C?

This falls under Problem II., of right angled plane trigonometry, page 172; and by making A B radius, the proportion will be

Is to height of the object A B 175 ft.-BE 5 ft.=170 Log. = 2.230449 So is the angle of elevation A D E=37:20: Log. co-tangent= 10. 117637

To the horizontal dist. DE=CB=222. 89 Log. =

2.348086

PROBLEM III.

To find the Height of an inaccessible Object, as A B.

RULE.

G

D

E

B

At any convenient points, as C and D (these being in the same vertical plane with AB), observe the angles of elevation AFE and AGE; and measure the distance CD: then, because the exterior angle AFE is equal to the two interior and opposite angles AG F and GAF (Euclid, Book I., Prop. 32), if from the angle AFE the angle AGE be subtracted, the remainder will be the angle GAF. Now, in the oblique angled triangle A GF, given the side GFDC, and the angles A and G; to find the side AF: and, in the right angled triangle AFE, given the hypothenuse AF, found as above, and the angle AFE; to find the perpendicular AE: to which let the height of the observer's eye above the horizontal plane be added, and the sum will be the height of the object A B.

Example.

In the above diagram let the angle of elevation at CAFE be 49:28, and, after receding 200 feet in the same vertical plane, to the point D, let the angle of elevation AGE be 31:20; now, admitting the height of the observer's eye above the horizontal plane = DG or BE to be 5 feet, it is required to determine the height of the object A B?

The angle AFE 49:28:

the angle AGF 31:20. the angle GAF 18:8.

Now, in the oblique angled triangle AGF, since the angles and one side are given, the side A F is found by oblique angled plane trigonometry, Problem I., page 177; and, in the right angled triangle A EF, since the hypothenuse AF is now known, and the angle at F given, the perpendicular AE is found by right angled plane trigonometry, Problem I., page 171. Hence,

Remark. If it be required to know the horizontal distance BC or BD, it may be readily determined by means of the last problem.

PROBLEM IV.

To find the Distance of an inaccessible Object which the Observer can neither advance towards nor recede from in its vertical Line of Direction.

RULE.

Let the point A be any inaccessible object, and B and C two stations from which the distance of that object is to be determined: measure the distance B C, and, with a sextant or other instrument, observe the horizontal angles A B C and ACB; then, in the triangle c

A

B

ABC, the angles and the side BC are given; to find the other two sides, viz., A B and A C.

Example.

Let the horizontal angle ABC, measured with a sextant, be 59:15, the angle ACB 42:45, and the measured base line BC 350 yards; required the respective distances A B and AC?

The angle ABC 59:15 + the angle ACB 42:45 102: ; and 180:102: 78%, the angle CAB.

Now, the angles and one side being thus known, the remaining sides are to be determined by oblique angled trigonometry, Problem 1., page 177. Hence the following proportions :

Remark. This problem will be found of very essential service to His Majesty's ships and vessels of war, on many hostile occasions: for, when it is intended that a squadron of those ships should cannonade a fort to effect, or batter a breach in the sea-defences of a town, the distance at which the ships should be placed, abreast of such fort or town, with the view of opening their fire to the greatest advantage, may be readily determined in the above manner. Thus, let two competent persons, provided with sextants, in two ships, observe the angles subtended between the fort and each ship respectively; and let the distance between the two ships be carefully ascertained, which is readily done by Problem II., page 530, provided the height of the masts be known; or it may be found by means of a boat sent from one ship to the other, with instructions to pull at an uniform rate : then, if the interval, per watch, be noted between the time of the boat's pulling off from one ship and that of her arrival at the other, and her velocity or hourly rate of sailing be duly determined by the log, the distance between those ships may be easily obtained by the rule of proportion.

Now, with the distance between the two ships as a base line, thus found, and the angles subtended between the fort and each ship, the respective distances of those ships from the fort may be very readily computed, agreeably to the principles of the present problem.

Note. The most convenient distance for commencing a cannonade, is about 300 yards; that is, about a cable and a quarter's length from the object at which the guns are directed. On such occasions, however, the captains of His Majesty's ships of war always make choice of a much closer position, provided there be a sufficient depth of water.