Note 2. When there are given two sides of a right-angled triangle, to find the third side; this is to be found by the property of the squares of the sides in theorem 34, Geom. viz. that the square of the hypothenuse, or longest side, is equal to both the squares of the two other sides together. Therefore, to find the longest side, add the squares of the two shorter sides together, and extract the square root of that sum; but to find one of the shorter sides, subtract the one square from the other, and extract the root of the remainder.

OF HEIGHTS AND DISTANCES, &c.

BY the mensuration and protraction of lines and angles, are determined the lengths, heights, depths, and distances of bodies or objects.

Accessible lines are measured by applying to them some certain measure a number of times, as an inch, or foot, or yard. But inaccessible lines must be measured by taking angles, or by such-like method, drawn from the principles of geometry.

When instruments are used for taking the magnitude of the angles in degrees, the lines are then calculated by trigonometry in the other methods, the lines are calculated from the principle of similar triangles, or some other geometrical property, without regard to the measure of the angles.

Angles of elevation, or of depression, are usually taken either with a theodolite, or with a quadrant, divided into degrees and minutes, and furnished with a plummet suspended from the centre, and two open sights fixed on one of the radii, or else with telescopic sights.

To take an Angle of Altitude and Depression with the Quadrant.

round there as a centre, till with one eye at D, the other being shut, you perceive the object a through the sights; then will the arc GH of the quadrant, cut off by the plumbline Bн, be the measure of the angle ABC as required.

The angle ABC of depression of any object A, below the horizontal line BC, is taken in the same manner; except that here the eye is applied to the centre, and the measure of the angle is the arc cs, on the other side of the plumb-line.

B

D

C

The following examples are to be constructed and calculated by the foregoing methods, treated of in Trigonometry.

EXAMPLE I

Having measured a distance of 200 feet, in a direct horizontal line, from the bottom of a steeple, the angle of elevation of its top, taken at that distance, was found to be 47° 30'; from hence it is required to find the height of the steeple.

Construction.

Draw an indefinite line; on which set off AC = 200 equal parts for the measured distance. Erect the indefinite perpendicular AB; and draw CB so as to make the angle c = 47° 30', the angle of elevation; and it is done. Then AB, measured on the scale of equal parts, is nearly 2181.

What was the perpendicular height of a cloud, or of a balloon, when its angles of elevation were 35° and 64°, as taken by two observers, at the same time, both on the same side of it, and in the same vertical plane; the distance between them being half a mile or 880 yards. And what was its distance from the said two observers ?

Construction.

Construction.

=

Draw an indefinite ground line, on which set off the given distance AB = 880: then A and B are the places of the observers. Make the angle a 35, and the angle B = +4° ; then the intersection of the lines at c will be the place of the balloon whence the perpendicular cp, being let fall, will be its perpendicular height. Then by measurement are found the distances and height nearly as follow, viz. ac 1631, BC 1041, DC 936.

Having to find the height of an obelisk standing on the top of a declivity, I first measured from its bottom a distance of 40 feet, and there found the angle, formed by the oblique plane and a line imagined to go to the top of the obelisk, 41°; but after measuring on in the same direction 60 feet farther, the like angle was only 23° 45'. What then was the height of the obelisk? Construction.

Construction.

Draw an indefinite line for the sloping plane or declivity, in which assume any point A for the bottom of the obelisk, from which set off the distance AC 40, and again CD = 60 equal parts. Then make the angle c = 41°, and the angle D = 23° 45′; and the point в where the two lines meet will be the top of the obelisk. Therefore AB, joined, will be its height.

Wanting to know the distance between two inaccessible trees, or other objects, from the top of a tower 120 feet high, which lay in the same right line with the two objects, I took the angles formed by the perpendicular wall and lines conceived to be drawn from the top of the tower to the botWhat tom of each tree, and found them to be 33° and 64°. then may be the distance between the two objects?

Construction.

Being on the side of a river, and wanting to know the distance to a house which was seen on the other side, I measured 200 yards in a strait line by the side of the river; and then, at each end of this line of distance, took the horizontal angle formed between the house and the other end of the line which angles were, the one of them 68° 2′, and the other 73o 15'. What then were the distances from each end to the

house?

;

Construction.

Draw the line AB = 200 equal parts. Then draw ac so as to make the angle a = 68° 2′, and вc to make the angle в = 73° 15'. So shall the point c be the place of the house required.

Calculation.