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Demonstr. AB being the given leg, in the right-angled triangle ABC; with the centre A, and any assumed radius AD, describe an arc DE, and draw DF perpendicular to AB, or parallel to BC. Now it is evident, from the definitions, that DF is the tangent, and AF the secant, of the arc DE, or of the angle A which is measured by that arc, to the radius AD. Then, because of the parallels BC, DF, it will be as AD: AB:: DF: BC:: AF: AC, which is the same as the theorem is in words.

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EXAMPLE 1.

In the right-angled triangle ABC,

Given the leg AB 162

Langle A 53° 7′ 49′′ To find AC and BC.

1. Geometrically.

Make AB 162 equal parts, and the angle A = 53° 7′ 48′′; then raise the perpendicular BC, meeting AC in C. So shall AC measure 270, and BC 216.

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Extend the compasses from 45° to 53°, on the tangents. Then that extent will reach from 162 to 216 on the line of numbers.

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Note. There is sometimes given another method for right-angled triangles, which is this:

ABC being such a triangle, make one leg AB radius, that is, with centre A, and distance AB, describe an arc BF. Then it is evident that the other leg BC represents the tangent, and the hypothenuse AC the secant, of the arc BF, or of the angle A.

In like manner, if the leg BC be made radius; then the other leg AB will represent the tangent, and the hypothenuse AC the secant, of the arc BG or angle C.

But if the hypothenuse be made radius; then each leg

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will represent the sine of its opposite angle; namely, the leg AB the sine of the arc AE or angle C, and the leg BC the sine of the arc CD or angle A.

And then the general rule for all these cases, is this, namely, that the sides of the triangle bear to each other the same proportion as the parts which they represent.

And this is called, Making every side radius.

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 some such method, drawn from the principles of geometry.

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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, 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 sides fixed on one of the radii, or else with telescopic sights.

To take an angle of altitude and depression with the quadrant.

Let A be any object, as the sun, moon, or a star, or the top of a tower, or hill, or other emi nence and let it be required to find the measure of the angle ABC, which a line drawn from the object makes with the horizontal line BC.

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Fix the centre of the quadrant in the angular point, and move it 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 plumb line BH, be the measure of the angle ABC as required.

The angle ABC of depression of any object A, 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 GH, on the other side of the plumb line.

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

EXAMPLE 1.

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, upon 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 Ċ = 47° 30′, the angle of elevation; and it is done. Then AB, measured on the scale of equal parts, is nearly 2184.

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

Construction.

Draw an indefinite ground line, upon which set off the given distance AB = 880; then A and B are the places of the observers. Make the angle A = 35o, and the angle B = 64o; and the intersection of the lines at C will be the place of the balloon; from whence the perpendicular CD, being let fall, will be its perpendicular height. Then, by measurement, are found the distances and height nearly as follows; viz. AC 1631, BC 1041, DC 936.

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Having to find the height of an obelisk standing on the top of a declivity, 1 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 further, the like angle was only 23° 45'. What then was the height of the obelisk ?

Construction.

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

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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 bottom of each

tree, and found them to be 33° and 64°. What then may be the distance between the two objects?

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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 straight 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 73° 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 BC to make the angle B = 73° 15'. So shall the point C be the place of the house required.

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