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To sin. op:

LP

Then, in the triangle APC, right-angled at P,
As the side
АС

232 log. 2.365488
.

90°

10.000000 So is the side AP

206:59

2.315109 To sin. op. | ACP

62° 56' 9.949621 Which taken from

90 00 leaves the LA 27 04

To sin. op.

LP

Again, in the triangle BPC, right angled at P,
As the side BC

174.07 log. 2.240724
90°

10'000000
So is side
BP
138.41

2:141168 To sin. op. 4 BCP

52° 40'

9.900444 which taken from

90 00 leaves the 4 B

37 20

O

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Also, the

2 АСР 620 56' added to

2 BCP 52 40 gives the whole ACB 115 36

So that all the three angles are as follow, viz. the 4 A 27° 4'; the LB 37° 20'; the 2c 115° 36'.

3. Instrumentally. In the first proportion.--Extend the compasses from 345 to 406, on the line of numbers; then that extent reaches, on the same line, from 58 to 68.2 nearly, which is the difference of the segments of the base.

In the second proportion.—Extend from 232 to 206, on the line of numbers; then that extent reaches, on the sinesg from 90° to 63.

In the third proportion.-Extend from 174 to 1384; then that extent reaches from 90° to 52oz on the sines.

EXAMPLE II.

In the plane triangle ABC,
AB 365 poles

LA 57° 12"
Given S AB
AC 154.33

Ans. LB 24 45
the sides
BC 309.86

C 93 3 To find the angles.

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EXAMPLE

EXAMPLE III.

In the plane triangle ABC,
Given S
AB 120

LA 570 28'
the sides
AC 1126

Ans. 2 B 57 57
BC 112

Lc 64 35
To find the angles.

The three foregoing theorems include all the cases of plane triangles, both right-angled and oblique. But there are other theorems suited to some particular forms of triangles, which are sometimes more expeditious in their use than the general ones; one of which, as the case for which it serves so frequently occurs, may be here taken, as follows:

THEOREM IV.

When a Triangle is Right-angled; any of the unknown parts may

be found by the following proportions : viz.
As radius
Is to either leg of the triangle;
So is tang. of its adjacent angle,
To its opposite leg;
And so is secant of the same angle,

To the hypothenuse.
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. Then it is evident, from
the definitions, that DF is the tangent, and

1) B
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 and :: AF : AC, which is the same
as the theorem is in words.

Note. The radius is equal, either to the sine of 90', or the tangent of 45°; and is expressed by 1, in a table of natural sines, or by 10 in the log. sines.

A

EXAMPLE I.

Given {

In the right-angled triangle ABC,
the leg AB 162}To find ac and Bc.

1. Geometricall

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.

2. Arithmetically. As radius

log. 10.000000 To leg AB 162

2.209515 So tang. L A 53° 7' 48"

10:124937 To leg. BC 216

2.334452 So secant LA 530 7' 48"

10*221848 To hyp. AC 270

2.431363 3. Instrumentally Extend the compasses from 45° to 53oy, on the tangents. Then that extent will reach from 162 to 216 on the line of numbers,

EXAMPLE II.

Given Şthe leg AB 180

F

In the right-angled triangle ABC,

Ans.

S AC 392.0146
Zihe 2 A 62° 4.0'

BC 348•2464
To find the other two sides.

Note. There is sometimes given another method for righta 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 br. 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 re present 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 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.

Then the general rule for all these cases is this, namely, that :he 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.

Note

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 a 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 trigo nometry: 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.

Let A be any object, as the sun, moon, or a star, or the top of a tower, or hill, or other eminence: and let it be required to find the measure of the angle abc, which a line drawn from the object makes above the horizontal line Bc.

Place the centre of the quadrant in the angular point, and move it

VOL. II.

!

G

round

с

R

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 BH, be the measure of the angle ABC as required. The angle ABC of depression of

C any object A, below the horizontal line bc, is taken in the same man- G ner; 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.

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'; 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 2185

Calculation.

As radius
TO AC 200
So tang. Lc 47° 30'
TO AB 218.26 required

10.000000
2.301030
10.037948
2:338978

А

EXAMPLE II.

What was the perpendicular height of a cloud, or of a balloon, when its angles of elevation were 350 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 froni the said two observers ?

Construction.

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