Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[blocks in formation]

3. A man travels from A to B, 3.7 miles, then turning a little to the left hand goes from в to c, which is 47 miles: at che observes that A and B make an angle of 29°.16′. What is the distance from A by the shortest cut?

Answer, 7 miles.

[ocr errors]

4. A man travels from A to B, 3.7 miles; returning in a mist, he loses his way, and going a little too much towards the left hand, comes to c, which is 4.7 miles from B. It now clearing up he could see both A and B, and observes that they make an angle of 299.16. Pray how far is he from A?,

Answer, 1.201 mile.

(N) CASE III. Given two sides and the angle contained be

tween them, to find the rest.

The side AB=98

Given The side BC=95·12

The angle B-114°.24/

[blocks in formation]
[blocks in formation]

1. Extend the compasses from 198·12 to 2.88 on the line of numbers, that extent will reach from 32°.48' to 0°.33' on the line of tangents. This is the method of working such examples as this, but so small an angle as 33' is not contained on the scale.

2. Extend from 33°.21' to 65°.36′ on the line of sines, that extent will reach from 98 to 162 on the line of numbers.

[blocks in formation]
[blocks in formation]

(P) CASE IV. Given the three sides, to find the angles.

The side AB=98

Given The side BC=95.12

The side AC162.34

Required the angles A,B,and C.

BY CONSTRUCTION.

Draw the longest side AC= 162.34 from a scale of equal parts; with AB 98 in your compasses (taken from the same scale) and one foot in a describe an arc; with

A

BC=95.12 in your compasses cross

A

B..

ED

it in B; then ABC is the triangle required. The angles measured by a scale of chords (S. 27.) will be A-32°.15', B=114°.24′, and c-33°.21'.,

BY CALCULATION. See Rule III.

Let E be the middle of the base AC, and DB perpendicular

[blocks in formation]

Then AC+ED=81·17+ 1·713=82.883=AD, the greater segment, and † soID=81•17—1·713=79.457=DC, the less segment.

AB=98

To find the angles in the right angled triangles ADC and CDB.

: radius, sine 90°

::AD=82-883

: sine DBA=57°•45′

=320.15 }

or co-sine DAB=

[blocks in formation]

Then DBA + DBC=57°.45′ + 56°.39′=114°.24′- ABC,

[merged small][ocr errors][merged small][merged small][merged small]
[merged small][ocr errors][merged small][merged small]
[blocks in formation]
[ocr errors][merged small][merged small][merged small]
[blocks in formation]

The remaining angles may be found by Rule I.

(Q) BY GUNTER'S SCALE, according to the first method.

1. Extend the compasses from 324-68 to 193.12 on the line of numbers, that extent will reach from 2.88 to 1.713 the distance of a perpendicular from the middle of the base.

2. Extend from 98 to 82.883 on the line of numbers, that extent will reach from 90° to 57°.45′ on the line of sines.

3. Extend from 95.12 to 79.457 on the line of numbers, that extent will reach from 90° to 56°.39' on the line of sines.

BY GUNTER'S SCALE, according to the second method.

1. Extend the compasses from half the sum of the three sides 177.73 to one of the containing sides AB=98, that extent will reach from Ac=162.34 the other containing side, to a fourth number 89.5, on the line of numbers.

[ocr errors]

2. Extend the compasses from this fourth number 89.5 to (the difference between the half sum of the three sides and the side opposite to the angles sought), 82.61 on the line of numbers, that extent will reach from 90° on the line of sines to the required angle 32°, on the line of versed sines, immediately under the line of sines.

This is derived from the proportions in the investigation of Gunter's Rule (X. 22, and note).

[blocks in formation]

SCHOLIUM.

(R) There are some authors and teachers of trigonometry, who make no distinction of cases between right and obliqueangled triangles, but divide the whole into three cases; because the three rules necessary for solving the problems that occur in oblique trigonometry, are sufficient for solving those which occur in right-angled trigonometry. For instance, Rule I. (E. 52.) will solve all the cases in right-angled triangles (except the 6th), and the first and second case in obliqueangled triangles: Rule II. (G. 53.) will solve the 6th case in right-angled triangles and the 3d case in oblique; and Rule III. (H. 54.) will solve the last case in oblique triangles.

CHAP. III.

THE APPLICATION OF PLANE TRIGONOMETRY TO THE MENSURATION OF DISTANCES, HEIGHTS, &c.

(S) The mensuration of distances and heights depends upon the rules of plane trigonometry already explained, together with the use of certain instruments for taking angles.

(T) Horizontal and vertical angles are usually measured with a theodolite furnished with one or two telescopes, and a vertical arc; and if the horizontal and vertical arcs of the instrument be described with a radius of not less than 34 inches, the observed angles may be measured to half a minute, or the 120th part of a degree.

(U) Angles which are oblique to the horizon are generally taken with a sextant; which must be held in such a position, that its plane may coincide with the two objects and the eye. When vertical angles are taken with this instrument, an artificial horizon must be used, and the reflected image of the object from the glasses of the sextant must be brought to coincide with the reflected image of the same object in the artificial horizon.

(V) Base lines are generally measured with rods, or the four pole Gunter's chain; but common tape of 50 or 100 feet in length is often preferred both for accuracy and expedition: especially if it be kept dry, and the ground be tolerably level.

(W) The use of instruments must be acquired under the direction of a person well skilled in their several adjustments, as

« ΠροηγούμενηΣυνέχεια »