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THEO. III.

Fig. 12.

In any right lined plane triangle ABD; the base AD, will be to the sum of the other sides, AB, BD, as the différence of those sides, is to the difference of the segments of the base, made by the perpendicular BE ; viz. the difference between AE and ED,

Produce B1), till BG=AB the lesser leg; and on B as a centre, with the distance BG or BA,

ibe a circle AGHF; which will cut B2, and AD in the points H and F; then it is plain, that GD will be the sum, and HD the difference of the sides AB and BD; also since AE=EF (by theo. 8. sect. 4.) therefore, FD is the difference of AE ED, the segments of the base ; but (by theo. 17. sect. 4.) AD: GD:: HD: FD; that is, the base is to the sum of the other sides, as the difference of those sides, is to the difference of the segments of the base. 2. E. D.

THEO. IV.

Fig. 13.

If to half the sum of two quantities, be added half their difference ; the sum will be the greatest of them ; and if from half the sum be subtracted half their difference ; the remainder will be the least of them.

Let the two quantities be represented by AB and BC: (making one continued line ;) whereof AB is the greatest, and BC the least; bisect the whole line AC in E; and make AD=BC; then

TRIGONOMETRY.

123

it is plain, that AC is the sum, and DB the difference of the two quantities; and AE or EC, their half sum, and DĖor EBtheir half difference. Now if to AE we add EB, we shall have AB, the greatest quantity; and if from EC we take EB, we shall have BC the least quantity. 2. E. D.

Cor. Hence, if from the greatest of two quantities, we take half the difference of them, the remainder will be half their sum ; or if to half their difference be added the least quantity, their sum will be half the sum of the two quantities.

OBLIQUE ANGLED TRIANGLES.

CASE I.

TWO sides, and an angle opposite to one of them given ; 10

find the other angles and side.

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In the triangle ABC, there is given AB 240, the angle A 46° 30', and BC 200 ; to find the angle C, being acute, the angle B, and the side AC.

1st. By Construction.

Draw a blank line, on which set AB 240, from a scale of equal parts ; at the point A, of the line AB, make an angle of 46° 30', by an indefinite blank line; with BC 200, from a like scale of equal parts that AB was taken, and one foot in B, describe the arc DC to cut the last blank line in the points D and C. Now if the angle C had been required obtuse, lines from D to B, and to A, would constitute the triangle"; but as it is required acute,

draw the lines from C to B and to A, and the triangle ABC is constructed. From a line of chords let the angles Band C be measured ; and AC from the same scale of equal parts that AB and BC were taken ; and you will have the answers required.

2d. By Calculalion.

This is performed by theo. 1. of this sect.

thus;

As BC

200 is to the sine of A = 46°. 30' So is AB

240

2.301030 9.860562 2.380211

12 240773 9.939743

60° 31'

to the sine of c,

180°—the sum of the angles A and C, will give the angle B, by cor. 1. theo. 5. sect. 4.

A 460. 30
C 60. 31

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Extend from 200 to 240, on the line of numbers; that distance will reach from 46° 30' on the line of sines, to 60° 31' for the angle C.

TRIGONOMETRY.

125

Extend from 46° 30', to 72° 59', on the line of sines ; that distance will reach from 200 to 263.7 on the line of numbers, for AC.

Note. The method by Natural Sines will be obvious from the foregoing analogies.

CASE II.

Two angles and a side given ; to find the other sides.

Pl. 5. fig. 15.

In the triangle ABC, there is the angle A 460 301 AB 230, and the angle B 37° 30', given to find AC and BC.

Ist. By Construction.

Draw a blank line, upon which set AB 230, from a scale of equal parts ; at the point A of the line AB, make an angle of 46° 30', by a blank line; and at the point B of the line AB make an angle of 37° 30', by another blank line : the intersection of those lines gives the point C, then the triangle ABC' is constructed. Measure AC and BC' from the same scale of equal parts that AB was taken ; and you have the answer required.

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2d. By Calculation.

By (cor. 1. theo.5. sect. 4.) 180°—the sum of the angles A and B=C.

A 46° 30'
B 37. 30

180°-849.00 =96° 00/=C.

By def. 27. sect. 4. The sine of 96o=the sine of 84°, which is the supplement thereof ; therrtore instead of the sine of 96°, look in the tables for the sine of 84".

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3d. By Gunter's Scale. Extend from 849 (which is the supplement of 96°) to 46° 30' on the sides; that distance will reach from 230 to 168, on the line of numbers, for BC.

Extend from 840 to 37o. 30', on the sines ; that extent will reach from 230 to 141, on the line of numbers, for AC.

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