Imágenes de páginas
PDF
EPUB

EBC-A=ACB,=12° 30', as before.

In the triangle ABC, find BC. Thus ;

1. As S.ACB: AB :: S. A: BC,=254.7

EBD-EAD=ADB,=12° 00'.

In the triangle ADB, find DB. Thus ;

2. S.ADB: AB :: S,DAB: DB,=207.4

CBE-DBE-CCD,=9° 00′

In the triangle CBD there is given, CB 254.7, DB 207.4, and the angle CBD 8° 00′;' to find DC.

This is performed by case 3, of oblique angular trigonometry, thus ;

3. As BC+BD : BC—BD : : T. of 1BDC+BCD T. of BDC-BCD,=55°.40'.

86° 00'+55° 40'-141° 40'-BDC.

86° 00′-55° 40′-30° 20′-BCD.

4. As S. BCD: BD:: S. CBD: DC, 57.15, the length of the tree.

To find DE, in the triangle DBE.

[ocr errors]

Say R. BD:: S.DBE: DE,=123.4, the height of the hill.

PROB. IX.

To find the height of an inaccessible object CD, on a hill BC, from ground that is not horizontal.

[merged small][graphic][subsumed][subsumed][subsumed][subsumed]

From any two points, as G and A, whose distance GA, is measured, and therefore given; let the angles HGD, BAD, BAC, and EAG, be taken; because GH is parallel to EA (by part. 2. theo. 3. sect. 1.) the angle HGA EAG; therefore EAG+HGD=AGD: and (by cor. 1. theo. 1. sect. 1.) 180-the sum of EAG and BAC=GAD; and (by cor. 1. theo. 5. sect. 1.) 180-the sum of the angles AGD and GAD=GDA: thus we have the angles of the triangle AGD and the side AG given; thence (by case 2, of obl. trig.) AD may be easily found. The angle DAB-CAB= DAC, and 90°-BAD-ADC and 180°-the sum of DAC and ADC=ACD: so have we the several angles of the triangle ACD given, and the side AD; whence (by case 2. of obl. trig.) CD may be easily found. We may also find AC, which with the an

:

The solutions of the several problems in heights and distances, by Gunter's scale, are omitted; because every particular stating has been already shewn by it, in the rectangular and oblique angular trigonometry.

OF DISTANCES.

ANY of the instruments used in surveying, will give you the angles or bearings of lines; which will be particularly shewn, when we come to treat of them.

PROB. I.

fig. 88.

B

Let A and B be two houses on one side of a river, whose distance asunder is 293 perches: there is a tower at C on the other side of the river, that makes an angle at A, with the line AB of 53° 20′; and another at B, with the line BA of 66° 20' required the distance of the tower from each house, viz. AC and BC.

This is performed by case 2. of oblique angled trigonometry, thus:

1. As S. C: AB:: S. A: BC,=270.5

PROB. II.

fig. 97,

C

B

Let B and C, be two houses whose direct distance asunder, BC, is inaccessible; however it is known that a house at A, is 252 perches from B, and 230 from C; and that the angle BAC, is 70°. What is the distance BC, between the two houses?

This is performed by case 3. of oblique angular trigonometry, thus;

1. As AB+AC: AB-AC:: T. of C+B:

T of C-B=3° 44′.

55°+3° 44' 58° 44'-C. 55°-3°-44-51° 16'=B.

2. As S. C: AB:: S. A: BC,=277.

[blocks in formation]
« AnteriorContinuar »