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second angle; from C, the point of intersection of the lines which form these angles, let fall the perpendicular CD; and that will be the height of the object required.

The external angle CBD of the triangle ABC is equal to the two internal opposite ones A and ACB (by theo. 4, sect. 4): wherefore if one of the internal opposite angles be taken from the external angle, the remainder will be the other internal opposite one. Thus,

CBD 55°—A 37°=ACB 18°. Therefore, in the triangle ABC we have the angles A and ACB with the side AB given, to find BC.

S.ACB: AB :: S.A: BC.

18° 87 37° 169.4. Having found BC, we have in the triangle BCD the angle CBD 55°; consequently BCD 35°, and BC 169.4, to find DC.

This is performed by the first case of right-angled trigonometry three several

ways.

Thus:
1. R: BC:: S.CBD: DC.
90° 169.4

55°

138.8,
the height required.
2. Sec. CBD: BC::T.CBD: DC.

55° 169.4 55° 138.8,
the height required.
3. Sec. BCD: BC::R: CD.
35°

169.4 90° 138.8,
the height required.
If BD, the breadth of the moat, were required, it may

also be found by three different statings, as in the first case of rightangled plane trigonometry.

PROBLEM VI.

Pl. 5. fig. 24. Let BC, a maypole, whose height is 100 feet, be broken at D, the upper part of which, DC, falls upon a horizontal plane, so that its extremity C is 34 feet from the bottom or foot of the pole.

Required the segments BD and DC.

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By Construction. Lay 34 feet from A to B; on B erect the perpendicular BC of 100 feet, and draw AC; bisect AC (by prob. 4, geom.)

with the perpendicular line EF; and from D, where it cuts the perpendicular BC, draw AD, which will be the upper segment, and DB will be the lower.

By cor. to lemma preceding theo. 7, geom., AD=DC; and (by the lemma) the angle C=CAD.

In the triangle ABC find C, as in Case 6 of right-angled trigonometry. Thus,

1. BC: R:: AB: T.C=GAD.

100 90° 34 18° 47'. By theo. 4, geom. The external angle ADB=37° 34', or to twice the angle C; i. e. to C and GAD.

Then in the triangle ABD there is ADB 37° 34', therefore
also its complement DAB 52° 26' and AB 34 given, to find
AD and PD.
By the second case of right-angled trigonometry:

2. S.ADB : AB::R: AD or DC.
37° 34' 34 90°

55.77.
BC-DC=BD.

100—55.77=44.23, required. These

may be had from other statings, as in the second case aforesaid.

PROBLEM VII.

Pl. 5. fig. 25.
To take the altitude of a perpendicular object on a hill, from a plane beneath it.

This is done at two stations. Thus,
Let the height DC of a windmill on a hill be required.

From any part of the plane whence the foot of the object can be seen, let angles be taken to the foot and top; measure thence any convenient distance towards the object, and at the end thereof take another angle to the top, and you have the proper requisites. Thus, First station. Angle to the foot DAB 21° 00'.

Angle to the top CAB 35° 00'.

Stationary distance AB 104 feet.
Second station. Angle to the top 48° 30'.

DC required.

By Construction. On an indefinite blank line lay the stationary distance AB 104 feet; from A set off the second, and from B the third given angle ; and from the intersecting point C of the line formed by

them let fall the perpendicular CE; from A set off the first angle, and the line formed by it will determine the point D. Thus have we the height of the hill, as well as that of the windmill.

The angle CBE-A=ACB, as in the last prob.
In the triangle ABC find AC thus,

S.ACB : AB :: S.ABC (or sup. of CBE): AC.
13° 30' : 104 :: 131° 30'

: 333.6. The angle CAE-DAE=CAD.

The angle ACD=AED+EAD, by theo. 4. In the triangle CAD find CD thus,

S.ADC: AC :: S.CAD : DC.

111° : 333.6 :: 14° : 86.46 required. CE, BE, or DE may be found by other various statings, as set forth in the first and second cases of right-angled trigonometry.

PROBLEM VIII.

Pl. 5. fig. 26. To find the length of an object that stands obliquely on the top of a hill, from a plane beneath.

Let CD be a tree whose length is required.
This is done at two stations.

Make a station at B, from whence take an angle to the foot and another to the top of the tree; measure any convenient distance backward to A, from whence also let an angle be taken to the foot and another to the top, and you have the requisites given. Thus: First station. Angle to the foot EBD=36° 30'.

Angle to the top EBC=44° 30'.

Stationary distance AB=104 feet. Second station. Angle to the foot EAD=24° 30'.

Angle to the top EAC=32° 00'. Let DC and DE be required.

The geometrical constructions of this and the next problem are omitted, as what has been already said, and the figures, are looked upon as sufficient helps.

EBC-A=ACB, or 44° 30'—32°=12° 30', as before. In the triangle ABC find BC thus,

1. S.ACB : AB :: S.A : BC.

12° 30' 104 32° 254.7.
EBD-EAD=ADB, or 36° 30'-24° 30'=12° 00'.

In the triangle ADB find DB thus,

2. S.ADB: AB :: S.DAB: DB.
12° 00'

104 24° 30' 207.4. CBE-DBE=CBD, or 44° 30'—-36° 30'=8° 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 as Case 3 of oblique-angled trigonometry. Thus, 3. BC+BD:BC-BD:: T. of (BDC+BCD): T. of 462.1 47.3

86° 00' (BDC-BCD). 55° 40'.

86° 00' +55° 40 =141° 40'=BDC.

86° 00'-55° 40'= 30° 20'=BCD. 4. S.BCD: BD:: S.CBD: DC.

30° 20' 207.4 8° 00' 57.15, length of the tree. To find DE in the triangle DBE, say

R. : BD:: S.DBE: DE.
90° 207.4 36° 30' 123.4, height of the hill.

PROBLEM IX.

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

Pl. 6. fig. 1. 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, geom.), the angle HGA=EAG; therefore EAG+HGD=AGD: and (by cor. 1, theo. 1, geom.) 180 the sum of EAG and BAD=GAD; and, by cor. 1, theo. 5, geom., 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. ang. trig.) AD may be easily found. The angle DAB-CAB=DAČ, 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 angle BAC will give CB the height of the hill.

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

G

2d. Of Distances.

The principal instruments used in surveying will give the angles or bearings of lines ; which was particularly shown when we treated of them.

PROBLEM I.

Pl. 6. fig. 2. 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 as Case 2 of oblique-angled trigonometry, thus,

1. S.C: AB :: S.A: BC.

60° 20' 293 53° 20' 270.5. 2. S.C: AB:: S.B: AC.

60° 20' 293 66° 20' 308.8.

PROBLEM II.

Pl. 6. fig. 11. 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 found to be 70°. What is the distance BC between the two houses ?

This is performed as Case 3 of oblique-angled trigonometry, thus, 1. AB+AC: AB-AC:: T. of }(C+B): T. of (C-B). 482 22

55° 00' 3° 44'. 55° +3° 44'=58° 44'=C. 55°/3° 44'=51° 16'=B.

2. S.C: AB :: S.A : BC.

58° 44' 252 70° 277.

PROBLEM III.

Pl. 6. fig. 3. Suppose ABC a triangular piece of ground, which by an old survey we find to be thus; AB 260, AC 160, BC 150 perches,

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