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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 BD.

By the second case of rectangular trigonometry.

2. As S.ADB : AB :: R: AD or DC,=55.77.

Hence, BC—DC=BD,=44. 25, as required.

PROB. VII.

To take the altitude of a perpendicular object on a

hill from a plane beneath it.

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Let the height DC, of a wind-mill 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

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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 of the lines 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 wind-mill.

The angle CBE-A=ACB, as in the last prob.

In the triangle ABC, find AC thus,

As S.ACB : AB :: S.ABC (or sup. of CBE): AC =333.6

The angle CAE-DAE=CAD.

The angle ADC=AED+EAD, by theo. 4.

In the triangle CAD, find CD thus,
As S.ADC : AC :: S.CAD : DC=86.46 required.

CE, BE, or DE, may be found by various other statings, as set forth in the first and second cases of rectangular trigonometry.

PROB. VIII.

To find the length of an object, that stands obliquely

on the top of a hill, from a plane beneath.

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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,

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 (BDC+BCD T. of į BDC—BCD,=550.40'.

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.

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

PROB. IX.

To find the height of an inaccessible object CD, ona

hill BC, from ground that is not horizontal.

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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

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