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PROB. II.

fig. 80.

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B

To fid the height of a perpendicular object, on an horizutal plane; by having the length of the shadow give.

Provide a d or staff, whose length is given, let that be set perpendicular, by the help of a quadrant thus; apply the ide of the quadrant, AC, to the rod, or staff; and whe the thread cuts 90° it is then perpendicular; the sale may be done by a carpenter's, or mason's plumb.

Having thus set the ad or staff perpendicular, measure the length of its shdow, when the sun shines, as well as the length of the shadow of the object, whose height is required; and you have the proper requisites given. Thus,

ab, the length of the shadow of the staff, 15 feet.
cb, the length of the staff, 10feet.

AB, the length of the shadowof the steeple, or object, 135 feet.

Required BC, the height of the object.

The triangles abc, ABC, are similar,thus; the angle b-B, being both right; the lines ac, AC are paral

lel, being rays, or a ray of the sun; whence the angle a=A (by part 3. theo. 3. sect. 1.) and consequently c=C. The triangles being therefore mutually equiangular are similar (by theo. 16. sect. 1.) it will be ab: bc: AB: BC,=90 the height, required.

The foregoing method is most to be dependd on; however, this is mentioned for variety's sake

PROB. III.

To take the altitude of a perpendiculo object, at the foot of a hill, from the hills side.

Turn the centre A of the quadrat, next your eye, and look along the side AC, or grside, to the top and bottom of the object; and notjg down the angles, measure the distance from th place of observation to the foot of the object. Tus,

Given,

Angle to the oot of the object, 55° † or

55° 15'.

Angle to the top of it, 31° or 31° 15'.
Distance the foot of it, 250 feet.

Required, the heint of the object.

Geometrically.

Draw the infite blank line AD, at any point in which A, make he angle EAB of 55° 15′ and EAC of 31° 15'; la 250 from A to B; from B, draw the perpendicula BE (by prob. 7. of geometry.) crossing AC in C; s will BC be the height of the object re

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ABE the complement of EAB to 90°, which is

34° 45'.

ACB the difference of the given angles 24° 00'.

The side AB. 250. Required BC.

This is performed as case 2. of oblique angular trigonometry. Thus,

180-the sum of ABE 34° 45', and CAB 24°00'= ACB 121° 15'. Then,

As S. ACB: AB:: S. CAB: BC, 119. height required.

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To take the altitude of a perpendicular object, on the top of a hill, at one station; when the top and bot tom of it can be seen from the foot of the hill.

As in prob. 1. take an angle to the top, and another to the bottom of the object; and measure from the place of observation to the foot of the object, and you have all the given requisites. Thus,

A tower on a hill.

Angle to the bottom, 48° 30'. Given, Angle to the top, 67° 00′.

Dist. to the foot of the object, 136 feet.

Required, the height of the object.

Geometrically

Make the angle DAB=48° 30', and lay 136 feet from A to B; from B, let fall the perpendicular BD; and that will be the height of the hill: produce BD upwards by a blank line: again, at A make the angle DAC-67°00' by a blank line, and from C where that crosses the perpendicular produced, draw the line CB, and that will be the height of the object required.

Let AC be drawn.

In the triangle ABC, there is given,

The angle ACD the complement of DAC=23° 00'.

CAB the difference between the two given angles= 18° 30'.

And the side AB 136. To find BC.

As S.C: AB:: S.CAB: BC,=110.4

If BD were wanted, it is easily obtained, by the

As R: AB:: S.DAB: DB=101.8

PROB. V.

To take an inaccessible perpendicular altitude, on an horizontal plane.

fig. 83.

This is done at two stations, thus ;

Let DC be a tower which cannot be approached, by means of a moat or ditch, nearer than B; at B,

take

an angle of altitude to C; measure any convenient distance backward to A, which note down: at A, take another angle to C; so have you the given requisites, thus;

(First angle, 55° 00'.

Given, Stationary distance, 87 feet.

Second angle, 37° 00′.

The height of the tower CD, is required.

Geometrically.

Upon an infinite} blank line, lay off the stationary distance 87, from A to B; from B set off your first; and from A, your second angle; from C, the point of intersection of the lines which form these angles,

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